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Title: Solutions to Functions of One Complex Variable by John B. Conway
Description: Solution for exercise question in Functions of One Complex Variable by John B. Conway
Description: Solution for exercise question in Functions of One Complex Variable by John B. Conway
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Solutions Manual for
Functions of One Complex Variable I, Second Edition 1
© Copyright by Andreas Kleefeld, 2009
All Rights Reserved2
1 by
John B
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The solutions manual is intented for all students taking a graduate level Complex Analysis course
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Conway
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Maybe students even study
for preliminary exams for their doctoral studies
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Of course, that is up to you
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de (LTEXpreferred
but not mandatory) in order to complete this solutions manual
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If you find typing errors or mathematical mistakes pop an email to kleefeld@tu-cottbus
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The recent
version of this solutions manual can be found at
http://www
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tu-cottbus
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html
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CONTRIBUTION
I thank (without special order)
Christopher T
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Michael
David Perkins
for contributions to this book
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1 The real numbers
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2 The field of complex numbers
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3 The complex plane
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4 Polar representation and roots of complex numbers
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1 Power series
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2 Analytic functions
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3 Analytic functions as mappings
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21
21
24
31
4
Complex Integration
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4
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7 Counting zeros; the Open Mapping Theorem
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8 Goursat’s Theorem
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1 Classification of singularities
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2 Residues
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3 The Argument Principle
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1 Definitions and examples of metric spaces
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6
7
The Maximum Modulus Theorem
6
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6
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6
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4 The Phragmén-Lindelöf Theorem
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1 The space of continuous functions C(G, Ω)
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2 Spaces of analytic functions
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3 Spaces of meromorphic functions
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4 The Riemann Mapping Theorem
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5 The Weierstrass Factorization Theorem
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6 Factorization of the sine function
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7 The gamma function
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8 The Riemann zeta function
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1 Runge’s Theorem
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2 Simple connectedness
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3 Mittag-Leffler’s Theorem
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1 Schwarz Reflection Principle
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2 Analytic Continuation Along a Path
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3 Monodromy Theorem
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4 Topological Spaces and Neighborhood Systems
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5 The Sheaf of Germs of Analytic Functions on an Open Set
9
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10 Harmonic Functions
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137
137
141
144
148
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11 Entire Functions
151
11
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151
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153
11
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158
12 The Range of an Analytic Function
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161
161
162
162
162
Chapter 1
The Complex Number System
1
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1
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Find the real and imaginary parts of the following:
√ 3
1 z−a
3 3 + 5i −1 + i 3
;
;
(a ∈ R); z ;
;
z z+a
7i + 1
2
Solution
...
Then
a)
√ 6
−1 − i 3 n 1 + i
;i ; √
2
2
n
for
2 ≤ n ≤ 8
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Find the absolute value and conjugate of each of the following:
3−i
i
−2 + i; −3; (2 + i)(4 + 3i); √
; (1 + i)6 ; i17
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It is easy to calculate:
a)
z = −2 + i,
|z| =
√
5,
z = −2 − i
¯
b)
z = −3,
|z| = 3,
c)
√
|z| = 5 5,
z = (2 + i)(4 + 3i) = 5 + 10i,
d
3−i
z= √
,
2 + 3i
|z| =
1 √
110,
11
e)
z=
1
3
i
=
+ i,
i + 3 10 10
z = −3
¯
|z| =
z = 5 − 10i
¯
3+i
z= √
¯
2 − 3i
1 √
10,
10
z=
¯
3
1
− i
10 10
f)
z = (1 + i)6 = −8i,
|z| = 8,
z = 8i
¯
g)
i17 = i,
|z| = 1,
Exercise 3
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¯
3
z = −i
¯
Solution
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⇒: If z is a real number, then z = x (y = 0)
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¯
¯
⇔: If z = z, then we must have x + iy = x − iy for all x, y ∈ R
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This means that z is a real number
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If z and w are complex numbers, prove the following equations:
|z + w|2
|z − w|2
|z + w|2
= |z|2 + 2Re(zw) + |w|2
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¯
+ |z − w|2 = 2 |z|2 + |w|2
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We can easily verify that z = z
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¯
¯
|z − w|2
z ¯
z
¯
z
¯
= (z − w)(z − w) = (z − w)(¯ − w) = z¯ − zw − w¯ + ww
2
2
2
2
¯
= |z| + |w| − zw − zw = |z| + |w| − zw − zw
¯ ¯
¯ ¯
zw + zw
¯
¯
¯
= |z|2 + |w|2 − zw − zw = |z|2 + |w|2 − 2
¯
2
= |z|2 + |w|2 − 2Re(zw) = |z|2 − 2Re(zw) + |w|2
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Exercise 5
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+ zn ; w = w1 w2
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|wn |; z = z1 +
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wn
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Not available
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Let R(z) be a rational function of z
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Solution
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a0
bm zm + bm−1 zm−1 +
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Let all coefficients of R(z) be real, that is
a0 , a1 ,
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, bm ∈ R
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a0
an zn + an−1 zn−1 +
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b
bm z
m−1 z
0
bm zm + bm−1 zm−1 +
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a0
=
bm zm + bm−1 zm−1 +
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a0
¯
¯
= R(¯)
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b0
¯
¯
1
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Prove (3
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Solution
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Then
||z| − |w||
= ||z − w + w| − |w||
≤ ||z − w| + |w| − |w||
= ||z − w||
= |z − w|
Notice that |z| and |w| is the distance from z and w, respectively, to the origin while |z − w| is the distance
between z and w
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Multiplying this by
¯
z
1
2 z
t = w = |w|2 · |w| · w
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w
w,
we get zw ·
¯
w
w
= |w|2 ·
z
w
≥ 0 if w
0
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Show that equality occurs in (3
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Solution
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Exercise 3
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Describe the set of points z satisfying
|z − a| − |z + a| = 2c
for every possible choice of a and c
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Solution
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1
...
Find the sixth roots of unity
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Start with z6 = 1 and z = rcis(θ), therefore r6 cis(6θ) = 1
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The following table gives a list of principle values of arguments and the corresponding
value of the root of the equation z6 = 1
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Calculate the following:
a) the square roots of i
b) the cube roots of i
c) the square roots of
√
3 + 3i
√
Solution
...
√
√
√ 2
Let z = 3 + 3i
...
Specifically,
n
n
√
√
4
z = 12 cos
π
3
+ 2kπ
+ i sin
2
π
3
3
√
3
= π
...
2
Therefore, the square roots of z, zk , are given by
√
√
√
4
4
1
z0 = 12 cos π + i sin π = 12 23 + 2 i
6
6
√
√
√
4
4
z1 = 12 cos 7π + i sin 7π = 12 − 23 − 1 i
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Exercise 3
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, an−1 are distinct nth
roots of unity
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What is the smallest value of k? What can be said if a and b are nonprimitive
roots of unity?
Solution
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Exercise 4
...
cosn−3 θ sin3 θ +
...
Not available
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Let z = cis 2π for an integer n ≥ 2
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+ zn−1 = 0
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The summation of the finite geometric sequence 1, z, z2 ,
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n
th
n
We want to show that z is an n root of unity
...
It follows that 1 + z + z +
...
Exercise 6
...
Solution
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Exercise 7
...
Solution
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Since zn =
rn (cos(nθ) + i sin(nθ)), we have
Re(zn ) = rn cos(nθ) ≥ 0
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Therefore, assume z
0, then r > 0
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This implies θ = 0 as we will show next
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If
π
θ ∈ (0, π/2), then there exists a k ∈ {2, 3,
...
If we choose n = k + 1, we have
k
π ≤ nθ <
(k + 1)π
k
which is impossible since cos(nθ) ≥ 0
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Then 2π − π/k ≤ θ < 2π − π/(k + 1) for some k ∈ {2, 3,
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5 Lines and half planes in the complex plane
Exercise 1
...
Find necessary and sufficient conditions in terms of β that Lβ be tangent to C at a
...
Not available
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6 The extended plane and its spherical representation
Exercise 1
...
7) and (6
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Solution
...
Exercise 2
...
Solution
...
, 2
, 2
|z|2 + 1 |z| + 1 |z| + 1
Φ(z1 ) = (0, 0, −1)
...
√
If z2 = 1 + i, then |z2 | = 2 and therefore
Φ(z2 ) =
2 2 1
, ,
...
3
3
√
If z3 = 3 + 2i, then |z3 | = 13 and therefore
Φ(z3 ) =
Thus, z3 = 3 + 2i corresponds to the point
3 2 6
7, 7, 7
3 2 6
, ,
...
Exercise 3
...
Solution
...
Thus
Φ(z) =
x2 − 1
2x
...
That means the unit circle x2 + z2 = 1 lying in the xz−plane corresponds
to the real axes in C
...
Thus
Φ(z) = 0,
2y y2 − 1
...
That means the unit circle y2 + z2 = 1 lying in the yz−plane
corresponds to the imaginary axes in C
...
Let Λ be a circle lying in S
...
Recall
from analytic geometry that
P = {(x1 , x2 , x3 ) : x1 β1 + x2 β2 + x3 β3 = l}
where (β1 , β2 , β3 ) is a vector orthogonal to P and l is some real number
...
Use this information to show that if Λ contains the point N then its projection on C is a straight line
...
Solution
...
Exercise 5
...
Let W be the point on S
corresponding to z + z
...
Solution
...
8
Chapter 2
Metric Spaces and the Topology of C
2
...
Show that each of the examples of metric spaces given in (1
...
6) is, indeed, a metric space
...
6) is the only one likely to give any difficulty
...
Solution
...
Exercise 2
...
We have
(a) A := {z ∈ C : |z| < 1}
Let z ∈ A and set εz = 1−|z| , then B(z, εz ) ⊂ A is open and therefore A = z∈A B(z, εz ) is open also
...
(b) B := {z ∈ C : z = x + iy, y = 0} (the real axis)
Let z ∈ C − B, then Imz
complement is open
...
Set εz =
|Imz|
2 ,
then B(z, εz ) ⊂ C − B
...
Thus B is not open
...
C is not open because if zn = 1 then z = rcis(θ) with r = 1 and any ε-ball around z would contain an
ε
element z := (1 + 4 )cis(θ)
...
Now fix x ∈ R − Q and let {xn }n be a rational sequence that converges to x
...
Now zm = 1 implies that sin(mx2π) = 0 which in turn means that mx ∈ Z contradicting the
choice x ∈ R − Q
...
Hence C is not closed
...
1) and it is not closed because zn := 1 −
sequence in D that converges to a point outside of D
...
1) and it is closed because its complement is open: If
z E and z real, then B(z, min{|x|,|x−1|} ) is contained in the complement of E, if z is imaginary then
2
B(z, |Imy| ) is completely contained in the complement of E
...
If (X, d) is any metric space show that every open ball is, in fact, an open set
...
Solution
...
Exercise 4
...
9c)
...
Not available
...
Prove Proposition 1
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Solution
...
Exercise 6
...
Solution
...
Exercise 7
...
6
...
6
...
2
Solution
...
i) d(z, z ) ≥ 0
...
)(1+|z
2
Obviously, d(z, ∞) = (1+|z|2 )1/2 ≥ 0 for all z ∈ C
...
We have 2|z − z | = 0 iff z = z
...
d(∞, ∞) =
)(1+|z
2
limz→∞ (1+|z|2 )1/2 = 0
...
iii) d(z, z ) = d(z , z)
...
Also d(z, ∞) = (1+|z|2 )1/2 =
2 )(1+|z|
)(1+|z
d(∞, z) by the symmetry of d(z, z ) in general
...
We have
2|z − z |
[(1 + |z|2 )(1 + |z |2 )]1/2
2|z − x|
2|x − z |
≤
+
2 )(1 + |x|2 )]1/2
[(1 + |z|
[(1 + |x|2 )(1 + |z |2 )]1/2
= d(z, x) + d(x, z )
d(z, z ) =
for all x, z, z ∈ C
...
First, we need the following equation
(z − z )(1 + x x) = (z − x)(1 + xz ) + (x − z )(1 + xz),
¯
¯
¯
10
(2
...
¯
Using (2
...
2)
where the last inequality follows by
|z + xz | ≤ (1 + |x|2 )1/2 (1 + |z |2 )1/2
¯
⇔ (1 + xz )(1 + x¯ ) ≤ (1 + |x|2 )(1 + |z |2 )
¯
z
and
|z + xz| ≤ (1 + |x|2 )1/2 (1 + |z|2 )1/2
¯
⇔ (1 + xz)(1 + x¯) ≤ (1 + |x|2 )(1 + |z|2 )
¯
z
since
(1 + xz)(1 + x¯) ≤ (1 + |x|2 )(1 + |z|2 )
¯
z
⇔ xz + x¯ ≤ |x|2 + |y|2
¯
z
⇔ 2Re(x¯) ≤ |x|2 + |y|2
z
which is true by Exercise 4 part 2 on page 3
...
2) by (1 + |x|2 )1/2 yields
|z − z |(1 + |x|2 )1/2 ≤ |z − x|(1 + |z |2 )1/2 + |x − z |(1 + |z|2 )1/2
...
|x|2 )1/2 (1
2
+ |z |2 )1/2 (1 + |z|2 )1/2
Exercise 8
...
Suppose G ⊂ X is open; show that G ∩ Y is open in
(Y, d)
...
Solution
...
In order to show that G1 is open in (Y, d),
pick an arbitrary point p ∈ G1
...
But then
BY (p; ) = BX (p; ) ∩ Y ⊂ G ∩ Y = G1
which proves that p is an interior point of G1 in the metric d
...
13a)
...
Then, for every p ∈ G1 , there exists an −ball
BY (p : ) ⊂ G1
...
G1 =
p∈G1
Since we can write
BY (p; ) = BX (p; ) ∩ Y,
we get
BY (p; ) =
G1 =
p∈G1
where G =
p∈G1
p∈G1
BX (p; ) ∩ Y =
p∈G1
BX (p; ) ∩ Y = G ∩ Y
BX (p; ) is open in (X, d)
...
9c since each BX (p; ) is open)
...
Do Exercise 8 with “closed” in place of “open
...
Not available
...
Prove Proposition 1
...
Solution
...
Exercise 11
...
For which
values of θ is {cis(kθ) : k a non-negative integer } dense in T?
Solution
...
2
...
The purpose of this exercise is to show that a connected subset of R is an interval
...
(b) Use part (a) to show that if a set A ⊂ R is connected then it is an interval
...
Not available
...
Show that the sets S and T in the proof of Theorem 2
...
Solution
...
Exercise 3
...
(b) X = [0, 1) ∩ 1 + n : n ≥ 1
...
a) Define X = {z : |z| ≤ 1} ∪ {z : |z − 2| < 1} := A ∪ B
...
Obviously A is path-connected and B is path-connected
...
Recall that a space is path-connected if for any two points x and y there exists a continuous function f from
the interval [0, 1] to X with f (0) = x and f (1) = y (this function f is called the path from x to y)
...
12
This function is obviously continuous, since f (1/3) = limt→ 1 − f (t) = limt→ 1 + f (t) = Re(x) ∈ X and
3
3
f (2/3) = limt→ 2 − f (t) = limt→ 2 + f (t) = Re(y) ∈ X
...
Therefore
3
3
X is path-connected and hence X is connected
...
Therefore X is not connected
...
, {1 + n }
...
The k − th component is given by
C − {A ∩ B} where
A = [2πk − 2π, 2πk)
and
B = {z = r cis θ : r = θ, 2πk − 2π ≤ θ < 2πk},
k ∈ {1, 2, 3,
...
Prove the following generalization of Lemma 2
...
If {D j : j ∈ J} is a collection of connected
subsets of X and if for each j and k in J we have D j ∩ Dk
then D = {D j : j ∈ J} is connected
...
Let D = j∈J D j and C = {D j : j ∈ J}
...
Thus, for each C ∈ C either C ⊂ A or C ⊂ B
...
If not, then there exist E, F ∈ C such that E ⊂ A and F ⊂ B
...
Therefore, all members of C are contained in either A or all B
...
Both contradicting the fact that A, B are assumed to be nonempty
...
Exercise 5
...
, zn in F with z0 = a, zn = b and d(zk−1 , zk ) < for 1 ≤ k ≤ n
...
Give an example to illustrate this
...
Not available
...
3 Sequences and completeness
Exercise 1
...
4
...
a) A set is closed iff it contains all its limit points
...
“⇐”: Assume S contains all its limit points
...
Let
x ∈ S c
...
13f)
...
“⇒”: Let S be closed and x be a limit point
...
If not, S c (open) would be an open
neighborhood of x, that does not intersect S
...
13f again) which contradicts the fact that x is
a limit point of S
...
Let A ⊂ X be a set
...
We want to show x ∈ A ∪ A
...
Suppose x A
...
13f) that for every > 0, B(x; ) ∩ A ∅
...
In particular, for every integer n there is a point xn in B(x; n ) ∩ A
...
2)
...
“⊇”: To show A ∪ A ⊆ A−
...
If x ∈ A, then x ∈ A− (A ⊂ A− )
...
Then there exists {xn } ⊂ A with limn→∞ xn = x
...
By
Proposition 1
...
Exercise 2
...
8
...
Not available
...
Show that diam A = diam A−
...
Not available
...
Let zn , z be points in C and let d be the metric on C∞
...
Also show that if |zn | → ∞ then {zn } is Cauchy in C∞
...
First assume that |zn − z| → 0, then
d(zn , z) =
2
(1 + |zn |2 )(1 + |z|2 )
|zn − z| → 0
since the denominator (1 + |zn |2 )(1 + |z|2 ) ≥ 1 is bounded below away from 0
...
We need to show that if zn → z in the d-norm, then |zn |
∞
because otherwise the denominator grows without bounds
...
Then
d (zn , z) =
4 |zn |2 − zn z − zn z + |z|2
¯ ¯
=
4 |zn |2 − 2Re(zn z) + |z|2
¯
=
4 1−
2
1 + |zn |2 1 + |z|2
1 + |z|2 |zn |2 + 1 + |z|2
2Re(zn z)
¯
|zn |2
1 + |z|2 +
|z|2
|zn |2
1+|z|2
|zn |2
+
if |zn |
0
4
0
...
Hence convergence in d-norm implies convergence
in |·|-norm for numbers zn , z ∈ C
...
Then clearly also 1 + |zn |2 → ∞ and
therefore d(zn , ∞) = √ 2 2 → 0
...
But
d(zn , ∞) → 0 implies
1+|zn |
1 + |zn |2 → ∞ which is equivalent to |zn | → ∞
...
Show that every convergent sequence in (X, d) is a Cauchy sequence
...
Let {xn } be a convergent sequence with limit x
...
Thus
d(xn , xm ) ≤ d(xn , x) + d(x, xm ) <
and therefore {xn } is a Cauchy sequence
...
Give three examples of non complete metric spaces
...
Consider
Solution
...
1,
1
n < x ≤1
It is obvious that the limit function f is discontinuous
...
1
Example 2: Let X = (0, 1] with metric d(x, y) = |x − y|, x, y ∈ X
...
Thus, the metric space is not complete
...
The sequence defined by x1 = 1,
y
xn+1 = x2n + x1n is a Cauchy sequence of rational numbers
...
Therefore,
the metric space is not complete
...
Put a metric d on R such that |xn − x| → 0 if and only if d(xn , x) → 0, but that {xn } is a Cauchy
sequence in (R, d) when |xn | → ∞
...
)
Solution
...
Exercise 8
...
Show that
{xn } must be convergent
...
Since {xnk } is convergent, there is a x such that xnk → x as k → ∞
...
Let > 0
...
Now, fix nk0 > M + N, then
d(xn , x) ≤ d(xn , xnk0 ) + d(xnk0 , x) <
2
+
2
= ,
∀n ≥ N
...
2
...
Finish the proof of Proposition 4
...
Solution
...
Exercise 2
...
, pn ) and q = (q1 ,
...
Let
R = [p1 , q1 ] ×
...
By definition
diam R = d(p, q) =
n
k=1
1
2
(qk − pk )
...
x∈R,y∈R
Obviously, R is compact, so we have
diam(R) = max d(x, y)
...
, xn ) ∈ R and y = (y1 ,
...
Then clearly, pi ≤ xi ≤ qi and pi ≤ yi ≤ qi for all
i = 1,
...
We also have
(yi − xi )2 ≤ (qi − pi )2 ,
∀i = 1,
...
(2
...
3) we obtain
n
n
d(x, y) =
i=1
(yi − xi )2 ≤
i=1
(qi − pi )2 = d(p, q)
...
, n
...
n
Exercise 3
...
× [an , bn ] ⊂ R and let > 0; use Exercise 2 to show that there are
rectangles R1 ,
...
If xk ∈ Rk then it follows that
k=1
Rk ⊂ B(xk ; )
...
Not available
...
Show that the union of a finite number of compact sets is compact
...
Let K =
n
i=1
Ki be a finite union of compact sets
...
λ∈Γ
Of course, {Gλ }λ∈Γ is an open cover of each Ki , i = 1,
...
, n
...
, n
...
Exercise 5
...
That is, {xn } ∈ X iff sup{|xn | :
n ≥ 1} < ∞
...
Show that for each x in X and
¯
> 0, B(x; ) is not totally bounded although it is complete
...
)
Solution
...
Exercise 6
...
16
Solution
...
Let > 0
...
9 d) p
...
, xn ∈ S such that
n
S ⊆
B(xk ; /2)
...
k=1
2
...
Prove Proposition 5
...
Solution
...
Exercise 2
...
Solution
...
The distance in X is d(x, y)
...
We have
ρ( f (x) + g(x), f (y) + g(y))
= | f (x) + g(x) − f (y) − g(y)| = | f (x) − f (y) + g(x) − g(y)|
≤ | f (x) − f (y)| + |g(x) − g(y)|
≤ Md(x, y) + Nd(x, y) = (M + N)d(x, y),
∀x, y ∈ X
...
Now, let f, g be both uniformly continuous, that is,
∀
1
> 0 ∃δ1 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| <
1
whenever d(x, y) < δ1
∀
2
> 0 ∃δ2 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| <
2
whenever d(x, y) < δ2
...
So, choosing =
δ = min(δ1 , δ2 ), we have shown that
1
+
2
and
∀ > 0 ∃δ > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| < whenever d(x, y) < δ
...
Exercise 3
...
Show that if f and g are bounded uniformly continuous (Lipschitz) functions from X into C then so is f g
...
Let f be bounded, that is, there exists a constant M1 > 0 with | f (x)| < M1 for all x ∈ X and let g
be bounded, that is, there exists a constant M2 > 0 with |g(x)| < M2 for all x ∈ X
...
So, there exists a constant M = M1 M2 with
| f (x)g(x)| ≤ M
∀x ∈ X
...
Now,
ρ( f (x)g(x), f (y)g(y)) = | f (x)g(x) − f (y)g(y)| =
= | f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|
≤
≤
≤
| f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)|
| f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)|
M1 N2 d(x, y) + M2 N1 d(x, y) = (M1 N2 + M2 N1 )d(x, y)
∀x, y ∈ X
...
Now, let f and g be both bounded and uniformly continuous, that is
∃
∃
∀
∀
M1 such that | f (x)| ≤ M1
M2 such that |g(x)| ≤ M2
1
2
∀x ∈ X
∀x ∈ X
> 0 ∃δ1 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| <
> 0 ∃δ2 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| <
18
1
2
whenever d(x, y) < δ1
whenever d(x, y) < δ2
...
We
have
ρ( f (x)g(x), f (y)g(y))
= | f (x)g(x) − f (y)g(y)| =
= | f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|
≤ | f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)|
≤ | f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)|
≤
M1
2
+ M2 1 ,
whenever d(x, y) < min(δ1 , δ2 )
...
Thus, f g is uniformly continuous and bounded
...
Is the composition of two uniformly continuous (Lipschitz) functions again uniformly continuous (Lipschitz)?
Solution
...
Exercise 5
...
Is this still true if we only assume that f is continuous? (Prove or give
a counterexample
...
Assume f : X → Ω is uniformly continuous, that is, for every > 0 there exists δ > 0 such that
ρ( f (x), f (y)) < if d(x, y) < δ
...
But then, by the uniform continuity, we have that
ρ( f (xn ), f (xm )) <
∀n, m ≥ N
whenever d(xn , xm ) < δ which tells us that { f (xn )} is a Cauchy sequence in Ω
...
Here is a counterexample: Let f (x) = 1 which is continuous
x
on (0, 1)
...
But
n
1
{ f (xn )} = { f ( n )} = {n} is obviously not Cauchy
...
x
To see that pick = 1
...
Assume there
exists such a δ
...
Let y = x + δ/2 and set x = δ/2,
then
δ/2
1
1 1
y−x
| f (x) − f (y)| = − =
=
= > 1,
x y
xy
δ/2· δ δ
that is no matter what δ < 1 we choose, we always obtain | f (x) − f (y)| > 1
...
1
x
cannot be
Exercise 6
...
14)
...
Use Exercise 5 to show that there
is a uniformly continuous function g : X → Ω with g(x) = f (x) for every x in D
...
Not available
...
Let G be an open subset of C and let P be a polygon in G from a to b
...
15
and 5
...
19
Solution
...
Exercise 8
...
8) to give another proof of Theorem 5
...
Solution
...
To show f is uniformly continuous
...
Since f is continuous, we have for all x ∈ X there is a δ x > 0 such that ρ( f (x), f (y)) < /2 whenever
d(x, y) < δ x
...
Since X is by assumption compact (it is also sequentially compact as stated in
Theorem 4
...
22), we can use Lebesgue’s Covering Lemma 4
...
21 to obtain a δ > 0 such that x ∈ X
implies that B(x, δ) ⊂ B(z; δz ) for some z ∈ X
...
Exercise 9
...
5
...
, zn in X with z0 = a,
zn = b, and d(zk−l , zk ) < for 1 ≤ k ≤ n
...
(Hint: Use Theorem 5
...
)
Solution
...
Exercise 10
...
Prove that if f (x) = g(x) for x in D then f = g
...
Solution
...
2
...
Let { fn } be a sequence of uniformly continuous functions from (X, d) into (Ω, ρ) and suppose
that f = u − − lim fn exists
...
If each fn is a Lipschitz function with
constant Mn and sup Mn < ∞, show that f is a Lipschitz function
...
Solution
...
20
Chapter 3
Elementary Properties and Examples of
Analytic Functions
3
...
Prove Proposition 1
...
Solution
...
Exercise 2
...
6
...
Not available
...
Prove that lim sup(an +bn ) ≤ lim sup an +lim sup bn and lim inf(an +bn ) ≥ lim inf an +lim inf bn
for bounded sequences of real numbers {an } and {bn }
...
Let r > lim supn→∞ an (we know there are only finitely many by definition) and let s > lim supn→∞
(same here, there are only finitely many by definition)
...
This however, implies that
r + s ≥ lim sup(an + bn )
...
n→∞
n→∞
n→∞
Let r < lim inf n→∞ an (we know there are only finitely many by definition) and let s < lim inf n→∞ (same
here, there are only finitely many by definition)
...
This
however, implies that
r + s ≤ lim inf (an + bn )
...
n→∞
n→∞
n→∞
Exercise 4
...
21
Solution
...
Let M = lim supn→∞ an
...
Then, by definition of the lim supn→∞ an = M, we obtain that an < s for infinitely many n’s which implies
that bn < s for all n and hence lim supn→∞ bn = m < s
...
But the infimum of all
these s’s is M
...
n→∞
n→∞
Exercise 5
...
Solution
...
Then by definition, we
have: ∀ > 0 ∃N > 0 such that ∀n ≥ N, we have |an − a| ≤ , that is a − ≤ an ≤ a +
...
This shows that
a − ≤ lim inf an ≤ a +
n→∞
=:m
and
a − ≤ lim sup an ≤ a +
...
Hence,
0≤ M−m≤2
...
n→∞
n→∞
Exercise 6
...
n=0
n=0
n=0
∞
n=0
Solution
...
We also have,
lim sup |bk |1/k = lim sup |ak |1/k = lim sup |a| = |a|
...
k→∞
k→∞
1,
R = |a|
∞,
2
2
R
a 0
...
n→∞ |a|
∞, |a| < 1
=
lim
22
∞
n=0
an zn , a ∈ C;
c) Now, bn = kn , k is an integer
0
...
n→∞
n→∞
n→∞
So
d) We can write
∞
n!
n=0 z
=
Thus,
∞
k=0
1 1,
= k 1
R=
|k| − k ,
ak zk where
0,
2,
ak =
1,
0,
k > 0, k integer
...
k→∞
k→∞
Therefore 1/R = 1 which implies R = 1
...
Show that the radius of convergence of the power series
∞
n=1
(−1)n n(n+1)
z
n
is 1, and discuss convergence for z = 1, −1, and i
...
)
Solution
...
t
...
else
To find the radius of convergence we use the root criterion and therefore need the estimates
√
√
1 ≤ n(n+1) n ≤ n n
for n ∈ N
...
For the second inequality
note that
⇔
⇔
n ≤ nn+1
1
1
n n(n+1) ≤ n n
√
√
n(n+1)
n ≤ nn
...
n
n
n
n(n+1)
Vague memories of calculus classes tell me that
√
n
n → 1, thus
1
R
= lim sup
√
an = 1, i
...
R = 1
...
n=1 n
If z = −1 we note that the exponents n(n + 1) are always even integers and therefore the series is the
same as in the previous case of z = 1
...
The expression in(n+1) will always be real, so if the series converges at z = i, it converges
to a real number
...
i
=
cn with n 1
−
n
if n mod 4 ∈ {2, 3}
n
n=1
n=1
k
n=0 ck
...
To verify this, note that
16k2 + 8k − 1
< 0, hence a)
(4k + 1)(4k + 2)(4k + 3)
1
1
S 4k+4 − S 4k = c4k+3 + c4k+4 = −
+
< 0, hence c)
4k + 3 4k + 4
S 4k+2 − S 4k+1 = c4k+2 < 0, hence d)
...
We remark that
{S 4k+l }k≥1 , l ∈ {0, 1, 2, 3} describe bounded and monotone subsequences that converge to some point
...
e
...
Therefore the power series converges also in the case of z = i
...
2 Analytic functions
Exercise 1
...
Solution
...
We have
f (z + h) − f (z)
h
=
=
¯
¯
|z + h|2 − |z|2 (z + h)(¯ + h) − z¯ z¯ + h¯ + zh + hh − z¯
z ¯
z
z
z
z
=
=
h
h
h
¯
h
z + h + z =: D
...
¯
1
...
Then h = h and thus
D=z+h+z
¯
h
=z+h+z
¯
h
24
and therefore, if the limit of D exists, its value has to be z + z
...
) Take the path along the imaginary axes, that is x = 0
...
¯
Because of the uniqueness of the limit of D, we must have
z + z = z − z ⇐⇒ z = −z ⇐⇒ z = 0,
¯
¯
¯
if the limit of D exists
...
Since z = 0, we have that D = h
and thus the limit of D is 0
...
Exercise 2
...
Docs this remain true if the requirement of positivity is dropped?
Solution
...
Then there exists a monotonic subsequence {ank } of {an } that converges
n→∞
to a
...
Hence, {ank bnk } is a subsequence of an bn that converges to ab
...
n→∞
n→∞
Hence, lim sup an bn ≥ b lim sup an
...
Then there exists a subsequence {ank } of {an } such that lim ank = a > 0
...
Hence lim sup an bn = ∞
...
n→∞
In both cases, we have established that ab ≤ lim sup an bn
...
Applying the inequality we have established replacing bn with
n→∞ bn
b
with an bn :
consider lim
lim sup an = lim sup
n→∞
n→∞
1
bn
and an replaced
1
1
(an bn ) ≥ lim sup an bn
...
n→∞
n→∞
It follows that ab = lim sup an bn as required
...
Let bn = 0, − 1 , 0, − 1 , 0, − 1 ,
...
Also let an = 0, −2, 0, −3, 0, −4
...
In this case,
n→∞
ab = 0
n→∞
1 = lim sup an bn
...
Show that lim n1/n = 1
...
Let n ∈ N
...
Then
a = n1/n
⇐⇒ log a = log n1/n
log n
⇐⇒ log a =
n
log n
⇐⇒ lim log a = lim
n→∞
n→∞ n
log(x)
log n
...
Thus, lim
= lim f (n) = 0 also
...
Now, let f (x) =
n→∞
n→∞
Exercise 4
...
Solution
...
2i
2
Exercise 5
...
14)
...
Not available
...
Describe the following sets: {z : ez = i}, {z : ez = −1}, {z : ez = −i}, {z : cos z = 0},
{z : sin z = 0}
...
Using the definition we obtain
{z : ez = i} =
{z : ez = −i} =
1
+ 2k πi ,
2
{z : ez = −1} = {(1 + 2k) πi} ,
1
− + 2k πi ,
2
{z : cos z = 0} =
and
{z : sin z = 0} = {kπ}
where k ∈ Z
...
Prove formulas for cos(z + w) and sin(z + w)
...
We have
cos(z) =
sin(z) =
26
eiz + e−iz
2
eiz − e−iz
...
Claim: cos(z) cos(w) − sin(z) sin(w) = cos(z + w)
...
2
...
Proof:
sin(z) cos(w) + cos(z) sin(w) =
=
=
=
=
Exercise 8
...
where is this function defined and analytic?
Solution
...
3 that tan z = cos z is analytic wherever cos z 0
...
Thus let
G≡
(2k + 1)π
k∈Z
...
If z ∈ G, then cos z = 0 so tan z is undefined on the non-extended
complex plane
...
Suppose that zn , z ∈ G = C − {z : z ≤ 0} and zn = rn eiθn , z = reiθ where −π < θ, θn < π
...
Solution
...
Exercise 10
...
20
...
Suppose that g and h are
analytic, g (ω) 0 for any ω, that f is continuous, h is one-one, and that they satisfy h(z) = g( f (z)) for z in
G
...
Give a formula for f (z)
...
Not available
...
Suppose that f : G → C is a branch of the logarithm and that n is an integer
...
27
Solution
...
Let n ∈ Z and consider en f (z)
...
CASE 1: Assume n > 0
...
CASE 2: Assume n < 0
...
CASE 3: Assume n = 0
...
1
Exercise 12
...
Solution
...
Thus
1
1
1
1
1
1
1
θ
θ
+ i sin
z 2 = eLogz 2 = e 2 Logz = e 2 (lnr+iθ) = e 2 lnr e 2 θi = e 2 lnr cos
2
2
Hence,
Re(z) = e 2 lnr cos
1
since e 2 lnr > 0 and cos
1
θ
2
...
Thus, the real part of the function z 2 is always positive
...
Let G = C − {z ∈ R : z ≤ 0} and let n be a positive integer
...
Solution
...
Thus, we can
write
f (z) = z1/n = elog(z)/n = e(Log(z)+2kπi)/n = eLog(z)/n · e2kπi/n
...
They correspond to
the n distinct powers of the expression ζ = e2πi/n
...
, n − 1 and therefore they are all constant multiples of each other
...
Suppose f : G → C is analytic and that G is connected
...
Solution
...
Since f : G → C is analytic, that is f is continuously differentiable
(Definition 2
...
By Theorem 2
...
That is,
∂u ∂v
=
∂x ∂y
∂u
∂v
=−
...
1)
v(z) ≡ 0
and therefore f (z) = u(z)
...
1) we obtain
∂u ∂u
=
=0
∂x ∂y
and thus u (z) = 0 (see reasoning of equation 2
...
23 on page 41)
...
Since G is
connected and f : G → C is differentiable with f (z) = 0 ∀z ∈ G, we have that f is constant
...
For r > 0 1et A = ω : ω = exp
1
z
where 0 < |z| < r ; determine the set A
...
Define the set S = {z : 0 < |z| < r} where r > 0
...
r
To find the image of A is the same as finding the image of T under ez
...
To proof the claim, we need to show that for w 0, the equation ez = w has a solution z ∈ T
...
We have to find a complex number z = x + iy such that x2 + y2 > 1 and
r
e x eiy = w = |w|eiθ
...
Using x = log |w| and k
0 such
that (log |w|)2 + (θ + 2kπ)2 > 1 , then we found z = x + iy
...
Find an open connected set G ⊂ C and two continuous functions f and g defined on G such
that f (z)2 = g(z)2 = 1 − z2 for all z in G
...
Not available
...
Give the principal branch of
√
1 − z
...
Not available
...
Let f : G → C and g : G → C be branches of za and zb respectively
...
Suppose that f (G) ⊂ G and g(G) ⊂ G and prove that both f ◦ g and
g ◦ f are branches of zab
...
Not available
...
Let G be a region and define G∗ = {z : z ∈ G}
...
29
Solution
...
By assumption f is analytic and therefore
u and v have continuous partial derivatives
...
Since f ∗ (z) = f ( x), we have f ∗ (z) = u∗ (x, y) + iv∗ (x, y) where u∗ (x, y) = u(x, −y)
and v∗ (x, y) = −v(x, −y)
...
x
y
x
y
y
x
Exercise 20
...
, zn be complex numbers such that Re zk > 0 and Re(zl
...
Show that log(z1
...
+ log zn , where log z is the principal branch of the logarithm
...
Let z1 ,
...
The proof will be
by induction
...
Let z1 , z2 ∈ C as above
...
But note arg(z1 z2 ) = arg z1 + arg z2 implies that −π < arg z1 + arg z2 < π
...
Let z1 ,
...
Then
log(z1
...
+ log zk−1 + log zk by the case when n = k − 1
Hence, this is true for all n such that zi , 1 ≤ i ≤ n, satisfy the restrictions
...
Then z1 z2 = 0 − 5i
...
Now,
log(z1 ) = ln |z1 | + i arg z1
log(z2 ) = ln |z2 | + i arg z2
√
√
Thus log z1 + log z2 = ln 5 + ln 5 + i( 3π ) = ln 5 + i( 3π ), but note that 3π
(−π, π) and log(z1 z2 ) =
2
2
2
ln 5 + i(− π )
...
Hence,
2
the formula is invalid
...
Prove that there is no branch of the logarithm defined on G = C − {0}
...
)
ˆ
ˆ
Solution
...
We use the notation Log for the principal
ˆ that is
part of the log on G,
Log(z) = log |z| + i arg(z)
where arg(z) ∈ (−π, π)
...
ˆ
Assume f (z) is a branch of the logarithm defined on G
...
This yields
on G
...
Since f is analytic in G, it is continuous at −1
...
3 Analytic functions as mappings
...
1
...
Solution
...
The image of {z = x + iy : x < 0, −π < y < π} under the exponential function is given by
{e x+iy : x < 0, −π < y < π} = {e x eiy : x < 0, −π < y < π}
= {e x (cos(y) + i sin(y)) : x < 0, −π < y < π}
= {r(cos(y) + i sin(y)) : 0 < r < 1, −π < y < π}
...
Since r varies between 0 and 1, we get that the image is a solid circle with radius 1, where the boundary
does not belong to it, the negative x-axis does not belong to it and the origin does not belong to is
...
Do exercise 1 for the set {z : |Im z| < π/2}
...
We have
π
π
< y < }
...
2
2
If r > 0 is fixed, then r(cos(y) + i sin(y)) describes a half circle with radius r centered at (0, 0) lying in the
right half plane not touching the imaginary axis
...
{e x+iy : x ∈ R, −
Exercise 3
...
Solution
...
Exercise 4
...
(Hint: use polar coordinates
...
Not available
...
Find the fixed points of a dilation, a translation and the inversion on C∞
...
In general, we have
az + b
...
To find a fixed point, we have to find
all z such that S (z) = z
...
Obviously z = 0 is a fixed point
...
c
To get a translation S (z) = z + b, we have that a = 1, c = 0, d = 1 and b ∈ R
...
In this case z + b = z, which is true if z = ∞
...
c
1
To get a inversion S (z) = z , we have that a = 0, b = 1, c = 1 and d = 0
...
In this case 1 = z which is equivalent to z2 = 1 and thus z = 1 and z = −1 are
z
fixed points
...
Evaluate the following cross ratios: (a) (7 + i, 1, 0, ∞) (b) (2, 1 − i, 1, 1 + i) (c) (0, 1, i, −1) (d)
(i − 1, ∞, 1 + i, 0)
...
We have
S (z) =
S (z) =
S (z) =
z − z3 z2 − z3
/
, if z2 , z3 , z4 ∈ C
z − z4 z2 − z4
z − z3
, if z2 = ∞
z − z4
z − z3
, if z4 = ∞
...
2)
(3
...
4)
a) By (3
...
2) we get
(2, 1 − i, 1, 1 + i) =
7+i−0
= 7 + i
...
2 − 1 − i 1 − i − 1 − i 1 − i −2i 1 − i
(1 − i)(i + 1)
2
c) By (3
...
0+1 1+1
2
1−i1+i
2
d) By (3
...
If T z =
az+b
cz+d ,
−2 i + 1 −2
i−1−1−i
=
=
(i + 1) = 1 + i
...
Solution
...
47
...
c
Exercise 8
...
Set z2 =
a
c
d−b
a−c ,
z3 = − b and
a
show that T (R∞ ) = R∞ iff we can choose a, b, c, d to be real numbers
...
Assume T (R∞ ) = R∞
...
Observe that this
cz + d
b
b
implies az0 = −b, so − ∈ R∞ and r1 ≡ ∈ R∞ as well
...
Let T z =
32
r2 ≡
d
∈ R∞
...
Then
c
az1 + b
=1
cz1 + d
az1 + b = cz1 + d
d b
c
= −
z1 1 −
a
a a
z1 z1 r2 r1
− − +
=0
c
a
a
c
z1 + r1 z1 + r2
=
c
a
z1 + r1
c
= ∈ R∞
...
Then = × = r2 r3 ∈ R∞
...
Now to prove the converse, assume T (R∞ )
R∞
...
14 we conclude that T (R∞ ) is some other circle in C∞
...
In other words, there must be some value zc ∈ R∞ for which T zc R∞
...
Then
azc + b
T zc =
czc + d
is clearly an element of R∞ , contradicting the observation that T zc is not real
...
Exercise 9
...
az+b
cz+d ,
find necessary and sufficient conditions that T (Γ) = Γ where Γ is the unit circle
Solution
...
z
T (z)T (z) = 1
⇐⇒
⇐⇒
⇐⇒
az + b az + b
(az + b)(¯ z + b)
a¯ ¯
= 1 ⇐⇒
=1
cz + d cz + d
(cz + d)(¯ z + d)
c¯ ¯
¯
¯
¯
az¯ z + b¯ z + azb + bb = cz¯ z + dcz + dcz + dd
a¯
a¯
c¯
¯¯ ¯
¯
¯ ¯ a
¯
z¯(a¯ − c¯ ) + z(ab − cd) + z(b¯ + dc) − dd = 0
...
Hence, we have the two conditions
¯
¯
ab − cd = 0
|a|2 + |b|2 = |c|2 + |d|2
...
Let c = λb, then ab − cd = 0 yields ab − λbd = 0 iff a = λd iff d = λ
...
Then
1
¯
λ
T (z) =
|a|2
= |a|2 + |b|2 ,
|λ|2
= λ, and so the form of the Möbius transformation is
az + b
,
¯
λ(bz + a)
¯
or
¯
T (z) = λ
where |λ| = 1
az + b
,
¯
bz + a
¯
where |λ| = 1
...
T (z) = eiθ
for some θ
...
Let D = {z : |z| < 1} and find all Möbius transformations T such that T (D) = D
...
Take an α ∈ D = {z : |z| < 1} such that T (α) = 0
...
Therefore T (α∗ ) = ∞
...
(It is easy to check that T (α) = 0 and T (α∗ ) = T ( α ) = ∞)
...
We have
T (z0 ) = K
eiθ − α
αeiθ − 1
¯
and therefore
eiθ − α e−iθ − α
¯
eiθ − α e−iθ − α
¯
|eiθ − α|
= |K|
...
We arrive at
T (z) = eiθ
z−α
,
αz − 1
¯
for some real θ
...
Show that the definition of symmetry (3
...
That is, show that if ω2 , ω3 , ω4 are also in Γ then equation (3
...
(Hint: Use Exercise 8
...
Let Γ be a circle in C∞ containing points z2 , z3 , z4 , w2 , w3 , and w4 , with all zi distinct and all wi
distinct
...
We know that
(z∗ , z2 , z3 , z4 ) = (z, z2 , z3 , z4 )
Recall that the above left-hand cross-ratio implies a unique Möbius transformation T for which T z2 = 1,
T z3 = 0, and T z4 = ∞
...
10, T maps Γ to R∞
...
or
z∗ = T −1 T z
(3
...
Again, S maps Γ to
R∞
...
(3
...
5) and (3
...
Thus we
observe that T S −1 = T S −1 so
T −1 T z = T −1 T S −1 S z = S −1 S z
...
Exercise 12
...
4
...
Not available
...
Give a discussion of the mapping f (z) = 2 (z + 1/z)
...
The function f (z) = 1 z + 1 = z 2z can be defined for all z ∈ C − {0} and therefore also in the
2
z
punctured disk 0 < |z| < 1
...
z1
z2
z1 z2
z1 z2
With the assumption 0 < |zi | < 1, i = 1, 2 the factor z1 z2 −1, is always nonzero and we conclude that z1 = z2 ;
hence f (z) is injective
...
To see this, write z = reiθ in polar
coordinates and let f (z) = w = a + ib, a, b ∈ R
...
r
r
For the real and imaginary part of w the following equations must hold
a
=
1
2
r+
1
r
cos θ
b
=
1
2
r−
1
r
sin θ
...
7)
If f (z) = w has imaginary part b = 0 then sin θ = 0 and |cos θ| = 1
...
For all other points the equations (3
...
Given any value of r ∈ (0, 1), the graph of f (reiθ ) as a function of θ looks like an ellipse
...
7) we see that
a
2
1
1
2 (r+ r )
+
b
1
1
2 (1− r )
2
= 1
...
7) that the graph of f (reiθ )
is a hyperbola and it degenerates to rays if z is purely real or imaginary
...
If cos θ 0 and sin θ 0 then cos θ − sin θ = 1
...
Suppose that one circle is contained inside another and that they are tangent at the point a
...
(Hint: first try
(z − a)−1
...
Using the hint, define the Möbius transformation T (z) = (z − a)−1 which sends the region G
between two lines
...
Hence, choose S (z) = cz + d where |c| = 1 such that
S (T (G)) = x + iy : 0 < y <
π
...
Finally, the Möbius transformation
z−1
z+1
maps the right half plane onto the unit disk (see page 53)
...
Doing
some simplifications, we obtain
c
e z−a +d − 1
f (z) = c +d
e z−a + 1
where the constants c and d will depend on the circle location
...
Can you map the open unit disk conformally onto {z : 0 < |z| < 1}?
Solution
...
Exercise 16
...
Can f be
one-one?
Solution
...
Exercise 17
...
Show that f is constant
...
Not available
...
Let −∞ < a < b < ∞ and put Mz = z−ia
...
Determine which of the regions A, B, C, D, E, F in Figure 1, are mapped by M
onto the regions U, V, W, X, Y, Z in Figure 2
...
We easily see that we have M(ia) = 0
...
Similarly we have M(ib) = ∞ and
therefore the region B and E which touch the line ib are mapped somehow to the region U and X which
touch ∞
...
Let us find out
...
Thus, the imaginary part of Mz is a positive number
multiplied by x(b − a) and therefore also positive
...
Because B and C meet at the line ia, we conclude that X and M(C) do, too
...
By a similar argument, we obtain that M maps A to Z and D to W
...
Let a, b, and M be as in Exercise 18 and let log be the principal branch of the logarithm
...
(b) Show that log(z − ic) is defined for Re z > 0 and any real number c; also prove that |Im log(z − ic)| < π
2
if Re z > 0
...
(d) Show that
b
dt
= i[log(z − ib) − log(z − ia)]
a z − it
(Hint: Use the Fundamental Theorem of Calculus
...
Solution
...
Exercise 20
...
Solution
...
Then
T (z) =
αz + β λaz + λb λ(az + b) (az + b)
=
=
=
= S (z)
...
⇒: Let S = T , that is S (z) = T (z)
...
γ
=
=
=
(3
...
9)
(3
...
9), then
⇐⇒
⇐⇒
⇐⇒
⇐⇒
λ3 c + λ1 d λ3 γ + λ1 δ
=
c+d
γ+δ
λ3 cγ + λ1 dγ + λ3 cδ + λ1 dδ = λ3 γc + λ1 δc + λ3 γd + λ1 δd
λ1 dγ + λ3 cδ − λ1 δc − λ3 γd = 0
λ1 (dγ − δc) − λ3 (dγ − δc) = 0
(λ1 − λ3 )(dγ − δc) = 0
...
b
If λ1 − λ3 = 0, then from (3
...
9), we get λ3 = a = d = λ1 which implies ad − bc = 0 a contradiction
c
(Not possible, otherwise we do not have a Möbius transformation)
...
(3
...
12)
Insert (3
...
12) into (3
...
d δ
d dλ
Insert (3
...
12) into (3
...
c γ
c cλ
Therefore, we have
α = λa, β = λb, γ = λc, δ = λd
...
Let T be a Möbius transformation with fixed points z1 and z2
...
Solution
...
We have
S −1 T S (S −1 z1 ) = S −1 T S S −1 z1 = S −1 T z1 = S −1 z1 ,
where the first step follows by the associativity of compositions, the second step since S S −1 = id and the
last step by T z1 = z1 since z1 is a fixed point of T
...
38
We have
S −1 T S (S −1 z2 ) = S −1 T S S −1 z2 = S −1 T z2 = S −1 z2 ,
where the first step follows by the associativity of compositions, the second step since S S −1 = id and the
last step by T z2 = z2 since z2 is a fixed point of T
...
The
composition of Möbius transformation is a Möbius transformation
...
(a) Show that a Möbius transformation has 0 and ∞ as its only fixed points iff it is a dilation,
but not the identity
...
Solution
...
We know that
Mz =
az + b
cz + d
b
with a, b, c, d ∈ C
...
But we also
d
a
az
know that M(∞) = ∞ so = ∞, meaning c = 0
...
Since both
c
d
b and c were 0, we know that both a and d are nonzero so α is likewise nonzero and finite
...
Then for any z ∈ C∞ , M˜ = z and M has infinitely many fixed
˜
z ˜
points, a contradiction
...
On the other hand, assume M is a dilation not equal to the identity
...
It is clear from this representation that M has fixed points at 0 and ∞ and,
of course, at no other points
...
Again, we know that
Mz =
az + b
cz + d
and we can see that c = 0 as before
...
Recall that if z is a fixed point of M, then z satisfies c˜2 + (d − a)˜ − b = 0, which in the case
˜
˜
z
z
b
...
Suppose that d a
...
Thus, it must be the case that d = a and
b
Mz = z +
...
Now assume M is a translation
...
Here, in the convention of
the text, we have a = 1, b = β, c = 0, and d = 1
...
Clearly, however, M has ∞ as a fixed
˜
point, so ∞ is the only fixed point of M
...
Show that a Möbius transformation T satisfies T (0) = ∞ and T (∞) = 0 iff T z = az−1 for
some a in C
...
Not available
...
Let T be a Möbius transformation, T the identity
...
(Hint: Use Exercises 21 and 22
...
Let T and S have the same fixed points
...
: Suppose T and S have two fixed points, say z1 and z2
...
Then
MS M −1 (0) = MS M −1 Mz1 = MS z1 = Mz1 = 0
and
MS M −1 (∞) = MS M −1 Mz2 = MS z2 = Mz2 = ∞
...
Similar,
we obtain MT M −1 (0) = 0 and MT M −1 (∞) = ∞ and therefore is also a dilation
...
: Suppose T and S have one fixed points, say z
...
Then
MS M −1 (∞) = MS M −1 Mz = MS z = Mz = ∞
...
Similar, we
obtain MT M −1 (∞) = ∞ and therefore is also a translation
...
Exercise 25
...
Solution
...
Exercise 26
...
(a) Let GL2 (C) = all invertible 2 × 2 matrices with entries in C and let M be the group
a b
of Möbius transformations
...
Show that ϕ is a group
cz+d
c d
homomorphism of GL2 (C) onto M
...
(b) Let S L2 (C) be the subgroup of GL2 (C) consisting of all matrices of determinant 1
...
What part of the kernel of ϕ is in S L2 (C)?
Solution
...
A simple calculation shows that this is true
...
This is equivalent to az + b = cz2 + dz and by
cz+d
comparing coefficients we obtain b = c = 0 and a = d
...
Note that the kernel is a normal subgroup of GL2 (C)
...
The kernel of the restriction is simply N ∩ S L2 (C) = {±I}
...
If G is a group and N is a subgroup then N is said to be a normal subgroup of G if S −1 T S ∈ N
whenever T ∈ N and S ∈ G
...
Prove that the group M of Möbius transformations is a simple group
...
Not available
...
Discuss the mapping properties of (1 − z)i
...
Not available
...
For complex numbers α and β with |α|2 + |β|2 = 1
uα,β =
¯
αz − β
βz + α
¯
U = {uα,β : |α|2 + |β|2 = 1}
and let
(a) Show that U is a group under composition
...
What is its kernel?
−β α
(l)
(d) If l ∈ {0, 1 , 1, 3 ,
...
For uα,β = u in U define T u : Hl → Hl ,
2
2
(l)
(l)
(l)
by (T u f )(z) = (βz + α)2l f (u(z))
...
(c) Show that
Solution
...
Exercise 30
...
(b) Find all Möbius transformations S (z) that map D onto D and such that f (S (z)) = f (z) when |z| < 1
...
Not available
...
1 Riemann-Stieltjes integrals
Exercise 1
...
Show that γ is of bounded variation and V(γ) =
γ(b) − γ(a)
...
Let γ : [a, b] → R be a monotone, non-decreasing function
...
< tm = b}, for all n ∈ N, γ(tn ) ≥ γ(tn−1 )
...
Let P = {a = t0 < t1 <
...
Then,
m
n=1
|γ(tn ) − γ(tn−1 )| = [γ(tm ) − γ(tm−1 )] + [γ(tm−1 ) − γ(tm−2 )] +
...
+ [γ(t2 ) − γ(t1 )] + [γ(t1 ) − γ(a)]
= γ(b) − γ(a)
Since γ(a), γ(b) ∈ R, γ(b) − γ(a) < ∞
...
And
also v(γ; P) = sup{v(γ; P) : P a partition of [a, b]} = V(γ) = γ(b) − γ(a)
...
Prove Proposition 1
...
Solution
...
(a) If P and Q are partitions of [a, b] and P ⊂ Q then v(γ; P) ≤ v(γ; Q)
...
Let P = {a = t0 < t1 <
...
Suppose Q = P ∪ {x}, where tk < x < tk+1
...
The smallest partition set for Q \ P
contains the trivial partition composed of {a, b}
...
(b) If σ : [a, b] → C is also of bounded variation and α, β ∈ C, then αγ + βσ is of bounded variation and
42
V(αγ + βσ) ≤ |α|V(γ) + |β|V(σ)
...
< tm = b} be a partition
...
Then,
m
v(αγ + βσ; P) =
k=1
m
=
k=1
m
=
k=1
m
≤
k=1
m
=
k=1
m
=
k=1
|(αγ + βσ)(tk ) − (αγ + βσ)(tk−1 )|
|αγ(tk ) + βσ(tk ) − αγ(tk−1 ) − βσ(tk−1 )|
|(αγ(tk ) − αγ(tk−1 )) + (βσ(tk ) − βσ(tk−1 ))|
|αγ(tk ) − αγ(tk−1 )| + |βσ(tk ) − βσ(tk−1 )|
m
|αγ(tk ) − αγ(tk−1 )| +
|α| |(γ(tk ) − γ(tk−1 ))| +
m
= |α|
k=1
|βσ(tk ) − βσ(tk−1 )|
k=1
m
k=1
m
|γ(tk ) − γ(tk−1 )| + |β|
|β| |(σ(tk ) − σ(tk−1 ))|
k=1
|σ(tk ) − σ(tk−1 )|
= |α|v(γ; P) + |β|v(σ; P)
≤ |α|sup{v(γ; P) : P a partition of [a, b]} + |β|sup{v(σ; P) : P a partition of [a, b]}
= |α|V(γ) + |β|V(σ)
<∞
The result is two-fold, V(αγ + βσ) ≤ |α|V(γ) + |β|V(σ) and it follows αγ + βσ is of bounded variation by
|α|V(γ) + |β|V(σ)
...
Prove Proposition 1
...
Solution
...
Exercise 4
...
8 (Use induction)
...
Not available
...
Let γ(t) = exp((−1 + i)/t−1 ) for 0 < t ≤ 1 and γ(0) = 0
...
Give a rough sketch of the trace of γ
...
Not available
...
Show that if γ : [a, b] → C is a Lipschitz function then γ is of bounded variation
...
Let γ : [a, b] → C be Lipschitz, that is ∃ a constant C > 0 such that
|γ(x) − γ(y)| ≤ C|x − y|,
∀x, y ∈ [a, b]
...
< tm = b} of [a, b], we have
m
v(γ; P) =
k=1
m
|γ(tk ) − γ(tk−1 )| ≤
k=1
m
C|tk − tk−1 | = C
k=1
|tk − tk−1 | = C(b − a) =: M > 0
...
< tm = b} of [a, b]
m
v(γ; P) =
k=1
|γ(tk ) − γ(tk−1 )| ≤ M
...
Show that γ : [0, 1] → C, defined by γ(t) = t + it sin 1 for t
t
not rectifiable
...
0 and γ(0) = 0, is a path but is
Solution
...
Exercise 8
...
Express γ and σ as paths and calculate
f and σ f where f (z) = |z|2
...
For γ we have γ(t) = (1 − t) + it for 0 ≤ t ≤ 1
...
Therefore,
1
γ
|z|2 dz =
0
((1 − t)2 + t2 )(−1 + i)dt
1
= (−1 + i)
0
= (−1 + i)
(2t2 − 2t + 1)dt
2 3 2
t −t +t
3
1
0
2
−1+1
= (−1 + i)
3
2
2
=− +i
...
So γ1 (t) = i and γ2 (t) = −1
...
Define γ : [0, 2π] → C by γ(t) = exp(int) where n is some integer (positive, negative, or zero)
...
z
Solution
...
Thus
2π
z−1 dz =
γ
2π
e−int ineint dt =
0
0
Exercise 10
...
zn dz for every integer n
...
Clearly, γ(t) = eint is continuous and smooth on [0, 2π] (It is the unit circle)
...
0
−1
2π
zn dz
γ
2π
eint ieit dt = i
=
0
ei(n+1)t dt = t
0
ei(n+1)t
i(n + 1)
2π
0
2π
1
1
=
ei(n+1)t =
ei(n+1)2π − 1
0
n+1
n+1
1
1
[cos((n + 1)2π) − i sin((n + 1)2π)1] =
[1 − i· 0 − 1]
=
n+1
n+1
= 0
...
Let γ be the closed polygon [1 − i, 1 + i, −1 + i, −1 − i, 1 − i]
...
Solution
...
Thus
γ
1
dz
z
1
=
−1
1
=
−1
1
−1
−1
1
1
1
i dt +
dt +
i dt +
1 + it
t+i
−1 + it
1
1
1
−1
1
1
1
dt + i
dt
+
−
t+i t−i
−1 + it
−1 1 + it
1
−t + i + t + i
−1 + it − 1 − it
dt + i
dt
−1 (t + i)(t − i)
−1 (1 + it)(−1 + it)
1
1
1
1
dt − 2i
dt
= 2i
2+1
2
−1 −1 − t
−1 t
1
1
1
1
= 2i
dt + 2i
dt
2+1
t
t2 + 1
−1
−1
1
π
π
1
− −
dt = 4i [arctan(z)]1 = 4i
= 4i
−1
4
4
t2 + 1
−1
1
−1
1
dt
t−i
=
Exercise 12
...
2
dz where γ : [0, π] → C is defined by γ(t) = reit
...
Solution
...
Let > 0, then
|I(r) − 0| = |I(r)| =
γ
π
eiz
dz =
z
0
it
eire
rieit dt = i
reit
π
0
π
it
eire dt ≤
it
eire
dt
...
38 eq
...
13), so
it
it
eire = eRe(ire ) = eRe(ir cos(t)+ir·i sin(t)) = eRe(ir cos(t)−r sin(t)) = e−r sin(t)
...
Since e−r sin(t) is continuous, so δ1 > 0 can be chosen such that
δ1
e−r sin(t) dt <
0
3
,
∀r > 0
...
Similar, δ2 > 0 can be chosen such that
π
e−r sin(t) dt <
π−δ2
46
3
Finally, because of the Lebesgue’s dominated convergence Theorem
π−δ2
lim
r→∞
π−δ2
e−r sin(t) dt =
δ1
δ1
we have
π−δ2
lim
r→∞
lim e−r sin(t) dt
r→∞
e−r sin(t) dt <
δ1
3
...
r→∞
1
Exercise 13
...
Solution
...
Exercise 14
...
Solution
...
Exercise 15
...
16 is an equivalence relation
...
Not available
...
Show that if γ and σ are equivalent rectifiable paths then V(γ) = V(σ)
...
Not available
...
Show that if γ : [a, b] → C is a path then there is an equivalent path σ : [0, 1] → C
...
Not available
...
Prove Proposition 1
...
Solution
...
Exercise 19
...
Solution
...
Exercise 20
...
Solution
...
Exercise 21
...
Solution
...
Recall then that F1 = f = F2
...
Then for all z ∈ G,
h (z) = F1 (z) − F2 (z) = f (z) − f (z) = 0
Since G is open and connected with h (z) = 0 for all z ∈ G, then h is a constant, say c: h = c
...
47
Exercise 22
...
γ
G
...
Since
Solution
...
Then f has the antiderivative F(z) =
1−n
F (z) = f (z) where n ≥ 2 (note F is undefined for n = 1)
...
Since z1 = z2 , F(z1 ) = F(z2 )
...
Exercise 23
...
Let f and g be analytic in G and let γ be a
rectifiable curve from a to b in G
...
Solution
...
Then
b
fg +
f g
γ
(def)
γ
b
f (γ(t))g (γ(t)) dγ(t) +
=
a
f (γ(t))g(γ(t)) dγ(t)
a
b
=
f (γ(t))g (γ(t)) + f (γ(t))g(γ(t)) dγ(t)
Prop
...
7 (a)
a
=
(def)
fg + f g =
γ
( f g)
γ
Since ( f g) is continuous ( f and g are analytic) by Theorem 1
...
Hence,
γ
f g = f (b)g(b) − f (a)g(a) −
f g
...
2 Power series representation of analytic functions
Exercise 1
...
2) is continuous
...
Not available
...
Prove the following analogue of Leibniz
...
) Let G be an open set and let γ be a rectifiable curve in C
...
If
∂ϕ
∂z
exists for each (w, z) in {γ} × G and is continuous then g is analytic and
g (z) =
γ
∂ϕ
(w, z) dw
...
Not available
...
Suppose that γ is a rectifiable curve in C and ϕ is defined and continuous on {γ}
...
(w − z)n+1
g(n) (z) = n!
γ
Solution
...
Exercise 4
...
Prove that
an (z − a)n have radius of convergence 1 and suppose that
lim
r→1−
an rn = A
...
)
(b) Use Abel’s Theorem to prove that log 2 = 1 − 1 + 1 −
...
Not available
...
Give the power series expansion of log z about z = i and find its radius of convergence
...
Assume a ∈ C is not zero, then
1
1
1
=
=
z a+z−a a
1
a+z−a
a
=
1 1
1
=
a 1 + z−a
a
a
∞
n=0
(−1)n
(z − a)n =
an
∞
n=0
(−1)n
(z − a)n ,
an+1
where the radius of convergence is |a|
...
The convergence radius is |i| = 1
...
Give the power series expansion of z about z = 1 and find its radius of convergence
...
From the definition of numbers with rational exponents za = exp(a log z) we see that the square
root cannot be analytic on all of C
...
Note that the logarithm is undefined at zero
...
49
Exercise 7
...
Solution
...
Clearly f (z) is analytic on C and
¯
B(0; 1) ⊂ C
...
13
...
z2
¯
b) Let a = a, r = r, n = 0, and f (z) = 1
...
Now use Corollary
2
...
f (0) (a) =
0!
2πi
γ
1
1
dz ⇐⇒ 1 =
2πi
(z − a)1
γ
1
dz ⇐⇒
z−a
γ
1
dz = 2πi
...
Then f (z) = cos(z) and f (z) = − sin(z)
...
38) and B(0; 1) ⊂ C
...
13
...
z3
d) We have that
f (z) =
log(z)
zn
is analytic in the disk B(1; 1 ) where n ≥ 0
...
Obviously γ(t) = 1 + 1 eit ⊆ B(1; 1 )
...
15, f
2
2
has a primitive and therefore γ f = 0
...
zn
Exercise 8
...
15 holds if the disk B(a; R) is replaced
by a half plane
...
Not available
...
Evaluate the following integrals:
(a)
ez − e−z
dz where n is a positive integer and γ(t) = eit , 0 ≤ t ≤ 2π;
zn
γ
(b)
dz
γ
z−
1 n
2
where n is a positive integer and γ(t) =
1
+ eit , 0 ≤ t ≤ 2π;
2
(c)
γ
dz
where γ(t) = 2eit , 0 ≤ t ≤ 2π
...
a) Let a = 0, r = 1, n = m − 1 and
f (z) = ez − e−z
f (z) = ez + e−z
f (z) = ez − e−z
...
...
¯
Clearly f (z) is analytic on C and B(0; 1) ⊂ C
...
13
...
...
(m−1)
f
(z) = 0
...
¯
Clearly f (z) is analytic on C and B( 1 ; 1) ⊂ C
...
13
...
,m > 1
1
B
Az + Ai + Bz − Bi
A
!
+
=
=
z2 + 1 z − i z + i
z2 + 1
which implies
A+B = 0
Ai − Bi
(4
...
(4
...
1), we get A = −B inserted into (4
...
Thus
dz
...
6
...
Clearly f (z) is analytic on C
...
Then
f (z) =
1
2πi
γ
f (w)
1
dw ⇐⇒ 1 =
w−i
2πi
γ
1
dw ⇐⇒
w−i
γ
1
dz = 2πi,
z−i
where | − i − 0| < 2 ⇐⇒ 1 < 2
...
6
...
Clearly f (z) is analytic on C
...
Then
f (z) =
1
2πi
γ
f (w)
1
dw ⇐⇒ 1 =
w+i
2πi
γ
1
dw ⇐⇒
w+i
γ
1
dz = 2πi,
z+i
where |i| < 2 ⇐⇒ 1 < 2
...
z + i 2i
2i
dz
= 0
...
Clearly f (z) is analytic on C and B(0; 1) ⊂ C
...
13
...
z
e) Let a = 1, r = 1/2, n = m − 1 and
f (z) = z1/m
1 1/m−1
f (z) =
z
m
1 1
− 1 z1/m−2
f (z) =
m m
...
...
·
− (m − 2) z1/m−(m−1)
...
Now use Corollary 2
...
2
f (m−1) (1) =
γ
z1/m
dz
(z − 1)m
(m − 1)!
1
1 1
− 1 ·
...
Evaluate
2 < r < ∞
...
Let f (z) =
(m − 1)!
2πi
2πi 1 1 − m
z1/m
dz =
(z − 1)m
(m − 1)! m
m
z1/m
2πi
dz =
(z − 1)m
(m − 1)!mm−1
z2 +1
γ z(z2 +4)
γ
z1/m
dz
(z − 1)m
1 − 2m
1 − (m − 1)m
·
...
dz where γ(t) = reit , 0 ≤ t ≤ 2π, for all possible values of r, 0 < r < 2 and
z2 + 1
...
=
z
z − 2i z + 2i
z(z2 + 4)
1
z2 + 1
1
1
1
3
3
1
dz =
dz +
dz +
dz
...
6,
dz =
4 γ z
8 γ z − 2i
8 γ z + 2i
z(z2 + 4)
γ
γ z
1
2πi f (0) = 2πi where f ≡ 1 as f ≡ 1 is analytic on B(0, 2) and γ = 0 + reit with 0 < r < 2
...
Similarly,
analytic on B(0, 2) and γ for 0 < r < 2 is in B(0, 2)
...
15,
z − 2i
γ
1
1
z2 + 1
π
dz = 2πi = i
...
It follows for 0 < r < 2,
2 + 4)
4
2
γ z + 2i
γ z(z
so γ (t) = ireit
...
0
For the second integral, we use Proposition 2
...
Clearly f (z) is analytic on C and B(0; 2 < r < ∞) ⊂ C
...
Similarly,
f (2i) =
1
2πi
γ
f (w)
dw ⇐⇒
w − 2i
γ
1
dz = 2πi
...
Find the domain of analyticity of
f (z) =
1
1 + iz
log
;
2i
1 − iz
also, show that tan f (z) = z (i
...
, f is a branch of arctan z)
...
3
...
)
Solution
...
Exercise 12
...
These numbers are called Euler’s constants
...
(E10 = 50521 and E12 = 2702765)
...
It is easily seen that
sec z = 1 +
∞
k=1
E2k 2k
z ,
(2k)!
54
(for some E2 , E4 ,
...
1
1
since sec(0) = cos(0) = 1 = 1 and sec z = cos z is an even function
...
2
Since 1 = cos z sec z, we can multiply the series for sec z and cos z together to obtain
∞ n
2i
E2k
z
...
In addition, we have E2 = 1, E4 = 5, E6 = 61 and E8 = 1285
...
Find the series expansion of
f (z) = ez z and let
−1
ez −1
z
about zero and determine its radius of convergence
...
What is the radius of convergence? Show that
0 = a0 +
n+1
1
a1 + · · · +
n+1
n
an
...
The numbers
2
B2n = (−1)n−1 a2n are called the Bernoulli numbers for n ≥ 1
...
Solution
...
We determine the largest R such that
z
=
ez − 1
∞
k=0
Bk k
z
k!
for 0 < |z| < R
...
1
if x = 0
Furthermore the function f has essential singularities exactly at z = 2πk, k ∈ Z − {0}
...
The uniqueness of the Laurent expression together with the
55
formula for the coefficients in the expansion (Theorem 2
...
72) give the desired result
...
Also
−1
z
1
compute B1
...
Consider the equation ez z = ∞ Bk zk for
k=0 k!
−1
(ez −1)2
z ∈ ann(0; 0, 2π)
...
z=
l! k=0 k!
l=1
Divide by z
0, then
1=
=
∞
∞
∞
Bk
Bk
zl
zl−1
=
l! k=0 k! l=0 (l + 1)! k=0 k!
∞
l=1
∞
n
zn−m
Bm m
z
m! (n + 1 − m)!
n=0 m=0
=
∞
n
zn
n=0
=
∞
Bm
m!(n + 1 − m)!
m=0
n
n=0
zn
n+1
Bm
(n + 1)! m=0 m
= B0 +
∞
n
n=1
Since B0 = 0 conclude that
0=
zn
n+1
Bm
...
This can only happen if the inner summation equals zero for all n ∈ N
...
For the second claim, one only needs to show that f (z) + 2 z is an
even function
...
f (z) + z =
2
2(ez − 1)
2(ez − 1)
as well as t’
1
−z(ez + 1) −z(1 + ez ) z(1 + ez )
1
f (−z) − z = −z
=
=
= f (z) + z
...
56
In part b) the values of B0 = 1 and B1 = − 1 have been computed already
...
It remains to compute B2 , B4 , B6 and B8
...
k
n+1
k=0
This gives
1
3
1
1
B2 = − (B0 + 3B1 ) = − 1 −
= ,
3
3
2
6
1
1
B4 = − (B0 + 5B1 + 10B2 ) = − ,
5
30
1
1
B6 = − (B0 + 7B1 + 21B2 + 35B4 ) =
,
7
42
and
1
1
B8 = − (B0 + 9B1 + 36B2 + 126B4 + 84B6 ) = −
...
Find the power series expansion of tan z about z = 0, expressing the coefficients in terms of
Bernoulli numbers
...
)
2
2
Solution
...
We obtain
f (2iz) =
cos(z)
2iz(e2iz + 1) iz(eiz + e−iz )
=
=z
= z cot(z)
...
(2n)!
cot z − 1 tan z with 2z gives
2
z tan z = z cot z − 2z cot 2z
and therefore
Hence,
z tan z = 1 −
∞
n=1
4n B2n 2n
z − 1 −
(2n)!
tan(z) =
∞
n=1
∞
n=1
42n B2n 2n
z =
(2n)!
∞
n=1
4n (4n − 1)B2n 2n−1
z
...
(2n)!
4
...
Let f be an entire function and suppose there is a constant M, an R > 0, and an integer n ≥ 1
such that | f (z)| ≤ M|z|n for |z| > R
...
Solution
...
3 p
...
We have to show that f (k) (0) = 0 for k ≥ n + 1, then we are done
...
Note that γ is a closed and rectifiable curve
...
13 p
...
|z|k+1
Since r > R, |γ(t)| = |reit | = r > R, 0 ≤ t ≤ 2π
...
Hence, | f (k) (0)| = 0 ∀k ≥ n + 1 and thus
n
f (z) =
k=0
f (k) (0) k
z;
k!
a polynomial of degree less or equal n
...
Give an example to show that G must be assumed to be connected in Theorem 3
...
Solution
...
Exercise 3
...
Solution
...
Clearly f is analytic on C since it is entire
...
Now let
g be an entire function such that g(x) = e x for all x ∈ R
...
So, R ⊆ H
...
Hence, by Corollary 3
...
So f (z) = ez is the set of all
entire functions f such that f (x) = e x for all x ∈ R
...
Prove that ez+a = ez ea by applying Corollary 3
...
Solution
...
Exercise 5
...
8
...
Not available
...
Let G be a region and suppose that f : G → C is analytic and a ∈ G such that | f (a)| ≤ | f (z)|
for all z in G
...
58
Solution
...
Let a ∈ G such that
| f (a)| ≤ | f (z)|
∀z ∈ G
...
3)
Obviously, f (a) satisfies (4
...
Now, suppose f (a) 0 ∀a ∈ G
...
g(z) =
f (z)
Clearly, g ∈ A(G) (since f (a) 0 ∀a ∈ G) and
|g(z)| =
(def)
1
1
≤
= |g(a)|,
| f (z)| (4
...
According to the Maximum Modulus Theorem 3
...
79, g is constant and thus f is constant
...
Exercise 7
...
Solution
...
Exercise 8
...
Show that either f ≡ 0 or g ≡ 0
...
It is easy to verify the following claim:
If f (z) =constant in B(a; R) ⊂ G, then f (z) =same constant for all z ∈ G
...
Define g(z) = f (z) − c, then g(z) ∈ A(G)
...
So g(z) = f (z) − c ≡ 0 on G
...
Assume f (z) 0, ∀z ∈ G
...
Hence g(z) = 0 ∀z ∈ B(a; R) to fulfill f (z)g(z) = 0
∀z ∈ B(a; R)
...
g(z) = 0,
So g ≡ 0
...
0 ∀z ∈ G, then by the same reasoning, we obtain f ≡ 0
...
Let U : C → R be a harmonic function such that U(z) ≥ 0 for all z in C; prove that U is
constant
...
There exists an analytic function f : C → C such that
U(z) = Re( f (z)),
since U : C → R is assumed to be harmonic
...
For all z ∈ C,
define g(z) = e− f (z) ∈ A(C)
...
Thus, g(z) is bounded and an entire function,
and therefore g is constant according to Liouville’s Theorem 3
...
77 and hence f (z) is constant, too
...
59
Exercise 10
...
Solution
...
4
...
Prove Proposition 4
...
Solution
...
Exercise 2
...
Solution
...
Exercise 3
...
If γ(t) = Reit , 0 ≤ t ≤ 2π, show that γ p (z) dz = 2πin
...
Since p(z) is a polynomial of degree n, we can write
n
p(z) = c
k=1
(z − ak )
for some constant c (see Corollary 3
...
77) where a1 ,
...
By the product rule we
obtain
n
n
p (z) = c
k=1
(z − ak )
n
(z − al ) = c
l=1
l k
n
k=1 l=1
(z − al )
...
+
dz
z − a1 z − a2
z − an
(z − a1 )−1 dz +
γ
(z − a2 )−1 dz +
...
+ n(γ, an )2πi =
2πi = 2πin,
k=1
since by Definition 4
...
81 γ(t) = Reit , 0 ≤ t ≤ 2π or γ(t) = Re2πit , 0 ≤ t ≤ 1 is a closed and rectifiable
curve and a1 ,
...
In addition,
we have n(γ; ai ) = 1 ∀1 ≤ i ≤ n
...
p(z)
Exercise 4
...
Show that there is an
integer k such that γ z−1 dz = log r + iθ + 2πik
...
Not available
...
5 Cauchy’s Theorem and Integral Formula
Exercise 1
...
Prove that ϕ is continuous and for each fixed w, z → ϕ(z, w) is analytic
...
Not available
...
Give the details of the proof of Theorem 5
...
Solution
...
¯
Exercise 3
...
Let γ1 , γ2 , γ3 be curves whose traces are |z−1| = 1,
2
|z+1| = 1, and |z| = 2, respectively
...
Solution
...
Exercise 4
...
Solution
...
If γ is a closed rectifiable curve in G such that n(γ; w) = 0 ∀w ∈ G − C,
then γ f = 0
...
Define
f (z)− f (a)
, z a
...
(Check the power series of f (z) like in the proof of Theorem 3
...
78) By
Cauchy’s Theorem, we obtain
g=0
γ
⇒
a γ
⇒
⇒
⇒
Def
...
2 p
...
2πi γ z − a
γ
This is exactly Cauchy’s Integral Formula
...
Let γ be a closed rectifiable curve in C and a
{γ}
...
Solution
...
So there exists R > 0 such that
{γ} ⊂ B(a; R)
...
Therefore n(γ; w) = 0
∀w ∈ C\B(a; R) by Theorem 4
...
82
...
9 p
...
Further γ is a closed and rectifiable curve in G such that
∀w ∈ C\G,
n(γ; w) = 0,
61
then for a in B(a; R)\{γ}
1
(n − 1)!
dz
2πi
(z − a)n
γ
(n − 1)!
1
⇒ 0· n(γ; a) =
dz
2πi
(z − a)n
γ
f (n−1) (a)n(γ; a) =
since f (n−1) (a) = 0 if n ≥ 2
...
Exercise 6
...
Show | f (0)| ≤ 1
...
We can directly apply Cauchy’s estimate 2
...
73 with a = 0, R = 1, M = 1 and N = 1, that is
by assumption f is analytic on B(0; 1) and | f (z)| ≤ 1 for all z in B(0; 1) (since |z| < 1)
...
Exercise 7
...
Find
γ
1!· 1
= 1
...
Solution
...
By Theorem 4
...
82, n(γ; w) = 0 ∀w ∈ C\G since C\G belongs to the unbounded
component of G and γ is a closed and rectifiable curve in C
...
9 with G and f defined above such that n(γ; w) = 0 ∀w ∈ C\G and γ is a
closed rectifiable curve in G, then for a = 1 in G\{γ}
f (n−1) (1)n(γ; a) =
(n − 1)!
2πi
γ
z n
dz = 2πin,
z−1
⇒ n! =
⇒
γ
zn
(n − 1)!
(n − 1)!
dz =
2πi
(z − 1)n
2πi
γ
n
z
dz
z−1
γ
z
z−1
n
dz
since f (n−1) (1) = n! and n(γ; a) = 1
...
Let G be a region and suppose fn : G → C is analytic for each n ≥ 1
...
Solution
...
Let T be an arbitrary triangular path in G
...
We also have, that n(T ; w) = 0
∀w ∈ C\G, since w is an element of the unbounded component of G (Theorem 4
...
82)
...
85 with m = 1, we have
∀n ∈ N,
fn = 0,
T
62
(4
...
We also have
f = lim
n→∞
T
T
fn = lim 0 = 0
(4
...
7 p
...
1 p
...
Finally, we can apply Morera’s Theorem 5
...
86 to obtain that f is analytic in G, since f is continuous
and γ f = 0 for every triangular path T in G
...
Show that if f : C → C is a continuous function such that f is analytic off [−1, 1] then f is an
entire function
...
By Morera’s Theorem, it suffice to show that for every triangular path T in C, we have T f = 0,
since we already assume f is continuous
...
In this case, we obtain by Cauchy’s Theorem, that T f = 0,
since we can find an open neighborhood G containing T (is closed and rectifiable) such that f ∈ A(G) (by
assumption) and n(T ; w) = 0 ∀w ∈ C\G (by Theorem 4
...
82)
...
This single point of intersection P is a removable
singularity, since f is continuous
...
(Procedure: Translate
the triangle by ± i depending if it lies above or below the x-axis
...
Case 3: One edge of the triangle touches I
...
Let {T n } be a sequence of
triangles that are not intersecting I, but whose limit is the given T
...
7 and the second one by case 1
...
Then, we can always decompose the triangle T in three parts
...
Thus, T f = 0 again
...
In this instance, we can decompose
the triangle into 5 parts
...
Hence, T f = 0
...
Exercise 10
...
(This exercise was
taken from a paper by C
...
McCarthy, Amer
...
Monthly, 82 (1975), 390–391)
...
Not available
...
6 The homotopic version of Cauchy’s Theorem and simple connectivity
Exercise 1
...
Show that if γ is closed rectifiable curve in G and γ ∼ σ1 , then γ ∼ σ2
...
)
63
Solution
...
Since σ1 (t) ≡ a and σ2 (t) ≡ b, we
surely have that σ1 and σ2 are closed and rectifiable curves in G
...
So, now we show σ1 ∼ σ2 , that is, we have to find a continuous function Γ : [0, 1] × [0, 1] → G such that
Γ(s, 0) = σ1 (s) = a
and
Γ(s, 1) = σ2 (s) = b,
for
0≤s≤1
and Γ(0, t) = Γ(1, t) for 0 ≤ t ≤ 1
...
Clearly Γ is a continuous function
...
In addition, it satisfies Γ(s, 0) = a ∀0 ≤ s ≤ 1 and
Γ(s, 1) = b ∀0 ≤ s ≤ 1 and Γ(0, t) = tb + (1 − t)a = Γ(1, t) ∀0 ≤ t ≤ 1
...
Exercise 2
...
1 then the
curve γ0 (t) = e2πit , 0 ≤ t ≤ 1, is homotopic to the constant curve γ1 (t) ≡ 1 in the region G = C − {0}
...
Not available
...
3
...
11 gives an
equivalence relation on C
...
Not available
...
Let G = C − {0} and show that every closed curve in G is homotopic to a closed curve whose
trace is contained in {z : |z| = 1}
...
Not available
...
Evaluate the integral
dz
γ z2 +1
where γ(θ) = 2| cos 2θ|eiθ for 0 ≤ θ ≤ 2π
...
A sketch shows that the two zeros of z2 + 1 (they are ±i) are inside of the closed and rectifiable
curve γ (The region looks like a clover with four leaves )
...
2 p
...
Hence
γ
dz
= 0
...
Let γ(θ) = θeiθ for 0 ≤ θ ≤ 2π and γ(θ) = 4π − θ for 2π ≤ θ ≤ 4π
...
γ z2 +π2
Solution
...
Using
partial fraction decomposition yields
γ
dz
z2 + π2
=
=
Def 4
...
81
1
1
1
1
dz −
dz
2πi γ z − iπ
2πi γ z + iπ
n(γ; iπ) − n(γ; −iπ) = 0 − 1 = −1,
64
since iπ is not contained in the region generated by γ, so n(γ; iπ) = 0 and −iπ is contained in the region, so
n(γ; −iπ) = 1
...
2 + π2
γ z
i
1
Exercise 7
...
2
2
Find γ f
...
Not available
...
Let G = C − {a, b}, a b, and let γ be the curve in the figure below
...
(b) Convince yourself that γ is not homotopic to zero
...
Can you prove it?) Notice that this example shows that it is possible to have a closed curve γ in a region
such that n(γ; z) = 0 for all z not in G without γ being homotopic to zero
...
10 is false
...
Let γ be the path depicted on p
...
We can write it as a sum of 6 paths
...
Therefore, two integrals will be zero and we
will have another 2 pair of non-closed paths
...
The second pair begins at the middle crossing pair and goes around b in
opposite direction
...
If we integrate over the path around a
1
is equivalent to evaluate γ −γ z−a dz where γ1 (t) = a + reit , 0 ≤ t ≤ π and γ2 (t) = a + reit , π ≤ t ≤ 2π for
1
2
some r > 0
...
Exercise 9
...
Suppose γ0 ∼ γ1 , and Γ
satisfies (6
...
Also suppose that γt (s) = Γ(s, t) is smooth for each t
...
Solution
...
Exercise 10
...
dz
,
γ 1+z2
where γ is any closed rectifiable curve in C not passing
Solution
...
Using partial fraction decomposition and the definition of the winding number we obtain
γ
1
dz
=
1 + z2 2i
γ
Exercise 11
...
z+i
2i
dz where γ is one of the curves depicted below
...
)
Solution
...
9 p
...
In our case a = 0 and γ satisfies the above assumptions for exercise a), b) and c)
...
We have f (3) (0) = e0 + e−0 = 2 and using the formula above for k = 3
we obtain
1 2πi
ez − e−z
dz = 2πi· 2n(γ; 0) =
n(γ; 0)
...
Hence, n(γ; 0) = 1 for part a) and thus the result is
is 4πi
...
3
We have n(γ; 0) = 1 for part b) and thus the result
4
...
Show that if f : G → C is analytic and γ is a rectifiable curve in G then f ◦ γ is also a
rectifiable curve
...
Not available
...
Let G be open and suppose that γ is a closed rectifiable curve in G such that γ ≈ 0
...
(a) Show that {z : d(z, ∂G) < 1 r} ⊂ H
...
Solution
...
Exercise 3
...
Show that a is a zero of multiplicity m
iff f (m−1) (a) =
...
Solution
...
, m − 1 (we also have f (a) = 0) and f (m) (a)
f ∈ A(B(a; R)), we can write
∞
f (z) =
k=0
∞
=
k=0
ak (z − a)k
where
f (k) (a)
(z − a)k =
k!
= (z − a)m
∞
k=0
ak =
∞
k=m
0
...
Hence, a is a zero of multiplicity m
...
By definition, there exists g ∈ A(B(0; R)) such that
f (z) = (z − a)m g(z),
g(a)
0 (p
...
Then after a lengthy calculation, we obtain
n
f (n) (z) =
k=0
n
k
m!
(z − a)m−(n−k) g(k) (z),
(m − (n − k))!
66
n = 1, 2,
...
If z = a, then clearly f (n) (a) = 0 for m − (n − k) > 0 and f (n) (a) 0 for m − (n − k) = 0
...
Hence, f (n) (a) = 0 for n = 1, 2,
...
Exercise 4
...
Solution
...
Then f (a) = α where α is a constant
...
Then F(a) = f (a) − α = α − α = 0 and F (a) = f (a) = 0 by assumption
...
Then, we can use Theorem 7
...
98 to argue that there is an > 0 and δ > 0 such that for 0 < |ξ − a| < δ,
the equation f (z) = ξ has exactly m simple roots in B(a; ), since f is analytic in B(a; R) and α = f (a)
...
Since f (z) = ξ has exactly m (our case m ≥ 2) simple roots in B(a; ), we can find at least two distinct
points z1 , z2 ∈ B(a; ) ⊂ G such that
f (z1 ) = ξ = f (z2 )
contradicting that f is 1 − 1
...
Exercise 5
...
Show that f
is an open map iff f is a closed map
...
)
Solution
...
Exercise 6
...
(Hint: Consider the set F = {z : Im z = (Re z)−1 and Re z 0}
...
Not available
...
Use Theorem 7
...
Solution
...
4
...
67
Chapter 5
Singularities
5
...
Each of the following functions f has an isolated singularity at z = 0
...
a)
sin z
;
f (z) =
z
b)
cos z
;
f (z) =
z
c)
cos z − 1
f (z) =
;
z
d)
f (z) = exp(z−1 );
e)
f (z) =
log(z + 1)
;
z2
f)
f (z) =
cos(z−1 )
;
z−1
f (z) =
z2 + 1
;
z(z − 1)
g)
h)
f (z) = (1 − ez )−1 ;
i)
1
f (z) = z sin ;
z
68
j)
1
f (z) = zn sin
...
a) According to Theorem 1
...
103, z = 0 is a removable singularity, since
lim (z − 0)
z→∞
If we define,
sin(z)
sin(z)
= lim z
= lim sin(z) = 0
...
A simple computation using L’Hospital’s rule yields limz→0
b) We have (see p
...
1
z2k−1
= + analytic part
...
c) According to Theorem 1
...
103, z = 0 is a removable singularity, since
lim (z − 0)
z→∞
cos(z) − 1
cos(z) − 1
= lim z
= lim cos(z) − 1 = 1 − 1 = 0
...
A simple computation using L’Hospital’s rule yields limz→0
e) We know
∞
zk
(−1)k+1
...
So we can write
1
log(1 + z)
= 2
z2
z
∞
k=1
(−1)k+1
zk
=
k
∞
(−1)k+1
k=1
zk−2 1 1
= − +
k
z 2
∞
k=3
(−1)k+1
zk−2
k
and hence f (z) = log(1+z) has a pole of order 1 at z = 0 and the singular part is 1 (by Equation 1
...
8 p
...
2
z+1
z+1
g) First, we simplify z 2+1 = 1 + z(z−1)
...
7 and Definition 1
...
105)
...
38)
∞
z2k+1
(−1)k
sin(z) =
(2k + 1)!
k=0
so
1 2k+1
∞
1
z
=
f (z) = z sin = z (−1)k
z
(2k + 1)!
k=0
∞
(−1)
1 2k
z
k
k=0
(2k + 1)!
=1−
2
z−1
−
1
z
(by Equation
1
1
1
+
−
+
...
18c)
...
109
...
300)
...
Exercise 2
...
(z2 +z+1)(z−1)2
Solution
...
Exercise 3
...
17) from (1
...
Solution
...
1
Exercise 4
...
Solution
...
1 1
−
=
z
z
|1/z|<1and |z/2|<1 z
2
n+1
zn−1 =
∞
n=0
(1 < |z| < 2)
...
Exercise 5
...
Determine the singular part of f at each of these poles
...
Define zn = π + nπ
...
Hence, tan(z) has a simple
2
1
pole at each zn with residue −1 and with singular part − z−zn at each zn
...
If f : G → C is analytic except for poles show that the poles of f cannot have a limit point in
G
...
Not available
...
Let f have an isolated singularity at z = a and suppose f (z) 0
...
19) or
(1
...
19) holds if s > m and (1
...
Solution
...
Then it holds
obviously for any greater s’s
...
1)
lim(z − a)n+1 f (z) = 0
(5
...
Now, (z − a)n f (z) thus has a removable
singularity at z = a (by Theorem 1
...
103) applied to (5
...
2)
...
So
(z − a)n f (z) = (z − a)k h(z)
where h(z) is analytic at a and h(a)
Thus,
0
...
where m = n − k
...
Then there exists an integer n < 0 so that
lim |z − a|n | f (z)| = ∞
(5
...
Now (z − a)n f (z) thus has a pole at z = a
(by Definition 1
...
105 and (5
...
4 p
...
Hence
0,
s
s−n+n
s−n−l
lim(z − a) f (z) = lim(z − a)
f (z) = lim(z − a)
k(z) = ±∞,
z→a
z→a
z→a
k(a)
0,
lim |z − a| s | f (z)| =
∞,
z→a
where m = n + l
...
s>m
s
Exercise 8
...
Show: (a) m = 0 iff z = a is a removable singularity and
f (a) 0; (b) m < 0 iff z = a is a removable singularity and f has a zero at z = a of order −m; (c) m > 0 iff
r = a is a pole of f of order m
...
a) ⇒: Let m = 0
...
z→a
(5
...
2 p
...
⇐: Let z = a be a removable singularity and f (a) = limz→a f (z)
0
...
2 p
...
By Exercise 7, we obtain that there is an integer m such that
0, s > m
s
lim |z − a| | f (z)| =
∞, s < m
...
4)
...
Then by Exercise 7, we have
lim f (z) = 0
z→a
lim(z − a) f (z) = 0
...
Thus, f (z) has a removable
singularity at z = a and also the extended function has the value limz→a f (z) = 0 at z = a (p
...
2)
...
Therefore f (z) = (z − a) h(z) has a zero of order m at z = a
...
Then
f (z) = (z − a)m h(z) where h(z) is analytic in |z − a| < δ, δ > 0 and h(a) 0
...
Hence, the algebraic order of f at z = a is equal to −m (m > 0)
...
Then
lim(z − a) s f (z) = 0
z→a
for s > m by Exercise 7, that is
lim(z − a)m+1 f (z) = 0
z→a
and lim(z − a)m f (z)
z→a
0
by Exercise 7
...
Now f (z) = (z − a)−m h(z) where h ∈ A(B(a; δ)) and h(a) 0
...
⇐: Let f (z) have a pole of order m at z = a
...
Thus limz→a (z − a)m f (z) = g(a) 0 and
lim(z − a)m+1 f (z) = lim(z − a)g(z) = 0
...
Therefore, m > 0 is the algebraic order of f at z = a
...
A function f has an essential singularity at z = a iff neither (1
...
20) holds for any real
number s
...
⇒: We prove this direction by proving the contrapositive (P → Q ⇐⇒ ¬Q ⇒ ¬P)
...
19) or (1
...
Then by Exercise 7, there exists an integer m such that (1
...
20) holds if s < m
...
Hence z = a is not an essential singularity (by Definition 1
...
105)
...
Assume that f has an isolated
singularity at z = a which is not an essential singularity
...
3 p
...
Again by Problem 7 and 8, in either case there exists an m such that (1
...
20) holds
...
Exercise 10
...
Prove the following strengthened version
of the Casorati–Weierstrass Theorem
...
73
Solution
...
Exercise 11
...
Can you generalize this result?
Solution
...
Exercise 12
...
0
1
1
λ z−
2
z
for 0 < |z| < ∞, where
an zn +
n=1
(b) Similarly, show
exp
∞
= a0 +
= b0 +
∞
bn zn +
n=1
(−1)n
zn
π
0
cos(nt − λ sin t) dt
...
Not available
...
Let R > 0 and G = {z : |z| > R}; a function f : G → C has a removable singularity, a pole, or
an essential singularity at infinity if f (z−1 ) has, respectively, a removable singularity, a pole, or an essential
singularity at z = 0
...
(a) Prove that an entire function has a removable singularity at infinity iff it is a constant
...
(c) Characterize those rational functions which have a removable singularity at infinity
...
Solution
...
Exercise 14
...
Suppose that γ is a closed rectifiable
curve in G such that n(γ; a) = 0 for all a in C − G
...
Not available
...
Let f be analytic in G = {z : 0 < |z − a| < r} except that there is a sequence of poles {an } in G
with an → a
...
Solution
...
−1
1
and g(z) = 0 (t − z)−1 dt are
Exercise 16
...
Do they have any isolated singularities? Do they have any singularities that are not isolated?
Solution
...
Exercise 17
...
Show that if G | f (x + iy)|2 dx dy < ∞ then f
has a removable singularity at z = a
...
Not available
...
2 Residues
Exercise 1
...
Note that
∞
0
x2
since x4 +x2 +1 is even
...
(a + cos θ)2
x2 dx
1
=
x4 + x2 + 1 2
∞
−∞
x2 dx
x4 + x2 + 1
Let γR = [−R, R] ∪ CR where CR = Reit , 0 ≤ t ≤ π and R
0 (draw a sketch)
...
By the residual Theorem, we have
simple poles at a1 = e
2
R
R
2πi(Res( f, a1 ) + Res( f, a2 )) =
f =
−R
γR
x4
x2
dx +
+ x2 + 1
CR
z4
z2
dz
...
dz + 2πi
4 + z2 + 1
z
k=1
We have
f (z) dz =
CR
CR
z4
z2
dz ≤
+ z2 + 1
2
=
R
R4 − R2 − 1
CR
z4
z2
dz ≤
+ z2 + 1
CR
3
CR
|dz| =
|z4
|z|2
|dz| ≤
+ z2 + 1|
CR
πR
→ 0 as R → ∞
...
π
Res( f, ei 3 )
=
π
limπ (z − ei 3 )
z→ei 3
z4
2π
π
z2
1
= ei 3 limπ (z − ei 3 ) 4
i3
+ z2 + 1
z + z2 + 1
z→e
2π
=
=
π
ei 3
1
ei 3
=
=
3 + 2z
iπ
i 3π
i 2π
4e 3 + 2e 3
4e 3 + 2
z→ei 3 4z
√
√
1
1
1
1
+ 2 3i
1
1
2 + 2 3i
= 2 √
= √ +
√
1
1
4(− 2 + 2 3i) + 2
2 3i
4 3i 4
1
1
− √ i
...
2π
Res( f, ei 3 )
2π
lim (z − ei 3 )
=
2π
z→ei 3
z4
4π
2π
z2
1
= ei 3 lim (z − ei 3 ) 4
2+1
2π
+z
z + z2 + 1
z→ei 3
2π
4π
ei 3
1
ei 3
=
=
4π
2π
6π
2π 4z3 + 2z
4ei 3 + 2
4ei 3 + 2ei 3
z→ei 3
√
√
1
1
− 1 + 2 3i
− 1 + 2 3i
1
1
2
=
= 2 √
= √ −
√
1
1
4
4(− 2 − 2 3i) + 2
−2 3i
4 3i
1
1
= − − √ i
...
− √ i+ √ −
4 4 3
4
4 3i
2 3
3
3
Hence,
∞
0
x4
x2 dx
1 π
= √
...
Clearly γ f (z) dz = 0
since f is analytic on the region enclosed by γ
...
γr
γR
e−ix − 1
dx
...
2
R
R
76
−1
dx
The second term on the right hand side is π, since
1
eiz − 1
dz = − 2πi Res( f (z), 0) = −πi2 = π
2
2
z
lim
r→0
( f (z) can be written as
γr
i
z +analytic
part and therefore Res( f (z), 0) = i)
...
2
2
2
2
x
c) If a = 0, then
π
0
Assume a
0
cos 2θ dθ
1
=
1 − 2a cos θ + a2 2
Let z = eiθ
...
0
2
2
0
...
+ a2
dθ
=
= −
1
z2
1
dz
=
2 iz
4i
+a
2
1−a z+
|z|=1
1
4ia
1−
0
e2θi +e2θi
2
eθi +eθi
2a 2
dθ
...
z2 (z−a)(z− 1 )
a
It is easily obtained
d 2
z4 + 1
Res( f, 0) = lim z
2
z→0 dz z (z − a) z −
= lim
1
a
= lim d
z→0
dz
z4 + 1
(z − a) z −
1
−4z3 z2 − az − a z + 1 − (z4 − 1) 2z − a −
z→0
z2 − az − 1 z + 1
a
1
a
1
a
=a+
2
1
a
and
Res( f, a) = lim(z − a)
z→a
So
z4 + 1
|z|=1
Finally,
z2 (z − a) z −
π
0
1
a
z4 + 1
= lim
z4 + 1
a4 + 1
a4 + 1
...
+
2 − 1)
a a(a
a(a = 1)
1 − a2
z−
a2 a −
a2
1
cos 2θ dθ
4πia3
=π
=−
...
Verify the following equations:
a)
∞
0
(x2
b)
π
dx
= 3,
2 )2
+a
4a
∞
0
c)
∞
0
a > 0;
(log x)3
dx = 0;
1 + x2
π(a + 1)e−a
cos ax
dx =
,
4
(1 + x2 )2
if a > 0;
d)
π/2
0
π
dθ
=
,
1
2
a + sin θ 2 [a(a + 1)] 2
e)
∞
0
π
log x
dx = − ;
4
(1 + x2 )2
f)
∞
0
g)
∞
−∞
h)
if a > 0;
dx
π
= ;
1 + x2 2
eax
π
dx =
,
x
1+e
sin aπ
2π
if
0 < a < 1;
4π
log sin2 2θ dθ = 4
0
0
log sin θ dθ = −4π log 2
...
a) Let f (z) = (z2 +a2 )2 and γR = [−R, R] + CR where CR = Reit , 0 ≤ t ≤ π
...
So we have by the Residue Theorem
f (z) dz = 2πi Res( f, z1 )
γR
where
Res( f, z1 )
=
d
d
1
1
= lim
(z − ai)2
z→ai dz (z + ai)2
dz
(z − ai)2 (z + ai)2
−2
2
1
lim
=−
= 3
...
(R2 − a2 )2
78
CR
|dz|
Hence,
∞
0
eiaz
c) Let f (z) = (1+z2 )2
enclosed by γR
...
(x2 + a2 )2 2 2a3 4a3
it
and γR = [−R, R] + CR where CR = Re , 0 ≤ t ≤ π
...
−8i
4i
R
(a + 1)e−a π(a + 1)e−a
=
=
4i
2
f (x) dx +
−R
f (z) dz
CR
where the latter integral on the right hand side is zero since
f (z) dz
eiaz
|eiaz |
dz ≤
|dz| ≤
2 2
22
CR (1 + z )
CR |1 + z |
1
Rπ → 0 as R → ∞,
(R2 − 1)2
=
CR
=
CR
(|z|2
1
|dz|
− 1)2
since |eiaz | = e−Im(az) ≤ 1 since Im(az) ≥ 0 (a > 0 by assumption)
...
g) First of all substitute u = e x , so
∞
−∞
Now, let u = t2 , then
∞
0
Hence,
eax
dx =
1 + ex
ua−1
du = 2
1+u
∞
−∞
∞
0
∞
0
ua du
=
1+u u
t2a−2 t
dt = 2
1 + t2
eax
dx = 2
1 + ex
2a−1
∞
0
∞
0
ua−1
du
...
1 + t2
x2a−1
dx
...
Define log z on G = {z ∈ C : z 0, − π <
2
1+z
arg(z) < 3π } with log(z) = log |z| + iθ, θ = arg(z)
...
By the Residue
2
79
Theorem
= 2πRes( f, i) = 2πi lim(z − i)
f (z) dz
z→i
γ
= πi
Also
2a−1
−r
f (z) dz =
(2a−1) log i
= πe
π
π
= πe(2a−1)(0+i 2 ) = πeaπi e−i 2 = −πieaπi
...
f (x) dx +
γr
III
II
IV
I)
−r
R
f (x) dx
=
R
f (−x) dx =
−R
r
r
(−x)2a−1
dx =
1 + x2
R
(2a−1)iπ
= e
r
R
2aπi
= −e
r
e(2a−1) log x
dx = −e2aπi
1 + x2
x2a−1
dx
...
II) Similar to IV)
f (z) dz ≤
γr
πr2a
→ 0 as r → 0
r2 − 1
since a > 0
...
0
f (x) dx
0
−iπ
iπ
iπ
π
−πieaπi
=
=
=
=
...
Exercise 3
...
Solution
...
80
Exercise 4
...
Show that Res( f g; a) = g(a)Res( f ; a)
...
Not available
...
Use Exercise 4 to show that if G is a region and f is analytic in G except for simple poles at
a1 ,
...
, an such that γ ≈ 0 in G
...
Not available
...
Let γ be the rectangular path [n + 2 + ni, −n − 1 + ni, −n − 2 − ni, n + 1 − ni, n + i + ni] and
2
2
evaluate the integral γ π(z + a)−2 cot πz dz for a an integer
...
)
Solution
...
Exercise 7
...
Not available
...
Let γ be the polygonal path defined in Exercise 6 and evaluate
an integer
...
Solution
...
Exercise 9
...
Solution
...
81
2(−1)n a
a2 − n2
γ
π(z2 − a2 )−1 cot πz dz for a
Exercise 10
...
Show that
(±1) p (n+2p)!
(z2 ± 1)m dz p!(n+p)! , if m = 2p + n, p ≥ 0
1
=
0,
m+n+1
2πi γ z
otherwise
Solution
...
Exercise 11
...
12, consider an and bn as functions of the parameter λ and use Exercise 10 to
compute power series expansions for an (λ) and bn (λ)
...
)
Solution
...
Exercise 12
...
, am
...
k=1
1
(Res( f ; ∞) is defined as the residue of −z−2 f (z−1 ) at z = 0
...
) What can you say if f has infinitely many isolated
singularities?
Solution
...
Exercise 13
...
If γ(t) = Reit ,
0 ≤ t ≤ 2pi, evaluate γ [(z − a)(z − b)]−1 f (z) dz
...
Solution
...
5
...
Prove Theorem 3
...
Solution
...
¯
Exercise 2
...
Find the number of solutions
(counting multiplicities) of the equation f (z) = zn where n is an integer larger than or equal to 1
...
Not available
...
Let f be analytic in B(0; R) with f (0) = 0, f (0)
ρ = min{| f (z)| : |z| = R} > 0
...
Put
z f (z)
dz
f (z) − ω
where γ is the circle |z| = R
...
Solution
...
Exercise 4
...
82
Solution
...
Exercise 5
...
Solution
...
Exercise 6
...
(The letter
“H” stands for holomorphic
...
Others reserve the term
“analytic” for what many call the complete analytic function, which we will not describe here
...
Show that M(G), the
meromorphic functions on G, is a field
...
The question arises, is every meromorphic function on G the quotient of two analytic
functions on G? Alternately, is M(G) the quotient field of H(G)? The answer is yes but some additional
theory will be required before this answer can be proved
...
Not available
...
State and prove a more general version of Rouche’s Theorem for curves other than circles in
G
...
Not available
...
Is a non-constant meromorphic function on a region G an open mapping of G into C? Is it an
open mapping of G into C∞ ?
Solution
...
Exercise 9
...
Show that this solution must be real
...
Not available
...
Let f be analytic in a neighborhood of D = B(0; 1)
...
If | f (z)| ≤ 1 for |z| = 1, what can you say?
¯
Solution
...
Clearly g, h are analytic in a neighborhood of B(0, 1) since f is
analytic
...
Let γ = {z : |z| = 1}
...
By Roché’s Theorem, we get
Zg − Pg = Zh − Ph
...
Since h(z) = z has only one zero (z = 0) inside the unit circle, we get that
g(z) = f (z) − z
has exactly one zero inside the unit circle
...
If we choose f (z) = z, then | f (z)| ≤ 1 for |z| = 1 and therefore we have infinitely many fixed points
...
83
Chapter 6
The Maximum Modulus Theorem
6
...
Prove the following Minimum Principle
...
(See Exercise IV
...
6
...
Since f ∈ C(G) we have | f | ∈ C(G)
...
If a ∈ ∂G, then | f | assumes its minimum value on ∂G and we are done
...
But each Ai is a
region, so we can use Exercise IV 3
...
But f was assumed
to be non-constant, so f has to have a zero in G
...
Exercise 2
...
Show that
if there is a constant c ≥ 0 such that | f (z)| = c for all z on the boundary of G then either f is a constant
function or f has a zero in G
...
Assume there is a constant c ≥ 0 such that | f (z)| = c for all z ∈ ∂G
...
So
¯
∀z ∈ G
...
1)
¯
Since f ∈ C(G) which implies | f | ∈ C(G) and hence there exists an a ∈ G such that
| f (a)| ≤ | f (z)| ≤ c,
(6
...
6 either f is constant of f has a zero in G
...
(a) Let f be entire and non-constant
...
(b) Let p be a polynomial and show that each component of {z : |p(z)| < c} contains a zero of p
...
)
(c) If p is a polynomial and c > 0 show that {z : |p(z)| = c} is the union of a finite number of closed paths
...
Solution
...
Exercise 4
...
Show that there is a positive number
that for each polynomial p,
sup{|p(z) − z−1 | : z ∈ A} ≥
> 0 such
This says that z−1 is not the uniform limit of polynomials on A
...
Not available
...
Let f be analytic on B(0; R) with | f (z)| ≤ M for |z| ≤ R and | f (0)| = a > 0
...
Hint: If z1 ,
...
(Notation:
n
k=1
ak = a1 a2
...
)
Solution
...
¯
Exercise 6
...
Show that if
neither f nor g vanishes in B(0; R) then there is a constant λ, |λ| = 1, such that f = λg
...
Not available
...
Let f be analytic in the disk B(0; R) and for 0 ≤ r < R define A(r) = max{Re f (z) : |z| = r}
...
Solution
...
Exercise 8
...
Show that | f (z)| ≤ M, for
each z in G
...
We need to show
lim sup | f (z)| ≤ M
z→a
∀u ∈ ∂∞G
...
Instead of showing that
lim sup | f (zn )| ≤ M ⇒ lim sup | f (z)| ≤ M,
z→a
we show the contrapositive, that is
lim sup | f (z)| > M ⇒ lim sup | f (zn )| > M
...
But this implies lim supz→a | f (zn )| > M since zn → z as z → ∞, that is zn
gets arbitrarily close to z and since f is analytic, that is continuous, we have f (zn ) gets arbitrarily close to
f (z)
...
2 Schwarz’s Lemma
Exercise 1
...
By considering the
function g : D → D defined by
f (z) − a
g(z) =
1 − a f (z)
¯
where a = f (0), prove that
| f (0)| + |z|
| f (0)| − |z|
≤ | f (z)| ≤
1 + | f (0)| |z|
1 − | f (0)| |z|
for |z| < 1
...
We claim that |g(z)| ≤ 1 and g(0) = 0
...
1 − a f (0) 1 − aa
¯
¯
We also have
|g(z)| =
| f (z) − a|
≤1
|1 − a f (z)|
¯
since | f (z)| ≤ 1
...
⇐⇒
⇒
⇒
⇒
⇒
|g(z)| ≤ |z| for |z| < 1
| f (z) − a| ≤ |z|· |1 − a f (z)|
¯
|| f (z)| − |a|| ≤ | f (z) − a| ≤ |z|· |1 − a f (z)| ≤ |z| + |z|· |a|· | f (z)|
¯
|| f (z)| − |a|| ≤ |z| + |z|· |a|· | f (z)|
−|z| − |z|· |a|· | f (z)| ≤ | f (z)| − |a| ≤ |z| + |z|· |a|· | f (z)|
|a| − |z|
|a| + |z|
≤ | f (z)| ≤
...
1
Exercise 2
...
The answer is no
...
According
4
to the definition, we have
a
Eα = { f ∈ A(D) : | f (z)| ≤ 1, f (a) = α}
...
We know,
4
f (a) ≤
1 − |α|2
1 − |a|2
and therefore we must have
f
But f ( 1 ) =
2
2
3
1
2
≤
1−
1−
3 2
4
1 2
2
=
7
16
3
4
¯
= 0
...
86
=
7 3
7
¯
=
= 0
...
16 4 12
Exercise 3
...
(a) Show that Re f (z) > 0 for all z in D
...
What can be said if f (0) 1?
(c) Show that f (0) = 1, f also satisfies
(Hint: Use part (a))
...
Not available
...
Prove Caratheodory’s Inequality whose statement is as follows: Let f be analytic on B(0; R)
and let M(r) = max{| f (z)| : |z| = r}, A(r) = max{Re f (z) : |z| = r}; then for 0 < r < R, if A(r) ≥ 0,
M(r) ≤
R+r
[A(R) + | f (0)|]
R−r
(Hint: First consider the case where f (0) = 0 and examine the function g(z) = f (Rz)[2A(R) + f (Rz)]−1 for
|z| < 1
...
Not available
...
Let f be analytic in D = {z : |z| < 1} and suppose that | f (z)| ≤ M for all z in D
...
(b) If f (zk ) = 0 for 1 ≤ k ≤ n, each zk
0, and f (0) = Meia (z1 z2 ,
...
Solution
...
¯
Exercise 6
...
Find a
formula for f
...
)
Solution
...
¯
Exercise 7
...
Suppose that
1
1
f has a zero at z = 4 (1 + i) and a double zero at z = 1
...
Not available
...
Is there an analytic function f on B(0; 1) such that | f (z)| < 1 for |z| < 1, f (0) =
f (0) = 3 ? If so, find such an f
...
Not available
...
3 Convex functions and Hadamard’s Three Circles Theorem
Exercise 1
...
Show that f is logarithmically convex iff f (x) f (x) − [ f (x)]2 ≥ 0 for all x
...
Not available
...
Show that if f : (a, b) → R is convex then f is continuous
...
Since f is convex on (a, b) and if a < s < t < u < b, we have
f (u) − f (s)
f (u) − f (t)
f (t) − f (s)
≤
≤
t−s
u−s
u−t
(6
...
(6
...
Let a < s < t < u < b, t = λs + (1 − λ)u with λ = u−s ∈ (0, 1)
...
3), we get (rearrange terms and add/subtract a term)
⇐⇒
⇐⇒
t f (u) − t f (s) − s f (u) + s f (s) − u f (t) − u f (s) + s f (t) − s f (s) ≥ 0
(t − s)( f (u) − f (s)) − (u − s)( f (t) − f (s)) ≥ 0
f (t) − f (s)
f (u) − f (s)
≤
...
3), we get
⇐⇒
⇐⇒
u f (u) − s f (u) − u f (t) + s f (t) − u f (u) + u f (s) + t f (u) − t f (s) ≥ 0
(u − s)( f (u) − f (t)) − (u − t)( f (u) − f (s)) ≥ 0
f (u) − f (s)
f (u) − f (t)
≤
...
2)
...
Claim:
f (x) − f (x − δ)
f (z) − f (x)
f (x + δ) − f (x)
≤
≤
δ
z−x
δ
∀z ∈ (x − δ, x + δ)
...
4)
If the claim is true, (6
...
δ
δ
f
Taking the limit z → x, we have that f (x)−δ (x−δ) (z − x) → 0 and f (x+δ)− f (x) (z − x) → 0
...
88
Thus it remains to show the claim
...
2)
...
2) gives
f (x) − f (x − δ)
f (z) − f (x)
≤
δ
z−x
which gives the first inequality in (6
...
Applying the outer inequality in (6
...
4)
...
Then, the first inequality in (6
...
2) applied to the three points x − δ, x, z
...
x−x+δ
z−x
δ
z−x
The second inequality in (6
...
2) applied to x, z, x + δ
...
z−x
x+δ−x
z−x
δ
Thus, we have proved the claim
...
f (x) =
1, x = b
Clearly, f is convex on [a, b], but f is not continuous at x = b (but the function is continuous on (a, b))
...
Show that a function f : [a, b] → R is convex iff any of the following equivalent conditions is
satisfied:
f (u) u 1
(a) a ≤ x < u < y ≤ b gives det f (x) x 1 ≥ 0;
f (y) y 1
f (y)− f (x)
f (u)− f (x)
(b) a ≤ x < u < y ≤ b gives u−x ≤ y−x ;
(c) a ≤ x < u < y ≤ b gives f (u)− f (x) ≤ f (y)− f (u) ;
u−x
y−u
Interpret these conditions geometrically
...
Not available
...
Supply the details in the proof of Hadamard’s Three Circle Theorem
...
Not available
...
Give necessary and sufficient conditions on the function f such that equality occurs in the
conclusion of Hadamard’s Three Circle Theorem
...
Not available
...
Prove Hardy’s Theorem: If f is analytic on B(0; R) and not constant then
I(r) =
1
2π
2π
| f (reiθ )| dθ
0
is strictly increasing and log I(r) is a convex function of log r
...
(Note that r is fixed, so ϕ may depend on r
...
Not available
...
Let f be analytic in ann(0; R1 , R2 ) and not identically zero; define
I2 (r) =
1
2π
2π
| f (reiθ )|2 dθ
...
Solution
...
6
...
In the statement of the Phragmén-Lindelöf Theorem, the requirement that G be simply connected is not necessary
...
1 to regions G with the property that for each z in ∂∞G there is
a sphere V in C∞ centered at z such that V ∩ G is simply connected
...
Solution
...
Exercise 2
...
1 suppose there are bounded analytic functions ϕ1 , ϕ2 ,
...
∪ Bn such that condition (a) is satisfied and condition (b) is also satisfied
for each ϕk and Bk
...
Solution
...
, ϕn ∈ A(G), bounded and ϕk 0 by assumption
...
, n
...
, n
...
, n
(Corollary IV 6
...
Hence, gk (z) = exp{ηk log ϕk (z)} is an analytic branch of ϕk (z)ηk for ηk > 0 and
|gk (z)| = |ϕk (z)|ηk ∀ k = 1,
...
Define F : G → C by
n
−η
gk (z)κk k ;
F(z) = f (z)
k=1
then F is analytic on G and
n
|F(z)| ≤ | f (z)|
Hence,
k=1
n
−η
|ϕk (z)|ηk κk k
≤
|ϕk (z)|≤κk ∀k
| f (z)|
k=1
M,
lim sup | f (z)| = −ηk
(a),(b) Mκ
,
z→a
k
90
η −η
κk k κk k |ϕk (z)| = | f (z)| ∀z ∈ G
...
By the Maximum Modulus Theorem III, we have
−η
−η
|F(z)| ≤ max{M, Mκ1 1 ,
...
Thus,
n
| f (z)| = |F(z)|
n
k=1
η
|gk (z)|−1 κk k ≤
k=1
κk
ϕk (z)
ηk
−η
−η
max{M, Mκ1 1 ,
...
Letting ηk → 0 ∀k = 1,
...
Exercise 3
...
Also, suppose A < ∞ and a < 1 can be found such that
| f (z)| < exp[A exp(a|Re z|)]
for all z in G
...
Examine exp(exp z) to see that this is the best possible
growth condition
...
Not available
...
Let f : G → C be analytic and suppose M is a constant such that lim sup | f (zn )| ≤ M for each
sequence {zn } in G which converges to a point in ∂∞G
...
(See Exercise 1
...
Solution
...
5)
implies
lim sup | f (z)| ≤ M
for all
z→a
a ∈ ∂∞G
...
6)
If G would be a region together with the claim would give | f (z)| ≤ M ∀z ∈ G by the Maximum Modulus
Theorem III
...
But every open set can be written as a union of components
(components are open and connected)
...
Therefore, we can apply the Maximum Modulus Theorem III to each component once we have shown the
claim and additionally we have to show that the boundary of G is the same as taking the boundaries of the
components:
Clearly, since the components of G add up to G, we have that the boundary of the components must include
the boundary of G
...
Assume it would be the case
...
But this would imply that it must be connected to another component of G
contradicting the fact a component is a maximally connected subset
...
It remains to show the claim which we will do by showing the contrapositive
...
6), that is
∀M > 0 ∃a ∈ ∂∞G : lim sup | f (z)| > M
z→a
which is equivalent to
∀M > 0 ∃a ∈ ∂∞G : lim+ sup{| f (z)| : z ∈ G ∩ B(a; r)} > M
r→0
91
by definition
...
Let {zn } be a
sequence with limn→∞ = a and let rn = 2|zn − a|
...
Next, consider the
sequence αn = sup{| f (z)| : z ∈ F ∩ B(a; rn )}
...
In addition, {αn } is a nonincreasing
sequence, therefore limn→∞ αn > M + δ implies αn > M + δ ∀n
...
Taking lim sup in this inequality yields
lim sup | f (yn )| ≥ M + δ,
n→∞
that is
lim sup | f (yn )| ≥ M
n→∞
where limn→∞ yn = a, which gives “not (6
...
Hence (6
...
6)
...
Let f : G → C be analytic and suppose that G is bounded
...
Show that if limz→z0 |z − z0 | | f (z)| = 0 for every > 0 then
| f (z)| ≤ M for every z in ∂G
...
)
Solution
...
Exercise 6
...
2
Show that | f (z)| ≤ M for all z in G
...
Not available
...
Let G = {z : Re z > 0} and let f : G → C be analytic such that f (1) = 0 and such that
lim supz→w | f (z)| ≤ M for w in ∂G
...
Prove that
| f (z)| ≤ M
Hint: Consider f (z)
1+z
1−z
(1 − x)2 + y2
(1 + x)2 + y2
...
Not available
...
Chapter 7
Compactness and Convergence in the
Space of Analytic Functions
7
...
Prove Lemma 1
...
)
d(s,t)
Solution
...
d(s,t)
For s, t ∈ S the value µ(s, t) = 1+d(s,t) is well defined in the nonnegative real numbers because d is a metric
...
The metric d is
symmetric and therefore µ is also
...
Let therefore s, t, u ∈ S
...
(7
...
Thus if d(s, u) ≤
max{d(s, t), d(t, u)}, then also µ(s, u) ≤ max{µ(s, t), µ(t, u)} and by nonnegativity of µ also
µ(s, u) ≤ µ(s, t) + µ(t, u)
...
< min
,
1 + d(s, u)
1 + d(s, t) 1 + d(t, u)
Together with equation (7
...
To show that the metrics d and µ are equivalent on S , let O be on open set in (S , d) and let x ∈ O
...
Choose δ positive such that δ < 1+ε , then Bµ (x, δ) ⊂ Bd (x, ε)
...
Since this can be done for any element x ∈ O, open sets in (S , d) are open in (S , µ) and
vice-versa
...
Assume that {xn }n is a Cauchy sequence in (S , d)
...
By the above there is a δ small enough that Bd (x, δ) ⊂ Bµ (x, ε) and δ depends only on ε, but not
on x ∈ S
...
The opposite statement holds with a similar argument
...
Find the sets Kn obtained in Proposition 1
...
Solution
...
Set
Kn = B a, r −
1
n
where a ball of negative radius is to be understood as the empty set
...
n=1
b) G is an open annulus,
Let a ∈ C be the center of the annulus with radii r, R, 0 ≤ r < R < ∞ so that G = ann(a, r, R)
...
d) G = {z ∈ C : |Imz| < 1},
For this open set G define
Kn = {z = x + iy ∈ C : x, y ∈ R, |x| ≤ n} ∩ z = x + iy ∈ C : x, y ∈ R|y| ≤ 1 −
Although G is unbounded each Kn is bounded and closed and hence compact and G =
1
...
e) G = C − Z
...
Also Kn ⊂ int Kn+1 by
definition and G = ∞
...
Supply the omitted details in the proof of Proposition 1
...
94
Solution
...
Exercise 4
...
Show that F is totally
bounded
...
Not available
...
Suppose { fn } is a sequence in C(G, Ω) which converges to f and {zn } is a sequence in G which
converges to a point z in G
...
Solution
...
Exercise 6
...
e
...
Show that fn → f
...
By Proposition 1
...
Thus,
n=1
WLOG we can assume that G is compact
...
Clearly gn is continuous ∀n, since f
and fn are continuous by assumption
...
Since fn converges to f , gn converges pointwise to zero, that is
∀ > 0 ∃N ∈ N ∀x ∈ G : 0 ≤ gN (x) <
2
...
For any y ∈ B(x; r)
we have gN (y) < , that is
∀ > 0 ∀x ∈ G ∃N ∈ N ∃B(x; r) : ∀y ∈ B(x; r) : gN (y) <
...
These open neighborhoods will cover G
...
, xk such that
B(xi ; r) cover G for every value of
...
Therefore, let M = maxi Ni
...
Hence,
∀ > 0 ∃M ∈ N, ∀n ≥ M ∀x ∈ G : | fn (x) − f (x)| < ,
which shows fn → f
...
Let { fn } ⊂ C(G, Ω) and suppose that { fn } is equicontinuous at each point of G
...
Solution
...
2 we can find a sequence {Kn } of compact subsets of G such that ∞ Kn
...
Since fn is equicontiuous at each point of G, we have that
∀ > 0 ∃δ > 0 : d( fn (x), fn (y)) <
3
whenever |x − y| < δ
...
2)
Taking the limit as n → ∞, we also get
d( f (x), f (y)) <
3
whenever |x − y| < δ
...
3)
by the pointwise convergence ( f (z) = lim fn (z) ∀z ∈ G by assumption)
...
, yk ∈ G such that the open neighborhoods B(yi ; δ) cover G, since G is
95
compact
...
, k
...
Set N = maxi Ni , then we have
d( fn (yi ), f (yi )) <
3
∀n ≥ N
∀i = 1,
...
(7
...
2),(7
...
4) 3 3 3
where we choose yi such that |x − yi | < δ
...
Exercise 8
...
If fn (z) =
n=0
that fn → f in C(G; C)
...
n
k=0
ak zk , show
∞
n=−∞
an zn
...
a) Let G = B(0; R)
...
With the given function
f define an analytic function
g : G → C,
Let z ∈ G and define r :=
we have
|z|+R
2 (<
g(z) :=
∞
k=0
|ak |zk
...
5)
R) then z ∈ B(0; r)
...
Then for this fixed z and for any n ∈ N
n
n
| fn (z)| = |
k=0
ak zk | ≤
k=0
|ak ||z|k = g(|z|) < ∞
...
Next, we want to show equicontinuity at each point of G
...
Let r > 0 be such
that B(z0 ; r) ⊂ G
...
5) and the computations from above we have
for all w ∈ ∂B(z0 ; r) and for all n ∈ N
| fn (w)| ≤ g(|w|)
...
Thus there is a positive number M
such that |g(∂B(z0 , r))| ≤ M
...
r
|w−z0 |=r
fn (w)(z − z0 )
dw|
(w − z)(w − z0 )
By Arzela-Ascoli the sequence { fn } is normal in C(G) so there is a subsequence { fnk } that converges in C(G)
...
Hence every subsequence converges to the same limit, which shows that fn → f in C(G)
...
3) and { fn } ⊂ H(G) thus also the limit
function f ∈ H(G)
...
2 Spaces of analytic functions
Exercise 1
...
be elements of H(G) and show that fn → f iff for each closed rectifiable curve
γ in G, fn (z) → f (z) uniformly for z in {γ}
...
Not available
...
Let G be a region, let a ∈ R, and suppose that f : [a, ∞] × G → C is a continuous func∞
tion
...
Suppose that this integral does converge uniformly on compact subsets of G and that for each t in (a, ∞), f (t, · ) is analytic on G
...
Not available
...
The proof of Montel’s Theorem can be broken up into the following sequence of definitions and
propositions: (a) Definition
...
(b) If F ∈ C(G, C) is locally
Lipschitz then F is equicontinuous at each point of G
...
Solution
...
Exercise 4
...
Solution
...
Suppose that fn
f in H(G) then there must be a compact set K ⊂ G the convergence is not uniform on
K
...
By
compactness of K extract a subsequence of {zn }, say {znm } that converges to a point z0 ∈ K
...
On the set A of points of pointwise
convergence fn (z) → f and fn (z) → g
...
8)
already f = g on G which gives a contradiction on the set K and the point z0 because
| fnm (znm ) − f (znm )| → |g(z0 ) − f (z0 )| ≥ ε
...
Exercise 5
...
Solution
...
Exercise 6
...
Is the converse true?
Can you add something to the hypothesis that F is normal to insure that F is normal?
97
Solution
...
Exercise 7
...
Show
that if g is analytic on Ω and is bounded on bounded sets then {g ◦ f : f ∈ F } is normal
...
Not available
...
Let D = {z : |z| √ 1} and show that F ⊂ H(D) is normal iff there is a sequence {Mn } of positive
<
constants such that lim sup n Mn ≤ 1 and if f (z) = ∞ an zn is in F then |an | ≤ Mn for all n
...
Not available
...
Let D = B(0; 1) and for 0 < r < 1 let γr (t) = re2πit , 0 ≤ t ≤ 1
...
r
Solution
...
Then
≤
| f (z) − fn (z)| |dz|
sup | f (z) − fn (z)|
=
γr
sup | f (z) − fn (z)|· 2πr,
z∈γr
γr =re2πit ,0≤t≤1
γr
z∈γr
|dz|
∀0 < r < 1
...
Therefore,
γr
| f (z) − fn (z)| |dz| → 0
as n → ∞ for each r, 0 < r < 1
...
Let K be an arbitrary compact subset
r
of D = B(0; 1) and choose 0 < r < 1 such that K ⊂ B(0; r) and denote the shortest distance between an
arbitrary point in K and an arbitrary point on ∂B by δ > 0
...
(7
...
z−a
| f (z) − fn (z)|
1
|dz| ≤
|z − a|
2πδ
2π δ
| f (z) − fn (z)| |dz| <
=
...
6) 2πδ
∂B
Hence, fn (a) → f (z) uniformly ∀a ∈ K
...
10 b) yields that { fn } converges to f ∀ z ∈ D
...
Let { fn } ⊂ H(G) be a sequence of one-one functions which converge to f
...
Solution
...
Suppose f is not the constant function
...
Choose an arbitrary point z0 ∈ G and define the sequence gn (z) = fn (z) − fn (z0 )
...
In addition, we
98
have that gn never vanishes on G\{z0 }, since fn are assumed to be one-one functions
...
6 resulting from Hurwitz’s Theorem,
where the region is G\{z0 }
...
Since g(z) = f (z) − f (z0 )
is not identically zero on G\{z0 }, we must have g(z) = f (z) − f (z0 ) has no zeros in G\{z0 }
...
f (z0 )
Since z0 was an arbitrary point in G, we have shown that f is one-one on G
...
Suppose that { fn } is a sequence in H(G), f is a non-constant function, and fn → f in H(G)
...
Solution
...
Exercise 12
...
Solution
...
¯
Exercise 13
...
(b) Let G be a region and let M be a fixed positive constant
...
Show that F is normal
...
a) Let B(a; R) ⊂ G and f : G → C be analytic
...
0
Thus, letting g(z) = f (z)2 , we obtain
f 2 (a) =
2π
1
2π
f (a + reit )
2
dt
...
7)
0
and then integrate with respect to r yields
R
f 2 (a)r dr = f 2 (a)
0
R2
=
2 (7
...
0
So
| f (a)|2 ≤
R
0
2π
R
2
f (a + reiθ ) r dr dθ
...
Let F be the family of all functions f in H(G) such that
G
| f (z)|2 dx dy ≤ M
...
8)
To show that F is normal
...
153)
...
Then there is a compact set K ⊂ G such that sup{| f (z)| : z ∈ K, f ∈ F } =
∞
...
Therefore, ∃ zn ∈ K such that
| fn (zn )| ≥
n
...
9)
Every sequence has a convergence subsequence, say {znk }, with znk → z0 ∈ K since K is compact
...
Clearly z0 ∈ G since z0 ∈ K (K ⊂ G), hence ∃ R > 0 such that B(z0 ; R) ⊂ G (since
G is open)
...
In addition, we can
2
ˆ
ˆ
ˆ
find an R > 0 such that B(zn ; R) ⊂ B(z0 ; R ) because B(z0 ; R ) is open
...
| fn (z)|2 dx dy ≤
ˆ
(7
...
ˆ
π R
| fn (zn )| ≤
But by (7
...
2
This gives a contradiction if we let n → ∞
...
| fn (zn )| ≥
7
...
Prove Proposition 3
...
Solution
...
To show: There is a number δ > 0 such that B∞ (a; δ) ⊂ B(a; r)
...
Claim: Pick
2r
δ= √
1 + R2 1 + |a|2
to obtain
B∞ (a; δ) ⊂ B(a; R)
...
This is equivalent to
2|z − a|
⇐⇒
⇐⇒
1 + |z|2 1 + |a|2
2|z − a|
1+
|z|2
|z − a| < √
<δ
1+
r
|a|2
2r
< √
2 1 + |a|2
1+R
1+
R2
1 + |z|2
100
which implies (z ∈ B∞ (a; δ) ⇒ z ∈ B(0; R), that is |z| < R)
|z − a| < √
⇐⇒
So z ∈ B(a; r)
...
B∞ (a; δ) ⊂ B(a; r)
...
To show: There is a number r > 0 such that B(a; r) ⊂ B∞ (a; δ)
...
Proof of the claim: Let z ∈ B(a; r)
...
⇐⇒
So z ∈ B∞ (a; δ)
...
c) Let δ > 0
...
Claim: Choose K = B(0; r) where r >
4
δ2
− 1 to obtain
C∞ − K ⊂ B∞ (∞; δ)
...
)
101
Proof of the claim:
C∞ − K
C∞ − B(0; r)
=
=
=
r>
⊂
4
δ2
−1
{C − B(0; r)} ∪ {∞}
{C − {z ∈ C : |z| ≤ r}} ∪ {∞}
4
− 1 ∪ {∞}
C − z ∈ C : |z| ≤
2
δ
4
− 1 ∪ {∞}
z ∈ C : |z| >
2
δ
4
− 1 < |z| ∪ {∞}
z ∈ C :
2
δ
2
< δ ∪ {∞}
z ∈ C :
2
1 + |z|
=
=
=
{z ∈ C : d(∞, z) < δ} ∪ {∞}
=
B∞ (∞; δ)
...
d) Let K ⊂ C where K is compact
...
Let
B(0; r) such that B(0; r) ⊃ K where r > 0
...
So it suffices to show: There is a number δ > 0 such that
B∞ (∞; δ) ⊂ C∞ − B(0; r)
for a given r > 0
...
Proof of the claim:
B∞ (∞; δ)
=
=
⊂
δ≤ √ 2
r2 +1
=
=
=
=
{z ∈ C : d(∞, z) < δ} ∪ {∞}
2
< δ ∪ {∞}
z ∈ C :
2
1 + |z|
2
2
< √
z ∈ C :
∪ {∞}
2 + 1
2
r
1 + |z|
{z ∈ C : r ≤ |z|}} ∪ {∞}
{C ∪ {z ∈ C : |z| < r} ∪ {∞}
C∞ − {z ∈ C : |z| < r}
C∞ − B(0; r)
...
Exercise 2
...
Solution
...
7
...
Let G and Ω be open sets in the plane and let f : G → Ω be a continuous function which
is one-one, onto, and such that f −1 : Ω → G is also continuous (a homeomorphism)
...
Prove that
w ∈ ∂Ω
...
Not available
...
(a) Let G be a region, let a ∈ G and suppose that f : (G − {a}) → C is an analytic function
such that f (G − {a}) = Ω is bounded
...
If f is one-one,
show that f (a) ∈ ∂Ω
...
Solution
...
Exercise 3
...
Let a ∈ G ∩ R and suppose that f : G → D = {z : |z| < 1} is a one-one analytic function
with f (a) = 0, f (a) > 0 and f (G) = D
...
Show that f (G+ ) must lie entirely
above or entirely below the real axis
...
Not available
...
Find an analytic function f which maps {z : |z| < 1, Re z > 0} onto B(0; 1) in a one-one fashion
...
Not available
...
Let f be analytic on G = {z : Re z > 0}, one-one, with Re f (z) > 0 for all z in G, and f (a) = a
for some real number a
...
Solution
...
Let a ∈ G
...
Since, we have f : G −− −→ G (since Re f (z) > 0) and g :
−−−
1−1
1−1
−− −→ D so g−1 : D −− −→ G we can define h(z) = g( f (g−1 (z)))
...
We have
h(0) = g( f (g−1 (0)))
=
g−1 (0)=a
103
g( f (a)) = g(a) = 0
...
Thus, the hypothesis of Schwarz’s Lemma are satisfied, and hence, we get
|h (0)| ≤ 1
...
20 provided g (g−1 (z))
h (0)
= g ( f (g−1 (0))) f (g−1 (0))
0
...
Therefore
|h (0)| = | f (a)| ≤ 1
...
Exercise 6
...
Let f be a
one-one analytic mapping of G1 onto G2
...
Prove that for any one-one analytic
map h of G1 into G2 with h(a) = α it follows that |h (a)| ≤ | f (a)|
...
Define the function F(z) = f −1 (h(z))
...
Clearly F(z) is analytic since f and h are analytic, F(z) is one-one and F : G1 → G1 by
−−−
construction
...
Thus by Exercise 5, we get
|F (a)| ≤ 1
...
20 provided f ( f −1 (h(z)))
F (a) =
which is well-defines since f (a)
h (z)
0
...
Since
|F (a)| =
|h (a)|
≤1
| f (a)|
104
gives
|h (a)| ≤ | f (a)|
...
Exercise 7
...
Let ∆ = {ξ :
|ξ| < 1} and suppose that f is an analytic, one-one map of G onto ∆ with f (a) = 0 and f (a) > 0 for some
point a in G
...
Solution
...
Since g is one-one mapping from G onto ∆, we
clearly have g(a) = α, where α ∈ ∆ (α does not have to be zero)
...
1 − αz
¯
Clearly, g1 is a Möbius transformation mapping ∆ to ∆ and in addition g1 is one-one and analytic
...
Then, g2 is one-one and analytic and maps G onto ∆
...
But g2 (a) need not necessarily satisfy g2 (a) > 0
...
10)
0 ∀z ∈ G since g2 is one-one
...
Clearly g3 maps ∆ onto ∆
...
Finally, let
h(z) = g3 (g2 (z))
...
We have
h(a) = g3 (g2 (a))
= g3 (0) = 0
...
10)
And,
h (z) = (g3 (g2 (z))) = (g3 (g1 (g(z)))) = g3 (g1 (g(z)))g1 (g(z))g (z)
(1 − αg(z))· 1 − (g(z) − α)· (−α)
¯
¯
g (z)
...
2
2 )2
(1 − αα)
¯
(1 − |α|
1 − |α|2
Clearly, we can choose a θ ∈ [0, 2π) such that h (a) > 0 (depending on the sign of g (a)
...
For example: If g (a) < 0, then pick θ = π, so eiθ = −1 or
θ = −Arg(g (a)) will work, too
...
Thus,
f (z) = h(z) = g3 (g2 (z)) = eiθ g2 (z) = eiθ g1 (g(z)) = eiθ
g(z) − α
...
e + α f (z)
¯
Exercise 8
...
Solution
...
The inner boundary {z : |z| = r1 } of the domain is mapped to the inner boundary of
2
2
ann(0; r2 , R2 )
...
The inverse function is f −1 (z) = r2 z, an analytic and non-constant function also
...
Exercise 9
...
Solution
...
7
...
Show that
(1 + zn ) converges absolutely iff
(1 + |zn |) converges
...
Assume Re(zn ) > −1
...
6 p
...
∞ zn converges absolutely iff ∞ |zn |
n=1
k=1
converges (by definition)
...
n=1
n=1
Then ∞ log(1+|zn |) converges iff (1+|zn |) converges (by Proposition 5
...
165 applied to wn = 1+|zn |)
...
Proof of the claim: Let xn = |zn | for convenience
...
⇒: Assume |zn | converges, that is xn converges
...
So given > 0, we have
0 ≤ log(1 + |xn |) ≤ (1 + )xn
for sufficiently large n where the last inequality follows since limz→0 log(1+z) = 1 by Exercise 2
...
⇐: Assume log(1 + |zn |) converges, that is log(1 + xn ) converges
...
So given > 0, we have
(1 − )xn ≤ log(1 + xn )
for sufficiently large n where the last inequality follows since limz→0
comparison test, xn converges, that is |zn | converges
...
n=1
n=1
106
log(1+z)
z
= 1 by Exercise 2
...
Prove that limz→0 log(1+z) = 1
...
First note that,
f (z) =
log(1 + z)
z
has a removable singularity at z = 0, since
lim z f (z) = lim z
z→0
z→0
log(1 + z)
= lim log(1 + z) = log(1) = 0
z→0
z
(by Theorem 1
...
103)
...
165
(−1)k+1
k=1
zk
k
if |z| < 1
...
Hence,
∞
k
=1+
1 +
k z
lim f (z) = lim
(−1)
z→0
z→0
k + 1
k=1
∞
∞
k=2
(−1)k
k=1
(−1)k+2
zk−1
=1+
k
(limz→0 z)k
=1+
k+1
∞
(−1)k
k=1
∞
zk
k+1
0=1
k=1
and therefore
log(1 + z)
= 1
...
Let f and g be analytic functions on a region G and show that there are analytic functions
f1 , g1 , and h on G such that f (z) = h(z) f1 (z) and g(z) = h(z)g1 (z) for all z in G; and f1 and g1 have no
common zeros
...
Let f and g be analytic functions on the region G
...
Let {b j } be the zeros of g with multiplicity n j
...
Then f1 = f and g1 = g
...
Otherwise let {z j } be the set of common zeros of f and g with multiplicity m j = min(n j , n j )
...
Then there is an
analytic function h defined on G whose only zeros are at the points z j with multiplicity m j (by Theorem 5
...
170)
...
(since f1 and g1 have removable singularities at the z j ’s by construction;
the multiplicity of z j ’s in f and g is greater or equal than the multiplicity of z j ’s in h)
...
Exercise 4
...
Show that the infinite
converges in H(B(0; 1)) and that |B(z)| ≤ 1
...
)
(c) Find a sequence {an } in B(0; 1) such that (1 − |an |) < ∞ and every number eiθ is a limit point of {an }
...
Not available
...
Discuss the convergence of the infinite product
for p > 0
...
Not available
...
Discuss the convergence of the infinite products
∞
n=1
Solution
...
...
i
1
Let zn := 1 + n = rn eiθn with rn = 1 + n2 and θ = arctan
then ∞ rn = r and ∞ θn = θ
...
If there is z = reiθ such that
1
1
≥
n
2
∞
n1
∞
n=1 zn
= z,
1
n
is valid
...
i
The second infinite product is a product of real numbers and zn = |1 + n | =
1+
1
n2
≤ 1+
1 2
n2
implies that 1 ≤ zn ≤ 1 +
1
...
n2
The estimate
Thus for any integer N ∈ N
N
1≤
1+
n=1
N
zn ≤
1+
n=1
1
n2
where the last term is absolutely convergent by an example in class
...
With log(zn ) > 0
for all n, the comparison with the sum of logarithms (Proposition 5
...
Exercise 7
...
Solution
...
= exp − log(2n) + log(n + 1) = exp log
2n
2n
So,
∞
n=2
1−
1
n+1 1
=
...
For which values of z do the products (1 − zn ) and (1 + z2n ) converge? Is there an open
set G such that the product converges uniformly on each compact subset of G? If so, give the largest such
open
...
Consider the infinite product ∞ (1 − zn )
...
6 this converges abson=1
lutely if and only if ∞ −zn converges absolutely
...
If z is an nth root of unity for a positive integer n, then one of the factors in the infinite product becomes 0
and the product converges to 0, but not absolutely according to Definition 5
...
If there is no n such that z
is nth root of unity then no factor of the infinite product equals zero and there is a subsequence {(1 − znk }k
such that |1 − znk | > 1
...
We conclude that ∞ (1 + zn ) does not converge
...
The behavior of ∞ (1 + z2n ) is almost the same as the previous one
...
Also if |z| -and hence |w|- is greater than 1, then the
infinite product diverges
...
The maximal region G such that the convergence is uniform on compact sets is again the open unit disk
...
Use Theorem 5
...
Solution
...
Exercise 10
...
(a) Show that
∞
f (z)
fn (z)
k
fn (z) converges
n k
k=1
converges in H(G) and equals f (z)
...
Show that
f (z)
=
f (z)
∞
n=1
fn (z)
fn (z)
and the convergence is uniform over K
...
Not available
...
A subset J of H(G), G a region, is an ideal iff: (i) f and g in J implies a f + bg is in J for
all complex numbers a and b; (ii) f in J and g any function in H(G) implies f g is in J
...
If f ∈ H(G) let Z( f ) be the set of zeros of f counted according to
their multiplicity
...
If S ⊂ H(G) then Z(S) = ∩{Z( f ) : f ∈ S}, where the zeros
are again counted according to their multiplicity
...
(a) If f and g ∈ H(G) then f divides g (in symbols, f |g if there is an h in H(G) such that g = f h
...
(b) If S ⊂ H(G) and S contains a non-zero function then f is a greatest common divisor of S if: (i) f |g for
each g in S and (ii) whenever h|g for each g in S, h| f
...
c
...
S
...
c
...
S
...
c
...
(c) If A ⊂ G let J(a) = { f ∈ H(G) : Z( f ) ⊃ A}
...
(d) Let a ∈ G and J = J({a})
...
(e) Show that every maximal ideal in H(G) is a prime ideal
...
Solution
...
Exercise 12
...
Give the most elementary example possible (i
...
, choose the pn to be as small as possible)
...
Not available
...
Find an entire function f such that f (m + in) = 0 for all possible integers m, n
...
Solution
...
7
...
Show that cos πz =
∞
n=1
1−
4z2
(2n−1)2
...
We know by the double-angle identity of sine sin(2z) = 2 sin(z) cos(z) (this is proved easily by
z2
using the definition) or sin(2πz) = 2 sin(πz) cos(πz)
...
Hence
2πz
⇐⇒
2πz
∞
n=1
∞
n=1
1−
4z2
(2n)2
z2
1− 2
n
∞
n=1
∞
n=1
1−
1−
Thus,
cos(πz) =
∞
n=1
∞
4z2
z2
= 2πz
1 − 2 cos(πz)
(2n − 1)2
n
n=1
∞
4z2
z2
= 2πz
1 − 2 cos(πz)
...
Find a factorization for sinh z and cosh z
...
(2n − 1)2
Solution
...
175 Equation 6
...
2
n
nπ
n=1
Therefore,
∞
sinh(z) = z
1+
n=1
z2
n2 π2
...
176
cos(πz) =
∞
n=1
1−
4z2
⇒ cos(z) =
(2n − 1)2
Therefore,
cosh(z) =
∞
1+
n=1
Exercise 3
...
Solution
...
111
n=1
1−
4z2
...
Not available
...
Prove Wallis’s formula:
∞
πz
4
...
(2n − 1)2 π2
7
...
Show that 0 < γ < 1
...
57722
...
)
Solution
...
Exercise 2
...
Deduce from this that Γ( 1 ) =
2
√
π
...
Assume z is not an integer, then by Gauss’s Formula p
...
175 Equation 6
...
So
Γ(z)Γ(1 − z) = π csc(πz)
1
for z not an integer
...
Show:
√
√
1
= π· 1 = π
...
(Hint: Consider the function Γ(z)Γ(z + 2 )Γ(2z)−1
...
Following the hint define a function f on its domain G = {z ∈ C : z
2z−1 z
z 1
...
It suffices to show that f (x) satisfies the Bohr-Mollerup Theorem
...
Also f satisfies the functional equation of the Gamma-function:
x
x 1
2x
Γ +1
+
f (x + 1) = √ Γ
2 2
2
π
2x
x 1 x x
= √ Γ
+
Γ
π 2 2 2 2
x
x2 x−1 x 1
Γ
+
= √ Γ
2 2
2
π
= x f (x)
...
13, f agrees with Γ on the positive real numbers
...
Exercise 4
...
n
n
Solution
...
Exercise 5
...
Show that f = Γ
...
Not available
...
Show that
∞
Γ(z) =
n=0
for z
(−1)n
+
n!(z + n)
∞
e−t tz−1 dt
1
0, −1, −2,
...
Solution
...
(7
...
12)
1
Claim 1: Γ(z) given by (7
...
11), that is Γ(z) given by (7
...
Proof of Claim 1: We know from the book that Ψ(z) is analytic for Re (z) > 0
...
Thus Ψ(z) is analytic on C
...
Then
|tz−1 | = tRe(z)−1
≤
t−1
...
Hence, we have
1
1
|e−t tz−1 | ≤ |e−t |· |tz−1 | = e−t tRe(z)−1 ≤ e−t Ce 2 t = Ce− 2 t
1
and therefore Ce− 2 t is integrable on (1, ∞)
...
In summary,
∞
Ψ(z) =
1
Thus, Claim 2 is proved
...
∞
n=0
(−1)n
n!(z + n)
is analytic on C − {0, −1, −2,
...
and thus provides the analytic continuation
of Φ(z)
...
15 p
...
11) for z ∈ C − {0, −1, −2,
...
Show that
∞
0
Solution
...
2
1
2
cos(t2 ) =
0
√
= π (see Exercise 2 p
...
Thus
1
2
∞
√
1
π=Γ
=
2
∞
1
e−t t− 2 dt = 2a
0
2 2
e−a
t
dt
0
(because of Theorem 7
...
180)
...
Hence, we have
√
∞
π
2 2
...
13)
e−a t dt =
2a
0
Let a =
∞
1−i
√ ,
2
−
e
then a2 =
1−i
√
2
2
t2
dt
1−2i+i2
2
= −i and therefore
∞
=
∞
2
e−(−i)t dt =
0
0
0
∞
=
2
eit dt
∞
2
cos(t ) + i sin(t2 ) dt =
0
=
(7
...
2
Thus,
∞
∞
cos(t2 ) dt + i
0
1
1
π+i
2
2
1
2
sin(t2 ) dt =
0
1
π
2
which implies
∞
cos(t2 ) dt =
0
and
∞
sin(t2 ) =
0
1
2
1
2
1
π
2
1
π
...
Let u > 0 and v > 0 and express Γ(u)Γ(v) as a double integral over the first quadrant of the
plane
...
0
The function
B(u, v) =
Γ(u)Γ(v)
Γ(u + v)
is called the beta function
...
Let u > 0 and v > 0, then by Theorem 7
...
180 and three changes of variables (indicated by
“CoV”) we obtain
Γ(u)Γ(v)
∞
=
∞
e−s su−1 ds
0
=
∞
2
CoV:s=x2 ,t=y2
e−t tv−1 dt
0
∞
2
e−x x2u−2 x dx· 2
0
=
∞
4
∞
0
CoV:x=r cos(θ),y=r sin θ
∞
4
0
∞
4
0
4
=
−y2 2u−1 2v−1
x
y
1
4
2
2
e−r r2u−1 (cos θ)2u−1 r2v−1 (sin θ)2v−1 r dr dθ
2
e−r r2u+2v−1 (cos θ)2u−1 (sin θ)2v−1 r dr dθ
−r2 2u+2v−1
e
r
π/2
(cos θ)2u−1 (sin θ)2v−1 dθ
dr
0
CoV:t=r2
dx dy
0
∞
=
2
0
π/2
=
e−x
0
π/2
=
2
e−y y2v−2 y dy
0
∞
0
π/2
e−t tu+v−1 dt
0
(cos θ)2u−1 (sin θ)2v−1 dθ
0
π/2
=
(cos θ)2u−1 (sin θ)2v−1 dθ
...
0
Next, we show
1
tu−1 (1 − t)v−1 dt
...
Hence,
1
tu−1 (1 − t)v−1 dt
...
(1 + t)u+v
116
We have
1
B(u, v)
tu−1 (1 − t)v−1 dt
=
0
∞
=
t
CoV:s= 1−t
s
1+s
0
∞
u−1
1−
s
1+s
v−1
(1 + s)−2 ds
1
1
(1 + s)−2 ds
(1 + s)u−1 (1 + s)v−1
0
∞
1
su−1
(1 + s)−2 ds
(1 + s)u+v−2
0
∞
su−1
ds
...
(1 + t)u+v
Yes, the beta function can be generalized to the case when u and v are complex numbers with positive real
part
...
Thus it remains to show that
∞
B(u, v) =
0
tu−1
dt ∈ H(G × G)
(1 + t)u+v
where G × G ⊂ C × C with G = {z : Re(z) > 0}
...
∞
0
tu−1
dt =
(1 + t)u+v
1
0
tu−1
dt +
(1 + t)u+v
∞
1
tu−1
dt
...
This implies |tu−1 (1 + t)−u−v | is integrable on (0, 1)
...
(1 + t)u+v
II) We have
∞
n
|tu−1 (1 + t)−u−v | dt
n→∞
≤
n→∞
|tu−1 (1 + t)−u−v | dt
lim
=
1
≤
lim
lim
1
n
|tu−1 |· |(1 + t)−u−v | dt
1
n
n→∞
n
=
lim
n→∞
tRe(u)+Re(v)
lim
n→∞
t−1−Re(u)
1
1
t
n
≤
lim
n→∞
1
1
tRe(u)−1
1
n
=
tRe(u)−1 (1 + t)−Re(u)−Re(v) dt
1
1
t1+Re(v)
+1
1
t
+1
Re(u)+Re(v)
dt
1
dt
Re(u)+Re(v)
dt < ∞
where the last step follows since Re(v) > 0 and the previous one since Re(u) + Re(v) > 0, t ≥ 1 and
1
u−1
−u−v
therefore
| is integrable on (1, ∞)
...
This implies |t (1 + t)
1
( t +1)
any {γ} ⊂ G × G
∞
γ
1
∞
tu−1
dt dz =
(1 + t)u+v
which implies
1
γ
tu−1
dz dt = 0
(1 + t)u+v
∞
tu−1
dt ∈ H(G × G)
...
(1 + t)u+v
1
In summary
0
Exercise 9
...
Prove by induction and iterated
integrals that
1
αn = 2αn−1
0
(1 − t2 )(n−1)/2 dt
Solution
...
Clearly, V1 (r) = 2r, V2 (r) = πr2 , V3 (r) =
V1 (1) = 2, V2 (1) and V3 = 4 π
...
14)
(def)
Assume the claim is true, then we write
2
2
2
Bn (1) = {x ∈ Rn : |x| ≤ 1} = {(x1 , x2 ,
...
+ xn−1 + t2 ≤ 1}
...
, xn−1 ) ∈ Rn−1 : x1 + x2 +
...
14) we obtain
√
Vn−1
√
1 − t2
1 − t2 =
n−1
Vn−1 (1) = 1 − t2
1
−1
n−1
2
1 − t2
αn−1
...
Hence
Clearly, the ball Bn (1) is the union of all disjoint balls Bn−1
αn = Vn (1) =
n−1
2
1
1 − t2
αn−1 dt = 2αn−1
0
n−1
2
dt
which proves the formula above
...
We have by using a symmetry argument
√ 2 n−1 2
√2 2 √2 2 2
r −x1
r
Vn (r)
=
√ 2
− r2 −x1
√2
−r
r −x1
r
=
2n
0
0
=
n
0
0
√
1−y2 −y2
1
2
rn Vn (1)
...
(Note that this is kind of obvious, since we scale the unit ball in Rn by r in
each dimension to obtain the ball with radius r in Rn )
...
Using spherical coordinates simplifies the computation if preferred
...
Show that
αn =
πn/2
(n/2)Γ(n/2)
where αn is defined in problem 9
...
Not available
...
The Gaussian psi function is defined by
Ψ(z) =
Γ (z)
Γ(z)
(a) Show that Ψ is meromorphic in C with simple poles at z = 0, −1,
...
(b) Show that Ψ(1) = −γ
...
z
(d) Show that Ψ(z) − Ψ(1 − z) = −π cot πz
...
Solution
...
7
...
Let ξ(z) = z(z − l)π− 2 z ζ(z)Γ( 2 z) and show that ξ is an entire function which satisfies the
functional equation ξ(z) = ξ(1 − z)
...
The function ξ is analytic on C − ({−2k : k ∈ N0 } ∪ {1}) as a product of analytic functions
...
The zeta-function has a simple pole at z = 1, hence limz→1 (z − 1)ζ(z)
exists in C and the function ξ is well-defined at z = 1
...
For z = 0, limz→0 zΓ 2 z ∈ C
...
Therefore the function ξ is analytic at each point
...
3, p
...
e
...
2
(7
...
+
Γ
2
2 2
120
(7
...
17)
Then the following chain of equalities holds
...
15)
= z(z − 1)π− 2 + 2 Γ
−
2 π ζ(z)Γ(z) sin π(1 − z)
2 2
2
− 1 1−z
2
Γ(z)Γ
1
2
Γ
z
2
z
2
= ξ(z)π 2
−
1
π(1 − z)
2
sin
1
2 π(1
1
Γ(z) π csc
(7
...
17)
= ξ(z)π 2 21−z
Γ
1
2
+
z
2
π
1
2
Γ
1 z
+
2 2
= ξ(z)
...
Use Theorem 8
...
p−1 = ∞
...
Euler’s Theorem asserts that for real values x > 1, ζ(x) = ∞ 1−p−x , where pn are the prime
n=1
n
numbers
...
1
Now the argument goes as follows
...
All summands
n=1
are positive, therefore the sum converges absolutely
...
4, p
...
But this implies that
−
∞
n=1
∞
n=1
log 1 −
1
pn
∞
1
1
= log
= log ζ(1) < ∞
pn
1 − pn
n=1
log 1 −
which gives the desired contradiction because the log(ζ(z)) has a simple pole at z = 1
...
Prove that ζ 2 (z) =
∞ d(n)
n=1 nz
for Re z > 1, where d(n) is the number of divisors of n
...
Not available
...
Prove that ζ(z)ζ(z − 1) =
∞ σ(n)
n=1 nz ,
for Re z > 1, where σ(n) is the sum of the divisors of n
...
Not available
...
Prove that ζ(z−1) =
ζ(z)
which are relatively prime to n
...
Not available
...
Prove that ζ(z) = ∞ µ(n) for Re z > 1, where µ(n) is defined as follows
...
, pm and suppose that these primes are distinct
...
= km = 1 then let µ(n) = (−1)m ; otherwise let µ(n) = 0
...
Not available
...
Prove that ζζ(z) = −
m ≥ 1; and Λ(n) = 0 otherwise
...
Not available
...
(a) Let η(z) = ζ (z)/ζ(z) for Re z > 1 and show that limz→z0 (z − z0 )η(z) is always an integer for
Re z0 ≥ 1
...
(b) Show that for > 0
Re η(1 + + it) = −
∞
Λ(n)n−1(1+ ) cos(t log n)
n=1
where Λ(n) is defined in Exercise 7
...
(d) Show that ζ(z)
0 if Re z = 1 (or 0)
...
Not available
...
1 Runge’s Theorem
Exercise 1
...
14 if it is only assumed that E − meets each component of C∞ − G
...
Not available
...
Let G be the open unit disk B(0; 1) and let K = {z : 1 ≤ |z| ≤ 3 }
...
Remarks
...
Given a compact set K
contained in an open set G1 ⊂ G, can functions in H(G1 ) be approximated on K by functions in H(G)?
Exercise 2 says that for an arbitrary choice of K, G, and G1 this is not true
...
Exercise 3 is a lemma which is useful in proving
Exercise 4
...
Not available
...
Let K be a compact subset of the open set G and suppose that any bounded component D of
G − K has D− ∩ ∂G
...
Solution
...
Exercise 4
...
Solution
...
ˆ
Exercise 5
...
Not available
...
Let K be a compact subset of the region G and define KG = {z ∈ G : | f (z)| ≤ f K for all f in
H(G)}
...
ˆ G , C − G)
...
ˆ G ⊂ G1 ⊂ G and G1 is open then for every g in H(G1 ) and > 0 there is a function f in H(G) such
(d) If K
ˆ
that | f (z) − g(z)| < for all z in KG
...
)
ˆ G = the union of K and all bounded components of G − K whose closure does not intersect ∂G
...
Not available
...
2 Simple connectedness
Exercise 1
...
Show that G
ˆ
is simply connected
...
he result follows once it is established that every closed piecewise smooth curve in G is 0homotopic
...
Suppose otherwise then there are real r1 , r2 , t1 , t2 such that z = r1 eit1 = r2 eit2
...
Define
3
R : (−∞, 0) → G;
R(t) = 1 + et
...
Also define a function
Θ : [0, 1] × [0, 1] → G,
Θ(s, 0) = R(τ(s)),
Θ(s, 1) = R τ
1
2
Θ(0, u) = Θ(1, u)
,
∀ u ∈ [0, 1]
...
1 (p
...
14 (p
...
1
2
and by
Exercise 2
...
Solution
...
124
8
...
Let G be a region and let {an } and {bm } be two sequences of distinct points in G without limit
points in G such that an an bm for all n, m
...
Show that there is a meromorphic function f on G whose only poles and zeros are {an } and {bm }
respectively, the singular part at z = an is S n (z), and z = bm is a zero of multiplicity pm
...
Let G be a region and let {bm } be a sequence of distinct points in G with no limit point in G;
and let {pm } be a sequence of integers
...
15 p
...
Since g ∈ H(g) and {an } ∈ G, g has a Taylor series in a neighborhood B(an ; Rn ) of each an , that is
∞
gn (z) =
αk (z − an )k ∈ B(an ; Rn )
k=0
where αk =
sn (z) or
1 (k)
k! g (an )
...
Claim:
mn
B jk (z − an )− j−k
rn (z) =
j=1
(8
...
Proof of the claim:
∞
k=0
αk rn (z)(z − an )k
∞
=
mn
αk
j=1
k=0
∞
=
mn
αk
j=1
k=0
mn
=
j=1
where the last step follows by choosing
∞
B jk (z − an )− j−k (z − an )k
B jk (z − an )− j
(z − an )− j
∞
k=0
mn
αk B jk =
j=1
A jn
(z − an ) j
αk B jk = A jn
...
Let {an } be a sequence of distinct points without a limit point in G and such
that an bm for all n, m
...
1))
...
Set f = g· h
...
(Note that
the zeros do not cancel the poles since by assumption an bm ∀ n, m)
...
Let {an } be a sequence of points in the plane such that |an | → ∞, and let {bn } be an arbitrary
sequence of complex numbers
...
4)
converges absolutely for all r > 0 then
∞
k=n
r
an
kn
bn
z − an
(3
...
(b) Show that if lim sup |bn | < ∞ then (3
...
(c) Show that if there is an integer k such that the series
∞
n=1
bn
ak+1
n
(3
...
4) converges absolutely if kn = k for all n
...
Show that |an |−3 < ∞
...
6) with k = 2 converges absolutely
...
)
(e) Show that if the series (3
...
+ a
n
n
n
n
n=1
(f) Let ω and ω be two complex numbers such that Im(ω /ω) 0
...
but not w = 0, is convergent in M(C) to a
meromorphic function ζ with simple poles at the points 2nω + 2n ω
...
(g) Let P(z) = −ζ (z); P is called the Weierstrass pe function
...
Also show that
P(z) = P(z + 2nω + 2n ω )
for all integers n and n
...
Solution
...
126
Exercise 3
...
5
...
(a) Deduce from Exercises 2(a) and 2(b) that for any sequence {an } in C with lim an = ∞ and an 0 there
is a sequence of integers {kn } such that
∞
k −1
1
1 z
1 z n
1
h(z) =
z − a + a + a a +
...
The remainder of the proof consists of
showing that there is a function f such that h = f / f
...
(b) Let z be an arbitrary but fixed point in C − {a1 , a2 ,
...
} from 0 to z and h is the function obtained in part (a), then there is an integer m such that
γ1
h−
h = 2πim
...
Prove that for z
curve in C − {a1 , a2 ,
...
and γ any rectifiable
h
f (z) = exp
γ
defines an analytic function on C − {a1 , a2 ,
...
(That is, the value of f (z) is independent of
the curve γ and the resulting function f is analytic
...
}; show that z is a removable singularity of the function f defined in part
(c)
...
(e) Show that
∞
2
kn
z
z
+ 1 z +
...
7)
f (z) =
exp
1−
a
an
2 an
2 an
n
n=1
Remark
...
However this
would have meant that we must show that (3
...
The steps outlined in parts (a) through (d) give a proof of Weierstrass’s Theorem without introducing
infinite products
...
Not available
...
This exercise assumes a knowledge of the terminology and results of Exercise VII
...
11
...
Show that ( f, g) = 1 iff Z( f ) ∩ Z(g) =
...
)
(c) Let f1 ,
...
c
...
, fn }
...
, ϕn in H(G) such
that g = ϕ1 f1 +
...
(Hint: Use (b) and induction
...
If S ⊂ H(G) then let
J = ∩{S : S is an ideal of H(G) and S ⊂ J}
...
+ ϕn fn : ϕk ∈ H(G), fk ∈ S for 1 ≤ k ≤ n}
...
If S is finite then (S) is called a finitely generated ideal
...
(e) Show that every finitely generated ideal in H(G) is a principal ideal
...
Prove that if J = (S)
then Z(J) = Z(S) and that a proper principal ideal is fixed
...
Show that J is a fixed ideal in H(C) which
is not a principal ideal
...
Also show
that J = ( f ) if J is finitely generated
...
Show that there is a point a in G such that M = ((z − a))
...
Let J = { f ∈ H(G) : f (an ) = 0
for all but a finite number of the an }
...
(k) If J is a free ideal show that for any finite subset S of J, Z(S)
...
l) Let J be a proper free ideal; then J is a maximal ideal iff whenever g ∈ H(G) and Z(g) ∩ Z( f )
for
all f in J then g ∈ J
...
Not available
...
Let G be a region and let {an } be a sequence of distinct points in G with no limit point in G
...
Show that there is an analytic
(k)
(k)
function f on G such that f (an ) = k!An
...
Let h be a meromorphic function on G with poles at each an of order kn + 1 and with
singular part S n (z)
...
)
Solution
...
Exercise 6
...
√ √
2, 3,
...
We claim that the function
√
3 nz2 − 2z3
√
n3/2 (z − n)2
k=1
√
has the desired properties
...
Moreover f converges in M(C)
...
Then for |z| ≤ r and n ∈ N
√
√
√
3 nz2 − 2z3
r2 (3 n + n) 16r2
| 3/2
≤ 2 =: Mn
...
√
Next fix m ∈ N and consider the limit in equation limz→ √n (z − n)2 f (z) = 1,
√
√ 2
√ 2 ∞ 3 nz2 − 2z3
lim (z − m) f (z) = lim (z − m)
√
√
√
z→ m
z→ m
n3/2 (z − n)2
n=1
√
√
m)(3 nz2 − 2z3 ) 3m3/2 − 2m3/2
+
√
m3/2
z→ m
n3/2 (z − n)2
n=1
√
√
∞
(z − n)(3 nz2 − 2z3 )
+ lim
√
√
z→ m
n3/2 (z − n)2
m+1
m−1
= lim
√
(z −
= 0 + 1 + 0 = 1
...
By Proposition V 2
...
113, for a function
√
g(z) := (z − m)2 f (z), it has to be verified that g (z)|z= √m = 0
...
=
√
n3/2
n3/2 (z − m)
z=m
z=m
√
The residuum vanishes for all m, m ∈ N and therefore the function considered in this exercise has all the
required properties
...
1 Schwarz Reflection Principle
Exercise 1
...
e
...
Define a point w to be inside
γ if [a, w] ∩ {γ} = and let G be the collection of all points that are inside γ
...
(b) Let f : G− → C be a continuous function such that f is analytic on G
...
(c) Show that n(γ; z) = ±1 if z is inside γ and n(γ; z) = 0 if z G−
...
It is not necessary to assume that γ has such a point a as above; each part of this exercise remains
true if γ is only assumed to be a simple closed rectifiable curve
...
This is difficult to obtain
...
This is a very deep result of
topology
...
Not available
...
Let G be a region in the plane that does not contain zero and let G∗ be the set of all points z
such that there is a point w in G where z and w are symmetric with respect to the circle |ξ| = 1
...
3
...
)
(a) Show that G∗ = {z : (1/¯) ∈ G}
...
Show that f ∗ is analytic
...
Show that f = f ∗
...
Do the same thing for an arbitrary circle
...
Not available
...
Let G, G+ , G− , G0 be as in the statement of the Schwarz Reflection Principle and let f :
G+ ∪ G0 → C∞ be a continuous function such that f is meromorphic on G+
...
Show that there is a meromorphic function g : G → C∞ such that g(z) = f (z) for z in G+ ∪ G0
...
Not available
...
2 Analytic Continuation Along a Path
Exercise 1
...
, Dn } of open disks is called a chain of disks if D j−1 ∩ D j
for
1 ≤ j ≤ n
...
, Dn } is a chain
of disks and f j−1 (z) = f j (z) for z in D j−1 ∩ D j , 1 ≤ j ≤ n; then {( f j , D j ) : 0 ≤ j ≤ n} is called an analytic
continuation along a chain of disks
...
(a) Let {( f j , D j ) : 0 ≤ j ≤ n} be an analytic continuation along a chain of disks and let a and b be the centers
of the disks D0 and Dn respectively
...
j=0
(b) Conversely, let {( ft , Dt ) : 0 ≤ t ≤ 1} be an analytic continuation along a path γ : [0, 1] → C and let
a = γ(0), b = γ(1)
...
j=0
Solution
...
√
Exercise 2
...
Let γ(t) =
exp(2πit) and σ(t) = exp(4πit) for 0 ≤ t ≤ 1
...
(b) Find an analytic continuation {(gt , Bt ) : 0 ≤ t ≤ 1} of ( f0 , D0 ) along σ and show that [g1 ]1 = [g0 ]1
...
Not available
...
Let f be an entire function, D0 = B(0; 1), and let γ be a path from 0 to b
...
)
Solution
...
Exercise 4
...
Show that {( ft , Dt ) : 0 ≤ t ≤ 1} is also a continuation along γ
...
Not available
...
Suppose γ : [0, 1] → C is a closed path with γ(0) = γ(1) = a and let {( ft , Dt ) : 0 ≤ t ≤ 1} be
an analytic continuation along γ such that [ f1 ]a = [ f0 ]a and f0 0
...
Not available
...
Let D0 = B(1; 1) and let f0 be the restriction to D0 of the principal branch of the logarithm
...
Find a continuation {( ft , Dt ) : 0 ≤ t ≤ 1} along γ of
( f0 , D0 ) and show that [ f1 ]1 = [ f0 + 2πin]
...
Not available
...
Let γ : [0, 1] → C be a path and let {( ft , Dt ) : 0 ≤ t ≤ 1} be an analytic continuation along γ
...
Show that h( ft (z)) = z for all z in Dt and for all t
...
131
Solution
...
Exercise 8
...
Suppose that {( ft , Dt ) : 0 ≤
t ≤ 1} is an analytic continuation of f0 (z) = log z
...
Solution
...
9
...
Prove that the set T defined in the proof of Lemma 3
...
Solution
...
Exercise 2
...
If γ : [0, 1] → C is a path with γ(0) = a and
γ(1) = b and {( ft , Dt ) : 0 ≤ t ≤ 1} is an analytic continuation of ( f, D) along γ, let R(t) be the radius of
convergence of the power series expansion of ft at z = γ(t)
...
That is, if a second continuation {(gt , Bt )}
along γ is given with [g0 ]a = [ f ]a and r(t) is the radius of convergence of the power series expansion of gt
about z = γ(t) then r(t) = R(t) for all t
...
Find R(t)
...
Find R(t)
...
Solution
...
Exercise 3
...
Let γu (t) = Γ(t, u) and suppose that {( ft,u , Dt,u ) : 0 ≤ t ≤ 1} is an analytic continuation along γu
such that [ f0,u ]a = [ f0,v ]a for all u and v in [0, 1]
...
Show that either R(t, u) ≡ ∞ or R : [0, 1] × [0, 1] → (0, ∞) is a
continuous function
...
Not available
...
Use Exercise 3 to give a second proof of the Monodromy Theorem
...
Not available
...
4 Topological Spaces and Neighborhood Systems
Exercise 1
...
Solution
...
Exercise 2
...
Show that if f : X → Ω is a
continuous function then the restriction of f to Y is a continuous function of (Y, TY ) into (Ω, S)
...
Not available
...
Let X and Ω be sets and let {N x : x ∈ X} and {Mω : ω ∈ Ω} be neighborhood systems and let
T and S be the induced topologies on X and Ω respectively
...
(b) Let X = Ω = C and let Nz = Mz = {B(z; ) : > 0} for each z in C
...
Solution
...
Exercise 4
...
Show that a function f : X → Ω is open iff for each x in X
and U in N x there is a set ∆ in Mω (where ω = f (x)) such that ∆ ⊂ f (U)
...
Not available
...
Adopt the notation of Exercise 3
...
Show that {UY : y ∈ Y} is a neighborhood system for Y and the topology it induces on Y is TY
...
Not available
...
Adopt the notation of Exercise 3
...
(b) If U ∈ T and ∆ ∈ S, call the set U × ∆ an open rectangle
...
(c) Define p1 : X × Ω → X and p2 : X × Ω → X by p1 (x, ω) = x and p2 (x, ω) = ω
...
Furthermore if (Z, R) is a topological space show that a function
f : (Z, R) → (X × Ω, P) is continuous iff p1 ◦ f : Z → X and p2 ◦ f : Z → Ω are continuous
...
Not available
...
5 The Sheaf of Germs of Analytic Functions on an Open Set
Exercise 1
...
Solution
...
Exercise 2
...
If D is an open subset of G and f : D → C is a branch of the logarithm show that
[ f ]a ∈ F for all a in D
...
(Hint: Use Exercise 2
...
)
Solution
...
Exercise 3
...
Show that R is
homeomorphic to the graph Γ = {(z, ez ) : z ∈ G} considered as a subset of C × C
...
) State and prove an analogous result for branches of
z1/n
...
Not available
...
Consider the sheaf S(C), let B = {z : |z − 1| < 1}, let l be the principal branch of the logarithm
1
1
defined on B, and let l1 (z) = l(z) + 2πi for all z in B
...
(b) Find two disjoint open subsets of S(C) each of which
2
1
1
1
1
contains one of the points ( 2 , [l] 2 ) and ( 2 , [l1 ] 2 )
...
Not available
...
6 Analytic Manifolds
Exercise 1
...
That is, prove that if a ∈ X and U is an
open neighborhood of a then there is an open neighborhood V of a such that V − ⊂ U and V − is compact
...
Not available
...
Which of the following are analytic manifolds? What is its analytic structure if it is a manifold?
2
2
2
2
2
(a) A cone in R3
...
Solution
...
Exercise 3
...
3(b)
...
If (U, ϕ) ∈ Φ and h is
one-one on U let ∆ = h(U) and let ψ : ∆ → C be defined by ψ(ω) = ϕ ◦ (h/U)−1 (ω)
...
Prove that (Ω, Ψ) is an analytic manifold and h is an analytic function from X to Ω
...
Not available
...
Let T = {z : |z| = 1} × {z : |z| = 1}; then T is a torus
...
) If ω and ω are complex numbers such that Im (ω/ω )
0 then ω and
ω , considered as elements of the vector space C over R, are linearly independent
...
Define h : C → T by h(tω + tω ) = (e2πit , e2πit )
...
(Use Exercise 3
...
Let G = {tω + t ω : 0 < t < 1, 0 < t < 1} and Ω = {sζ + s ζ : 0 < s < 1, 0 < s < 1}; show that both σ
and τ are one-one on G and Ω respectively
...
) If Φτ and
Φσ are the analytic structures induced on T by τ and σ respectively, and if the identity map of (T, Φτ ) into
(T, Φσ ) is analytic then show that the function f : G → Ω defined by f = σ−1 ◦ τ is analytic
...
) (c)
Let ω = 1, ω = i, ζ = 1, ζ = α where Im α 0; define σ, τ, G, Ω and f as in part (b)
...
(Hint: Use the Cauchy-Riemann equations
...
Not available
...
(a) Let f be a meromorphic function defined on C and suppose f has two independent periods
ω and ω
...
Using
the notation of Exercise 4(a) show that there is an analytic function F : T → C∞ such that f = F ◦ h
...
4
...
)
(b) Prove that there is no non-constant entire function with two independent periods
...
Not available
...
Show that an analytic surface is arcwise connected
...
Not available
...
Suppose that f : C∞ → C∞ is an analytic function
...
(b) If f ∞, let a1 ,
...
Show that there are polynomials
p0 , p1 ,
...
(c) If f is one-one, show that either f (z) = az + b (some a, b in C) or f (z) =
a
z−c
+ b (some a, b, c in C)
...
Not available
...
Furnish the details of the discussion of the surface for
√
z at the end of this section
...
Not available
...
Let G = z : − π < Re z < π and define f : G → C by f (z) = sin z
...
Solution
...
9
...
Suppose that (X, ρ) is a covering space of Ω and (Ω, π) is a covering space of Y; prove that
(X, π ◦ ρ) is also a covering space of Y
...
Not available
...
Let (X, ρ) and (Y, σ) be covering spaces of Ω and Λ respectively
...
Solution
...
Exercise 3
...
Show that there is
an analytic structure Φ on X such that ρ is an analytic function from (X, Φ) to (Ω, ψ)
...
Not available
...
Let (X, ρ) be a covering space of Ω and let ω ∈ Ω
...
Solution
...
Exercise 5
...
If ω1 and ω2 are
points in Ω, show that ρ−1 (ω1 ) and ρ−1 (ω2 ) have the same cardinality
...
Show that f is a
˜
˜
˜
one-one map of ρ−1 (ω1 ) onto ρ−1 (ω2 )
...
Not available
...
In this exercise all spaces are regions in the plane
...
(Hint: If f (z1 ) = f (z2 ) let γ be a path in G from z1 to z2 and consider a certain analytic
continuation along f ◦ γ; apply the Monodromy Theorem
...
Show that if G1 is simply connected then there is an analytic function f : G1 → G2 such that (G1 , f ) is a
covering of G2 and f1 = f2 ◦ f
...
(c) Let (G1 , f1 ), (G2 , f2 ) and Ω be as in part (b) and, in addition, assume that both G1 and G2 are simply
connected
...
Solution
...
Exercise 7
...
Show that for every region Ω1 contained in Ω which is simply
connected there is an analytic function g : Ω1 → G such that f (g1 (ω)) = ω for all ω in Ω1
...
Not available
...
What is a simply connected covering space of the figure eight?
Solution
...
Exercise 9
...
Solution
...
Exercise 10
...
6
...
Solution
...
136
Chapter 10
Harmonic Functions
10
...
Show that if u is harmonic then so are u x =
∂u
∂x
and uy =
∂u
∂y
...
Let u be harmonic, that is
u xx + uyy = 0
...
1)
Since u : G → R is harmonic, u is infinitely times differentiable by Proposition 1
...
252
...
Now, we show that
u x is harmonic, that is
(u x ) xx + (u x )yy = 0
...
(10
...
Finally, we show that uy is harmonic, that is
(uy ) xx + (uy )yy = 0
...
(10
...
Exercise 2
...
Solution
...
(10
...
3 p
...
Thus, all the partials are continuous
(therefore u xy = uyx )
...
29)
...
Therefore, −uy is harmonic, too
...
b) We have to show that the Cauchy-Riemann equations are satisfied, that is
(u x ) x = −(uy )y
(u x )y = −(−uy ) x
...
2) since
(u x ) x = −(uy )y ⇐⇒ u xx + uyy = 0
...
Therefore
(u x )y = −(−uy ) x ⇐⇒ u xy = uyx ⇐⇒ u xy = uyx
...
Exercise 3
...
k,l=0
Show that p is harmonic iff:
(a) k(k − 1)ak,l−2 + l(l − 1)ak−2,l = 0 for 2 ≤ k, l ≤ n;
(b) an−1,l = an,l = 0 for 2 ≤ l ≤ n;
(c) ak,n−1 = ak,n = 0 for 2 ≤ k ≤ n
...
Not available
...
Prove that a harmonic function is an open map
...
)
Solution
...
0 for any z show that u = log | f | is harmonic on G
...
If f is analytic on G and f (z)
Solution
...
¯
Exercise 6
...
Show that
u(a) =
1
πR2
u(x, y) dx dy
...
Let D be a disk such that B(a; R) ⊂ D ⊂ G
...
From Cauchy’s Integral Formula, we have
f (a) =
1
2πi
γ
f (w)
1
dw =
w−a
2π
2π
f (a + reiθ ) dθ
0
where γ(t) = a + reit , 0 ≤ t ≤ 2π
...
0
Multiply by r, we get
f (a)r =
1
2π
2π
f (a + reiθ )r dθ
0
138
and then integrate with respect to r yields
R
R
f (a)r dr =
0
⇐⇒
⇐⇒
⇐⇒
0
r2
1
=
2
2π
1
f (a) =
πR2
1
f (a) =
πR2
R
2π
1
2π
f (a + reiθ )r dθ dr
0
2π
f (a + reiθ )r dθ dr
f (a)
0
2π
0
R
f (a + reiθ )r dr dθ
0
0
f (x, y) dx dy
...
Now
u(a) = Re( f (a)) =
1
πR2
Re( f (x, y)) dx dy =
¯
B(a;R)
Thus
u(a) =
Exercise 7
...
¯
B(a;R)
u(x, y) dx dy
...
2
Show that u is harmonic and limr→1− u(reiθ ) = 0 for all θ
...
7? Why?
2
Solution
...
Let f (z) = 1+z
...
By
1−z
Theorem III 2
...
Therefore
1 + z 2
u(z) = Im
1−z
is harmonic
...
We have
1 + reiθ 2
u(re ) = Im
1 − reiθ
1 + reiθ 2 1 − reiθ
= Im
1 − reiθ
1 − reiθ
(1 + reiθ )(1 − reiθ )
= Im
(1 − reiθ )(1 − reiθ )
iθ
2
2
(1 + reiθ − re−iθ − r2 )2
(1 − reiθ − re−iθ + r2 )2
(1 + 2ri sin θ − r2 )2
= Im
(1 − 2r cos θ + r2 )2
((1 − r2 ) + 2ri sin θ)2
= Im
(1 − 2r cos θ + r2 )2
= Im
(1 − r2 )2 + 4ri sin θ(1 − r2 ) − 4r2 sin2 θ
(1 − 2r cos θ + r2 )2
4r(1 − r2 ) sin θ
...
If we assume, that r = 1 and θ 2πk, k ∈ Z, then
lim− u(reiθ ) = u(eiθ ) =
r→1
4· 1· (1 − 12 ) sin θ
0
= 2
= 0
...
2 )2
r→1 (1 − r)4
(1 − 2r cos(2πk) + r
Thus,
lim u(reiθ ) = 0
...
7
...
Claim: lim supz→1 u(z) > 0
...
(1 − 2x + x2 + y2 )2
2
4y(−y2 ) −4y3
4
= 4 =−
y
y4
y
But − 4 → ∞ as y → 0−
...
Exercise 8
...
(a) Show that
∂2 U
∂U ∂2 U
∂ ∂U
∂2 U
∂2 u ∂2 u
+ 2 = r2 2 + r
r
+ 2
...
∂r ∂r
∂θ
(b) Let u have the property that it depends only on |z| and not arg z
...
Show that u is
harmonic iff u(z) = a log |z| + b for some constants a and b
...
Not available
...
Let u : G → R be harmonic and let A = {z ∈ G : u x (z) = uy (z) = 0}; that is, A is the set of zeros
of the gradient of u
...
Let u be harmonic and not constant
...
255, we know that u x and uy are harmonic
...
255, we know that f = u x − iuy is analytic
...
Clearly
A = {z ∈ G : u x (z) = uy (z) = 0} = B
since u x (z) − iuy (z) = 0 iff u x (z) = uy (z) = 0 (u x (z) and uy (z) are real)
...
10 p
...
This implies that A cannot
have a limit point in G
...
(In this case f ≡ 0 and by Theorem
3
...
78 A has a limit point in G)
...
State and prove a Schwarz Reflection Principle for harmonic functions
...
Not available
...
Deduce the Maximum Principle for analytic functions from Theorem 1
...
Solution
...
10
...
Let D = {z : |z| < 1} and suppose that f : D− → C is a continuous function such that both Re f
and Im f are harmonic
...
Using Definition 2
...
141
Solution
...
Clearly u and v are continuous functions mapping D to R
...
9 we have
u(reiθ ) =
π
1
2π
−π
Pr (θ − t)u(eit ) dt
for 0 ≤ r < 1 and all θ
(10
...
(10
...
3),(10
...
Thus, we have proved
f (reiθ ) =
1
2π
π
−π
f (eit )Pr (θ − t) dt
for all reiθ in D
...
⇒: Let f be analytic on D and assume n ≥ 1 (n integer)
...
Clearly g is
analytic on D
...
262
142
−π
f (eit )eint dt = 0 ∀n ≥ 1
...
We have for all z ∈ D
f (z) =
Part 1
=
=
=
π
1
2π
f (reiθ ) =
−π
∞
π
1
2π
f (eit )
−π
f (eit )Pr (θ − t) dt
r|n| ein(θ−t) dt
n=−∞
∞
π
1
r|n| einθ
2π n=−∞
f (eit )e−int dt
−π
−1
π
1
r|n| einθ
2π n=−∞
f (eit )e−int dt +
−π
1
2π
∞
π
f (eit )e−int dt
rn einθ
−π
n=0
=0 by assumption
=
1
2π
∞
π
f (eit )e−int dt
...
Thus f is analytic on D
...
In the statement of Theorem 2
...
Is the conclusion
of the theorem still valid? If not, what parts of the conclusion remain true?
Solution
...
Exercise 3
...
(b) If f : T → C is a continuous function define f˜ : D− → C by f˜(z) = f (z) for z in T and
1
f˜(reiθ ) =
2π
π
−π
f (eit )Pr (θ − t) dt
(So Re f˜ and Im f˜ are harmonic in D)
...
Show that for each r < 1 there
r
r
is a sequence {pn , (z, z)} of polynomials in z and z such that pn (z, z) → f˜ (z) uniformly for z in T
...
1
...
If f : T → C is a continuous function then there is a
sequence {pn (z, z)} of polynomials in z and z such that pn (z, z) → f (z) uniformly for z in T
...
Use part (c) to show that there is
a sequence {pn } of polynomials such that pn (t) → g(t) uniformly for t in [0, 1]
...
If g : [0, 1] → C is a continuous function then there is a
sequence {pn } of polynomials such that pn (t) → g(t) uniformly for t in [0, 1]
...
)
(f) Show that if the function g in part (e) is real valued then the polynomials can be chosen with real
coefficients
...
Not available
...
Let G be a simply connected region and let Γ be its closure in C∞ ; ∂∞G = Γ − G
...
(a) Show that ϕ(G) = D and ϕ(∂∞G) = ∂D
...
(c) Suppose that the function f in part (b) is not assumed to be continuous at ∞
...
Solution
...
Exercise 5
...
Suppose that u is a harmonic function on G0
such that limz→a u(z) exists and is equal to A
...
Solution
...
Exercise 6
...
u(x + iy) =
2 + (y − t)2
π −∞ x
Show that u is a bounded harmonic function on the right half plane such that for c in R, f (ic) = limz→ic u(z)
...
Not available
...
Let D = {z : |z| < l} and suppose f : ∂D → R is continuous except for a jump discontinuity at
¯
z = 1
...
5)
...
Let v be a harmonic conjugate of u
...
Not available
...
3 Subharmonic and superharmonic functions
Exercise 1
...
π
Solution
...
For all
ϕ(a + reiθ ) = ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ)2 + (β + r sin θ)2
= α2 + 2αr cos θ + r2 cos2 θ + β2 + 2βr sin θ + r2 sin2 θ
= α2 + β2 + 2αr cos θ + 2βr sin θ + r2
...
144
π
sin θ dθ +r2
−π
=0
Hence, ϕ ∈ Subhar(G)
...
Thus
1
2π
π
ϕ(a + reiθ ) dθ
−π
= α2 − β2 +
αr
π
π
cos θ dθ +
−π
r2
2π
π
−π
cos2 θ dθ −
−
π
sin θ dθ
−π
=π
=0
r2
2π
βr
π
=0
π
r2 r2
sin2 θ dθ = α2 − β2 + −
= α2 − β2 = ϕ(a)
...
c)
ϕ(a + reiθ )
= ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ)2 + (β + r sin θ)
= α2 + 2αr cos θ + r2 cos2 θ + β + r sin θ
...
2
Hence, ϕ ∈ Subhar(G)
...
Thus
1
2π
π
ϕ(a + reiθ ) dθ
−π
= α2 − β +
αr
π
π
cos θ dθ +
−π
r2
2π
=0
= α2 − β +
π
−π
cos2 θ dθ −
r
2π
π
sin θ dθ
−π
=π
2
r
≥ α2 − β = ϕ(a)
...
e)
ϕ(a + reiθ )
= ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ) + (β + r sin θ)2
= α + r cos θ + β2 + 2βr sin θ + r2 sin2 θ
...
= α + β2 +
2
Hence, ϕ ∈ Subhar(G)
...
Thus
1
2π
π
ϕ(a + reiθ ) dθ
−π
= α − β2 +
r
2π
π
−π
cos θ dθ −
βr
π
=0
π
−π
sin θ dθ −
r2
2π
π
sin2 θ dθ
−π
=0
=π
r2
= α − β2 −
≤ α − β2 = ϕ(a)
...
Exercise 2
...
(a) Show that Subhar(G) and Superhar(G) are closed subsets of C(G; R)
...
Not available
...
(This exercise is difficult
...
This defines a map T : (∂∞G; R) → Har(G) by T ( f ) = u f
...
e
...
(b) T is positive (i
...
, if f (a) ≥ 0 for all a in ∂∞G then T ( f )(z) ≥ 0 for all z in G)
...
Moreover, if { fn } is a sequence in C(∂∞G; R) such that fn → f uniformly then T ( fn ) →
T ( f ) uniformly on G
...
Is the converse true?
Solution
...
Exercise 4
...
11, suppose only that f is a bounded function on ∂∞G; prove
that the conclusion remains valid
...
If we define f : ∂∞G → R by f (z) = g(z) for z in ∂G and f (∞) = 0 then the conclusion
of Theorem 3
...
Of course there is no reason to expect that the harmonic function will have
predictable behavior near ∞ — we could have assigned any value to f (∞)
...
)
Solution
...
Exercise 5
...
5 is necessary
...
Not available
...
If f : G → Ω is analytic and ϕ : Ω → R is subharmonic, show that ϕ ◦ f is subharmonic if f
is one-one
...
Clearly f is continuous, since f : G → Ω is analytic and ϕ is continuous, since ϕ : Ω → R is
subharmonic
...
If we can show that for every bounded
¯
¯
region G1 such that G1 ⊂ G and for every continuous function u1 : G1 → R that is harmonic in G1 and
satisfies u(z) ≤ u1 (z) for z in ∂G1 , we have
u(z) ≤ u1
for z in G1 , then u is subharmonic (by Corollary 3
...
265)
...
We have to show u(z) ≤ u1 (z) ∀z ∈ G1
...
So define
¯
ϕ1 = u1 ◦ f −1 : Ω → R
¯
¯
with f −1 : Ω1 → G1
...
Since ϕ : Ω → R
is subharmonic, we can use Theorem 3
...
In fact ϕ−ϕ1 satisfies the Maximum Principle
˜
˜
on Ω1
...
Thus,
ϕ( f (z)) − ϕ1 ( f (z)) ≤ 0 ∀z ∈ ∂G1
...
Therefore, u = ϕ ◦ f is subharmonic on G
...
( f is locally one-one and onto and f −1 is analytic and hence u = ϕ ◦ f is locally subharmonic)
...
Then there is a neighborhood U of z0 and a
neighborhood V of w0 = f (z0 ) such that f : U → V is one-one and onto and f −1 : V → U is analytic
...
We can use Theorem 7
...
98 to
find > 0 and δ > 0 such that each w with |w − w0 | < δ has exactly one pre-image z with |z − z0 | <
...
By Theorem 7
...
98, f maps U one-one onto V
...
By the
Open Mapping Theorem (Theorem 7
...
99), f = ( f −1 )−1 is an open map, so f −1 is continuous from V to
U
...
7 p
...
We have
f −1 (w) =
1
2πi
|z−z0 |=
z f (w)
dz
...
147
10
...
Let G = B(0; 1) and find a barrier for G at each point of the boundary
...
Not available
...
Let G = C − (∞, 0] and construct a barrier for each point of ∂∞G
...
Not available
...
Let G be a region and a a point in ∂∞G such that there is a harmonic function u : G → R with
limz→a u(z) = 0 and lim inf z→w u(z) > 0 for all w in ∂∞G, w a
...
Solution
...
Exercise 4
...
9
...
Show that G has a barrier at a
...
)
Solution
...
10
...
(a) Let G be a simply connected region, let a ∈ G, and let f : G → D = {z : |z| < 1} be a
one-one analytic function such that f (G) = D and f (a) = 0
...
(b) Find the Green’s Functions for each of the following regions: (i) G = C − (∞, 0]; (ii) G = {z : Re z > 0};
(iii) G = {z : 0 < Im z < 2π}
...
Solution to part a):
Let G be simply connected, let a ∈ G, and let f : G → D = {z : |z| < 1} be a one-one analytic function such
that f (G) = D and f (a) = 0
...
Clearly, ga (z) has a singularity if | f (z)| = 0
...
By assumption f (z) = 0 iff z = a
...
This implies ga (z) has a singularity at a
...
1
...
This follows directly by Exercise 5 p
...
b) G(z) = ga (z)+log |z−a| is harmonic in a disk about a
...
Clearly ga (z) = − log | f (z)| is
harmonic on B(a; r)−{a}
...
255, we can argue again that log |z−a| is harmonic on B(a; r)−{a}
(choose f (z) = z − a and f (z) = 0 iff z = a)
...
But
G(z) = − log | f (z)| + log |z − a| = − log
|z − a|· | f˜(z)|
| f (z)|
= − log
= − log | f˜(z)|
|z − a|
|z − a|
where f (z) = (z − a) f˜(z) by assumption and f˜(z) is analytic on G with no zero
...
c) limz→w ga (z) = 0 for each w in ∂∞G
...
Hence
lim ga (z) = lim − log | f (w)| = − log 1 = 0
z→w
w→a
148
for each a in ∂∞G
...
Solution to part b) ii):
Clearly, G is a simply connected region (not the whole plane)
...
Note that h has to satisfy the following assumptions to use Part 1 a):
a) h : G → D one-one and analytic
b) h(a) = 0
c) h(G) = D
...
160 Theorem 4
...
It remains to find h:
Claim: h : G → D given by
h(z) = f (g(z))
where
g(z) =
z − g(a)
z−1
and f (z) =
z+1
1 − g(a)z
works
...
Proof of the Claim:
We will invoke the Orientation Principle (p
...
Then
g(z) =
z−1
z+1
(see p
...
From p
...
Hence h(z) = f (g(z)) maps G onto D
...
Exercise 2
...
Prove that if ψ is a
positive superharmonic function on G − {a} with lim inf z→a [ψ(z) + log |z − a|] > −∞, then ga (z) ≤ ψ(z) for
z a
...
Not available
...
This exercise gives a proof of the Riemann Mapping Theorem where it is assumed that if G is
a simply connected region, G C, then: (i) C∞ − G is connected, (ii) every harmonic function on G has a
harmonic conjugate, (iii) if a G then a branch of log(z − a) can be defined
...
Let u(z) = ga (z) + log |z − a| and let v be the harmonic conjugate of u
...
(So f is analytic in G
...
Prove that for 0 < r < 1,
Cr = {z : | f (z)| = r} consists of a finite number of simple closed curves in G (see Exercise VI
...
3)
...
Prove that f is one-one on Gr
...
(b) Let G be a simply connected region with G C, but assume that G is unbounded and 0, ∞ ∈ ∂∞G
...
Show that l is one-one on G and l(z) α + 2πi for any z in
G
...
(Show that l omits all values in a neighborhood of α + 2πi
...
Solution
...
Exercise 4
...
If
f : ∂G → R is continuous and g(z, a) = ga (z) is the Green’s Function on G with singularity at a, show that
h(a) =
f (z)
γ
∂g
(z, a) ds
∂n
(5
...
(Note: these concepts are not discussed in this book but the formula is sufficiently interesting so as to merit
presentation
...
)
(b) Show that if G = {z : |z| < 1} then (5
...
5)
...
Not available
...
1 Jensen’s Formula
Exercise 1
...
We have
2π
f (m) (0)
r
1
+ m log r = −
log
+
m!
|ak |
2π
k=1
r(z−ak )
r2 −¯ k z
a
0
...
maps B(0; r) onto itself and takes the boundary to the boundary
...
r(z − ak )
Then F ∈ H(G) and F has no zeros in B(0; r) and |F(z)| = | f (z)| if |z| = r
...
Also
g(z) ≡
=
=
f (z)
=
zm
∞ f (i) (0)zi
i=0
i!
zm
∞ f (i) (0)zi
i=m
i!
zm
∞
i=m
f (m) (0) 0 f (m+1) (0) 1
f (i) (0)zi−m
=
z +
z +
...
r
...
|ak |
2π
r
1
=
|ak | 2π
log | f (reiθ )| dθ
0
which is equivalent to
n
log
1
f (m) (0)
r
+
+ m log r = −
log
m!
|ak | 2π
k=1
2π
0
log | f (reiθ )| dθ
which proves the assertion
...
Let f be an entire function, M(r) = sup{| f (reiθ )| : 0 ≤ θ ≤ 2π}, n(r) = the number of zeros of f
in B(0; r) counted according to multiplicity
...
¯
Solution
...
Suppose a1 ,
...
, bm are the zeros of f in B(0; 2r) − B(0; r)
...
, an , b1 ,
...
Since f (0) = 1 0,
we have by Jensen’s formula
n
0 = log 1 = log | f (0)| = −
2r
−
|ak |
log
k=1
m
log
k=1
2r
1
+
|bk |
2π
2π
0
log | f (2reiθ )| dθ
which implies
n
m
2r
log
|ak |
k=1
−
=
log
k=1
2r
|bk |
+
1
2π
2π
0
log | f (2reiθ )| dθ
≥ 0 since r ≤ |bk | < 2r
2π
1
log | f (2reiθ )| dθ
2π 0
2π
1
log |M(2r)|
dθ = log |M(2r)|
...
|ak |
But
n
log
k=1
2r
|ak |
n(r)
n(r)
2r
2r
= log
|ak |
|ak |
k=1
k=1
n(r)
n(r)
n
r
2r
2
= log 2n(r) + log
= log
|ak |
|ak |
k=1
k=1
=
log
≥ n(r) log(2)
152
since 0 < |ak | < r implies
r
|ak |
n(r) r
k=1 |ak |
> 1 ∀k which means
> 1 so the logarithm is greater than zero
...
¯
Exercise 3
...
Evaluate
1
2π
2π
log | f (reiθ )| dθ
...
Not available
...
(a) Using the notation of Exercise 2, prove that
r
0
n(t)
dt =
t
n
log
k=1
r
|ak |
where a1 ,
...
(b) Let f be meromorphic without a pole at z = 0 and let n(r) be the number of zeros of f in B(0; r) minus
the number of poles (each counted according to multiplicity)
...
t
Solution
...
Exercise 5
...
(a) If {an } are the non-zero zeros of f in D counted according to multiplicity, prove that (1 − |an |) < ∞
...
5
...
(b) If f has a zero at z = 0 of multiplicity m ≥ 0, prove that f (z) = zm B(z) exp(−g(z)) where B is a Blaschke
Product (Exercise VII
...
4) and g is an analytic function on D with Re g(z) ≥ − log M (M = sup{| f (z)| :
|z| < 1})
...
Not available
...
2 The genus and order of an entire function
Exercise 1
...
n→∞
(Hint: Use Cauchy’s Estimate
...
Define M(r) = max{ f (reiθ ) : 0 ≤ θ ≤ 2π} = max|z|=r | f (z)|
...
6 p
...
By Cauchy’s estimate p
...
1)
where cn =
f (n) (0)
n!
...
n
r (11
...
2)
1
Define, r = n µ+1 λ, where λ is a positive constant, and choose λ > e for convenience
...
2) can be
written in the form
αλµ+1 n
µ+1
µ+1
e
1
eαnλ
eαnλ
n µ+1
...
Since n! < nn , for n ≥ 2, we have
1
(n!) µ+1 |cn | <
for n > N or
lim cn (n!)1/(µ+1) = 0
...
Let f1 and f2 be entire functions of finite orders λ1 and λ2 respectively
...
Show that λ = max(λ1 , λ2 ) if λ1 λ2 and give an example which
shows that λ < max(λ1 , λ2 ) with f 0
...
Not available
...
Suppose f is an entire function and A, B, a are positive constants such that there is a r0 with
| f (z)| ≤ exp(A|z|a + B) for |z| > r0
...
Solution
...
Exercise 4
...
Solution
...
n=0
By Exercise 5 e) p
...
n log n
λ
ncn zn−1 =
n=1
∞
n=0
154
(n + 1)cn+1 zn
...
λ
Hence, λ = λ and therefore f also has order λ
...
Let f (z) =
α = lim inf
n→∞
− log |cn |
n log n
(a) Show that if f has finite order then α > 0
...
)
(b) Suppose that 0 < α < ∞ and show that for any > 0, < α, there is an integer p such that |cn |1/n <
n−(α− ) for n > p
...
Take r sufficiently large so that
N > p and show that
∞
n=N+1
r
nα−
N
n
r
< 1 and
n=p+1
nα−
n
< B exp (2r)1/(α−
)
where B is a constant which does not depend on r
...
(e) Prove that f is of finite order iff α > 0, and if f has order λ then λ = α−1
...
Not available
...
Find the order of each of the following functions: (a) sin z; (b) cos z; (c) cosh
∞
−an n
z where a > 0
...
)
n=1 n
√
z; (d)
1
Solution
...
Note that sinh2 z = cosh2 z − 1 and that cos x is bounded for real values x,
so
| cos z|2 = cos2 x + sinh2 y ≤ cosh2 y ≤ cosh2 |z|
...
1
This implies that the order λ ≤ 2
...
Seeking contradiction assume that
2
there is > 0 and r > 0 such that
√
1
| cos z| ≤ exp(|z| 2 − )
(11
...
√
n
Let z = −n2 for an n ∈ N with n > max{ r, 4, 21/(2 ) } then in particular (i) |z| > r, (ii) − ln 2 + n > 2 , and
1
1
(iii) 2 > n2 and
| f (z)| = | cos
√
n
1
1
1
1
z| = |en + e−n | > en = e− log 2+n > e 2 > exp(n−2 n) = exp(n2( 2 − ) ) = exp(|z| 2 − )
...
3)
...
2
Exercise 7
...
Solution
...
Since f1 and
f2 are entire functions of finite order λ1 and λ2 , respectively, we have for > 0,
M1 (r) < er
λ1 + /2
M2 (r) < er
λ2 + /2
(11
...
5)
for sufficiently large r = |z| by Proposition 2
...
Hence
M(r)
= max | f (z)| = max | f1 (z)· f2 (z)|
|z|=r
|z|=r
≤ max | f1 (z)|· max | f2 (z)|
|z|=r
=
|z|=r
M1 (r)M2 (r)
= er
<
er
λ1 + /2
er
λ2 + /2
by (11
...
5)
λ1 + /2
+rλ2 + /2
≤
λ≤max(λ1 ,λ2 )
e2r
λ+ /2
< er
λ+
provided r is sufficiently large
...
Exercise 8
...
Let ρ = inf{a : |an |−a < ∞}; the
number called the exponent of convergence of {an }
...
(b) If ρ = the exponent of convergence of {an } then for every > 0, |an |−(ρ+ ) < ∞ and |an |−(ρ− ) = ∞
...
} be the non-zero zeros of f counted according
to multiplicity
...
(Hint: See the proof of (3
...
)
156
(d) Let P(z) = ∞ E p (z/an ) be a canonical product of rank p, and let ρ be the exponent of convergence of
n=1
{an }
...
(Hint: If λ is the order of P, ρ ≤ λ; assume that |al | ≤ |a2 | ≤
...
Choose N such that |an | ≤ 2|z| if n ≤ N and |an | > 2|z| if n ≤ N + 1
...
7) to show that for some > 0
∞
z
an
log E p
n=N+1
< A|z|ρ+
...
Use this to prove that
N
z
an
log E p
n=1
< C|z|ρ+
for some constant C independent of z
...
Not available
...
Find the order of the following entire functions:
(a)
f (z) =
∞
n=1
(1 − an z),
b)
f (z) =
∞
n=1
0 < |a| < 1
1−
z
n!
c)
f (z) = [Γ(z)]−1
...
b) Considering g(z) := ∞ 1 − n! one notes that g(z) is already in form of the canonical
m=1
product with the entire function in the exponent being constantly zero and p = 0
...
The simple zeros of g(z) are at z = n! and from Exercise 8a) it follows that λ ≤ 1
...
It suffices to show that the exponent of convergence ρ = 0 and to employ Exercise 8d)
...
(n − (2k − 1))· 2m
...
By choice of n, n ≤ 1
2
2m
and n − k + 1 > 2m which justifies the individual estimates by n
...
We conclude that ρ = inf{m : the left sum converges} =
0
...
157
11
...
Let f be analytic in a region G and suppose that f is not identically zero
...
Show that ∂h − i ∂h = ff on G0
...
Let f be analytic in a region G and suppose that f is not identically zero
...
Let f = u(x, y) + iv(x, y) = u + iv
...
We have by p
...
22 and 2
...
(C-R) 2
2
2
Therefore,
u x − iuy
f
=
...
Using the chain rule, we
u
v
ux + 2
vx
u2 + v2
u + v2
and
hy = hu uy + hv vy =
(11
...
u + iv
158
Compare this with (11
...
∂x
∂y
f
Exercise 2
...
7 and show that if λ1
λ2 then λ = max(λ1 ; λ2 )
...
Not available
...
(a) Let f and g be entire functions of finite order λ and suppose that f (an ) = g(an ) for a
sequence {an } such that |an |−(λ+1) = ∞
...
(b) Use Exercise 2
...
(c) Find all entire functions f of finite order such that f (log n) = n
...
} and no other zeros
...
a) Since f and g are entire functions of finite order λ, also F = f − g is an entire function of finite
order λ (see Exercise 2 p
...
Suppose that f (an ) = g(an ) for a sequence {an } such that |an |−(λ+1) = ∞
...
Since F 0, the sequence {an } are the zeros of F,
because
F(an ) = f (an ) − g(an ) = 0
since f (an ) = g(an ) by assumption
...
289), F has finite genus
µ ≤ λ since F ∈ H(C) with finite order λ
...
By the definition of the rank (p
...
1), we have
∞
n=1
|an |−(p+1) < ∞,
where an ’s are the zeros of F
...
Hence F ≡ 0, and therefore we
n=1
n=1
obtain f = g
...
286)
...
Assume F 0 ( f g), we will derive a contradiction
...
Let ρ be the exponent of convergence of {an }, then
ρ≤λ
by Exercise 8 c) p
...
Hence, according to Exercise 8 b) p
...
159
Therefore by the comparison test, we also have
∞
n=1
|an |−(λ+
)
Hence ∞ |an |−(λ+ ) < ∞ contradicting the fact
n=1
assertion f = g
...
|an |−(λ+ ) = ∞
...
1 Bloch’s Theorem
Exercise 1
...
Solution
...
e
...
Choose a with |a| = r0 , | f (a)| = K(r0 ) = 1−r0
...
ρ0
2ρ2
0
For z ∈ B(0, ρ0 ) define g(z) = f (z + a) − f (a)
...
It follows that
|g(z)| = |
γ f (w) dw| ≤
1
|z| ≤ 1 =: M
...
In this setting Lemma 1
...
6M
4·6
24
This statement rewritten for f yields
f (B(a, ρ0 )) ⊃ B f (a),
1
...
Suppose that in the statement of Bloch’s Theorem it is only assumed that f is analytic on D
...
) Do the same
for Proposition 1
...
Solution
...
161
12
...
Show that if f is a meromorphic function on C such that C∞ − f (C) has at least three points
then f is a constant
...
First consider an entire function f that misses three points in C∞ , one on which is ∞
...
By Little Picard’s Theorem the function f
must be a constant
...
Hence ∞ ∈ f (C) and by assumption there
are distinct a, b, c ∈ C − f (C)
...
Again using Little Picard’s
Theorem conclude that g is a constant function
...
Thus g is a nonzero constant function and
f (z) = g(z)−1 + a is also a constant function and the result is established
...
For each integer n ≥ 1 determine all meromorphic functions f and g on C with a pole at ∞
such that f n + gn = 1
...
Not available
...
3 Schottky’s Theorem
No exercises are assigned in this section
...
4 The Great Picard Theorem
Exercise 1
...
If it is assumed that this integral takes on
certain values for certain numbers a, does this imply anything about the nature of the singularity at z = 0?
Solution
...
Exercise 2
...
Solution
...
Exercise 3
...
1 equals F together with the constant functions
∞, 0, and 1
...
Not available
Title: Solutions to Functions of One Complex Variable by John B. Conway
Description: Solution for exercise question in Functions of One Complex Variable by John B. Conway
Description: Solution for exercise question in Functions of One Complex Variable by John B. Conway