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FIRST YEAR CALCULUS
W W L CHEN
c


W W L Chen, 1982, 2008
...

It is available free to all individuals, on the understanding that it is not to be used for financial gain,
and may be downloaded and/or photocopied, with or without permission from the author
...


Chapter 1
THE NUMBER SYSTEM

1
...
The Real Numbers
The purpose of the first four sections of this chapter is to discuss a number of the properties of the
real numbers
...
We do not propose to discuss here these properties in
great detail, and shall only give a brief introduction
...

The first collection of properties of R is generally known as the Field axioms
...

FIELD AXIOMS
...

(A2) For every a, b, c ∈ R, we have a + (b + c) = (a + b) + c
...

(A4) For every a ∈ R, there exists −a ∈ R such that a + (−a) = 0
...

(M1) For every a, b ∈ R, we have ab ∈ R
...

(M3) For every a ∈ R, we have a1 = a
...

(M5) For every a, b ∈ R, we have ab = ba
...

Remark
...
In the terminology of group theory, not usually covered in first
Chapter 1 : The Number System

page 1 of 20

First Year Calculus

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W W L Chen, 1982, 2008

year mathematics, we say that the set R forms an abelian group under addition, and that the set of all
non-zero real numbers forms an abelian group under multiplication
...
The property (D) is called the Distributive law
...

ORDER AXIOMS
...

(O2) For every a, b, c ∈ R satisfying a > b and b > c, we have a > c
...

(O4) For every a, b, c ∈ R satisfying a > b and c > 0, we have ac > bc
...
Clearly the Order axioms as given do not appear to include many other properties of the
real numbers
...
For example,
suppose that x > 0
...
It follows from (O3) and (A3)
that 0 = x + (−x) > 0 + (−x) = −x, giving −x < 0
...
2
...
However, this definition does not bring out some of the main properties of the set N
in a natural way
...

Definition
...

(N2) If n ∈ N, then the number n + 1, called the successor of n, also belongs to N
...

(WO) Every non-empty subset of N has a least element
...
The condition (WO) is called the Well-ordering principle
...
However, these two conditions alone are insufficient to
exclude from N numbers such as 5
...
Now, if N contained 5
...
5, 3
...
5, 1
...
5, −0
...
5, −2
...
, and so would not have a least element
...
This is achieved by the condition
(WO)
...
The following two forms
of the Principle of induction are particularly useful
...
4
...
Suppose that the statement p(
...

Then p(n) is true for every n ∈ N
...
Suppose that the statement p(
...

Then p(n) is true for every n ∈ N
...

Example 1
...
1
...
+ n =

n(n + 1)
2

(1)

for every n ∈ N
...
Then clearly p(1) is true
...
+ n =

n(n + 1)

...
+ n + (n + 1) =

n(n + 1)
(n + 1)(n + 2)
+ (n + 1) =
,
2
2

so that p(n + 1) is true
...

Example 1
...
2
...
+ n2 =

n(n + 1)(2n + 1)
6

(2)

for every n ∈ N
...
Then clearly p(1) is true
...
+ n2 =

n(n + 1)(2n + 1)

...
+ n2 + (n + 1)2 =

so that p(n + 1) is true
...

Example 1
...
3
...
To do so, let p(n) denote
the statement
(n ≤ 3) or (3n > n3 )
...
Suppose now that n > 3 and p(n) is true
...

It follows that (note that we are aiming for (n + 1)3 = n3 + 3n2 + 3n + 1 all the way)
3n+1 > 3n3 = n3 + 2n3 > n3 + 6n2 = n3 + 3n2 + 3n2 > n3 + 3n2 + 6n
= n3 + 3n2 + 3n + 3n > n3 + 3n2 + 3n + 1 = (n + 1)3 ,
so that p(n + 1) is true
...

Chapter 1 : The Number System

page 3 of 20

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First Year Calculus

W W L Chen, 1982, 2008

Example 1
...
4
...
To do so, let θ ∈ R be fixed, and let p(n) denote the statement (3)
...
Suppose now that p(n) is true, so that
(cos θ + i sin θ)n = cos nθ + i sin nθ
...
It now follows from the Principle of induction (Weak form) that (3) holds for
every n ∈ N
...
2
...
Consider the sequence x1 , x2 , x3 ,
...


(4)

We shall prove by induction that
xn = 2n+1 + 3n−1

(5)

for every n ∈ N
...
Then clearly p(1), p(2) are both true
...

Then
xn+1 = 5xn − 6xn−1 = 5(2n+1 + 3n−1 ) − 6(2n−1+1 + 3n−1−1 )
= 2n (10 − 6) + 3n−2 (15 − 6) = 2n+2 + 3n ,
so that p(n + 1) is true
...

Example 1
...
6
...
Then n is representable as a product of primes
...

First of all, clearly p(1) is true
...
Suppose now that n > 2 and that p(m) is true for every 1 ≤ m < n
...
If n is a prime, then it is obviously
representable as a product of primes
...
By our induction hypothesis, both n1 and n2 are representable as
products of primes, so that n must be representable as a product of primes, whence p(n) is true
...


1
...
Completeness of the Real Numbers
The set Z of all integers is an extension of the set N of all natural numbers to include 0 and all numbers
of the form −n, where n ∈ N
...

Chapter 1 : The Number System

page 4 of 20

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First Year Calculus

W W L Chen, 1982, 2008

We see that the Field axioms and Order axioms hold good if the set R is replaced by the set Q
...
A good illustration is the following well known result
...
No rational number x ∈ Q satisfies x2 = 2
...
Suppose that pq −1 has square 2, where p ∈ Z and q ∈ N
...
Then p2 = 2q 2 is even, so that p is
even
...
Then q 2 = 2r2 is even, so that q is even, contradicting that
assumption that p and q have no common factors apart from ±1
...


√

2 does not belong to Q
...
A non-empty set S of real numbers is said to be bounded above if there exists a number
K ∈ R such that x ≤ K for every x ∈ S
...

COMPLETENESS AXIOM
...
Then there is a real number M ∈ R such that M ≤ K for every upper bound K of the set S, and
that M > L for any real number L that is not an upper bound of S
...
The crucial assertion is that this number M is a real number
...
We can take K = 2 or K = 52000
...


1
...
Further Discussion on the Real Numbers
In this optional section, we shall first of all demonstrate the equivalence of the condition (WO) and the
two forms of the Principle of induction
...
Our first step is to show that the condition (WO) is equivalent to the Principle of induction
(strong form) (PIS)
...
Then the subset
S = {n ∈ N : p(n) is false}
of N is non-empty
...
If n0 = 1, then clearly (PIS1) does not hold
...

((PIS) ⇒ (WO)) Suppose that a non-empty subset S of N does not have a least element
...
Then p(1) is true, otherwise 1 would be the least element of S
...
, n belongs
to S
...
It now
follows from (PIS) that S does not contain any element of N, contradicting the assumption that S is a
non-empty subset of N
...

((PIS) ⇒ (PIW)) Suppose that (PIW1) and (PIW2) both hold
...
On the other hand, if p(m) is true for all m ≤ n, then p(n) is true in particular,
so it follows from (PIW2) that p(n + 1) is true, and this gives (PIS2)
...

Chapter 1 : The Number System

page 5 of 20

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First Year Calculus

W W L Chen, 1982, 2008

((PIW) ⇒ (PIS)) Suppose that (PIS1) and (PIS2) both hold for a statement p(
...
), where q(n) denotes the statement
p(m) is true for every m ≤ n
...
) imply respectively the two conditions
(PIW1) and (PIW2) for the statement q(
...

We next discuss the completeness of the real numbers in greater detail
...

COMPLETENESS AXIOM
...
Then there is a real number M ∈ R satisfying the following two conditions:
(S1) For every x ∈ S, the inequality x ≤ M holds
...

Remark
...
It is also easy to deduce that
if S is a non-empty set of real numbers and S is bounded below, then there is a unique real number
m ∈ R satisfying the following two conditions:
(I1) For every x ∈ S, the inequality x ≥ m holds
...

Definition
...
The real number m satisfying conditions (I1) and (I2) is
called the infimum of the non-empty set S, and denoted by m = inf S
...
Recall that there is √
rational number which satisfies the equation x2 = 2
...
We now want to show that it is a real number
...

Clearly the set S is non-empty, since 0 ∈ S
...
Hence S is a non-empty set of real numbers and S is bounded above
...
We now claim
that M 2 = 2
...
Then it follows from axiom (O1) that M 2 < 2 or M 2 > 2
...

If M 2 < 2, then we have
2

2

2

(M + ) = M + 2M  +  < 2




2 − M2
whenever  < min 1,

...

If M 2 > 2, then we have
(M − )2 = M 2 − 2M  + 2 > 2

whenever  <

M2 − 2

...

Note that M 2 = 2 and M is a real number
...

page 6 of 20

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First Year Calculus

W W L Chen, 1982, 2008

1
...
The Complex Numbers
It is easy to see that the equation x2 + 1 = 0 has no solution x ∈ R
...

Define the number i by i2 + 1 = 0
...
1
...
Note that the number a + 0i, where a ∈ R, behaves like the real number a
...
What we have said in the last paragraph basically amounts to the following
...
We have the addition rule
(a + bi) + (c + di) = (a + c) + (b + d)i,
and the multiplication rule
(a + bi)(c + di) = (ac − bd) + (ad + bc)i
...

For the division rule, suppose that c + di 6= 0, so that c 6= 0 or d 6= 0, whence c2 + d2 6= 0
...

It follows that
a = cx − dy,
b = cy + dx
...

c + di
c + d2
c + d2
The special case a = 1 and b = 0 gives
1
c − di
= 2

...

c + di
(c + di)(c − di)
c + d2
It is also useful to note that in has exactly four possible values, with i2 = −1, i3 = −i and i4 = 1
...
Suppose that z = x + yi, where x, y ∈ R
...
The real number y is called the imaginary part of z, and denoted by y = Imz
...

Example 1
...
1
...

1−i
1−i
(1 − i)(1 + i)
2
2 2
Hence
Re

(1 + 2i)2
7
=−
1−i
2

and

Im

(1 + 2i)2
1
=
...
5
...
We have 1 + i + i2 + i3 = 0 and 5 + 7i2003 = 5 − 7i
...
Suppose that z = x + yi, where x, y ∈ R
...

PROPOSITION 1B
...
Then
Rez =

z+z
2

z−z

...
Write z = x + yi, where x, y ∈ R
...

PROPOSITION 1C
...
Then
z+w =z+w

and

zw = z w
...
Write z = x + yi and w = u + vi, where x, y, u, v ∈ R
...

Chapter 1 : The Number System

page 8 of 20

Proof
...
Then
z + w = (x + u) + (y + v)i = (x + u) − (y + v)i = (x − yi) + (u − vi) = z + w
and
zw = (x + yi)(u + vi) = (xu − yv) + (xv + yu)i = (xu − yv) − (xv + yu)i = (x
−
− vi) = z w
c yi)(u
W W L Chen, 1982, 2008

First Year Calculus

as required
...
6
...
6
...

We
sometimes
vertex opposite the vertex (0, 0) in a parallelogram with (x, y) and (u, v) also as vertices
...

law
...
Suppose that z = x + yi, where
x, y ∈ R
...

shown
...

r =x +y
...


x = r cos θ

and

y = r sin θ
...
Suppose that z = x + yi, where x, y ∈ R
...
Suppose that z = x + yi, where|z|x,=y ∈ x
R
...
On the other hand,
|z| = x2 + y 2
called an argument of z, and denoted by arg z
...
given
On zthe
other
any
number
θ ∈ Rwesatisfying
the equations
(7) is
Remarks
...

Clearly
can add any
integer multiple
called
an
argument
of
z,
and
denoted
by
arg
z
...
We sometimes call a real number θ ∈ R the principal argument of z if
θ satisfies the equations (7) and −π < θ ≤ π
...
However,
Chapter 1 : The Number System
page 9 of 20
even
with this restriction on θ, it is not meaningful to write
"y#
(8)
θ = tan−1

...
Clearly the equations

c


First Year Calculus

W W L Chen, 1982, 2008

Remarks
...
Clearly we can add any integer multiple
of 2π to θ without affecting (7)
...
Note that it follows from (7) that y/x = tan θ
...


(8)

To see this, draw first of all√the complex number z = 1 + i on the Argand diagram
...
Furthermore, we have tan θ = 1
...
Clearly the equations (7) are satisfied with r = 2 and
θ = −3π/4
...
Note that the equation θ = tan−1 1 has two solutions
for θ in the range −π < θ ≤ π
...
Then
|x| =

√


x2

=

x
if x ≥ 0,
−x if x < 0,

and this is simply the absolute value of the real number x
...
The best advice is always to place the complex number z on the
Argand diagram and determine first of all a suitable range for θ
...
Once such a suitable range is
determined, the equation (8) will have a unique solution θ within this range
...
Suppose that z = x + yi 6= 0, where x, y ∈ R
...
Then we say that the pair (r, θ) form the polar
coordinates of z
...
(1) In view of (7), we have z = r(cos θ + i sin θ)
...
However, this is presupposing that we have understood the
exponential function with complex exponents
...
6
...
Suppose that z = 1 + i
...
Note also that

√ 
π
π
2 cos + i sin

...

Example 1
...
2
...

3
3
Try to draw the Argand diagram
...

PROPOSITION 1D
...

(b) For every z, w ∈ C, we have |zw| = |z||w|
...

Chapter 1 : The Number System

page 10 of 20

Example 1
...
2
...

3
3
Try to draw the Argand diagram
...


c


W W L Chen, 1982, 2008

PROPOSITION 1D
...

Proof
...
Then zz = (x + yi)(x − yi) = x2 + y 2
...

(c) For every z, w ∈ C, we have |z + w| ≤| z| + |w|
...
Then zw = (xu − yv) + (xv + yu)i, so that
Proof
...
Then zz = (x + yi)(x − yi) = x2 + y 2
...

(b) Write z = x + yi and w = u + vi, where x, y, u, x ∈ R
...

|zw|2 = (xu − yv)2 + (xv + yu)2 = (x2 + y 2 )(u2 + v 2 ) = |z|2 |w|2
...

z + w = 0
...
Then

  that z + w #= 0
...
Suppose

 z  now

|z| +
|w| is trivial
|z| if z +|w|

% + % w %
=
+
= %


%
%
%
|z + w|
|z +
+ w| % z +z w % % z +
|z|
|w|
|z|w| |z |w|
w w %%
=
+
=%
% + %

|z + w|
|z + w|
|z + w|w z + w
z
zz + w % w
$ = Re1 = 1
...

z+w
z+w
z+w z+w
The result follows immediately
...
!
Remark
...
Proposition
Proposition 1D(c)
1D(c) is
is known
known as
as the
the Triangle
Triangle inequality
...
It
It can
can be
be understood
understood easily
easily from
from the
the
diagram
below:
diagram below:

w

|w|

|z+w|

z
|z|

0
Chapter 1 : The Number System
10 of 19
The
inequality follows on noting that the sum of the lengths of two sides of a triangle is atpage
least
the
length of the third side
...

Let us use polar coordinates instead
...
Then
zw = rs(cos θ + i sin θ)(cos φ + i sin φ)
= rs((cos θ cos φ − sin θ sin φ) + i(cos θ sin φ + sin θ cos φ))
= rs(cos(θ + φ) + i sin(θ + φ))
...
On the other
hand, it is not difficult to show that
z
r
= (cos(θ − φ) + i sin(θ − φ))
...
6
...
Suppose that z = 1 + i and w = −1 − i 3
...

zw = 2 2 cos
4
3
4
3
12
12
Note also that
√
√
√
zw = (1 + i)(−1 − i 3) = ( 3 − 1) − i( 3 + 1),
so that

cos

π 2π
−
4
3

√


=

3−1
√
2 2


and

sin

π 2π
−
4
3

√


=−

3+1
√
...

w
2
4
3
4
3
12
12
2
Example 1
...
4
...
Then repeated application of (9) yields







√
√
5π
3π
5π
3π
5
z = 4 2 cos
+ i sin
= 4 2 cos −
+ i sin −

...

Our last example suggests the following important result
...
(DE MOIVRE’S THEOREM) Suppose that n ∈ N and θ ∈ R
...

Proof
...

Remarks
...
2
...

(2) In the notation eiθ = cos θ + i sin θ, de Moivre’s theorem is the observation that einθ = (eiθ )n
...
6
...
We have
cos 3θ + i sin 3θ = (cos θ + i sin θ)3 = cos3 θ + 3i cos2 θ sin θ + 3i2 cos θ sin2 θ + i3 sin3 θ
= (cos3 θ − 3 cos θ sin2 θ) + i(3 cos2 θ sin θ − sin3 θ)
...


Remark
...

Chapter 1 : The Number System

page 12 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

1
...
Finding Roots
Let us try to find the square roots of the complex number a + bi, where a, b ∈ R
...

We may assume that b 6= 0, otherwise the solution is trivial
...


(12)

It follows from (11) and (12) that
x2 + y 2 =

p
a2 + b2 ,

where the square root is non-negative
...

2

(13)

Note that the equations (13) generally yield two solutions for x and two solutions for y
...
It follows that
√

s
a + bi = ± 

a+

s

√

b
a2 + b2
+i
2
|b|

−a +


a2 + b2 
,
2

√

where the square roots are non-negative
...
As we
have shown earlier, it is more convenient to do multiplication of complex numbers in polar coordinates,
so let us attempt to find roots using polar coordinates
...
Consider the equation
z n = c,
where n ∈ N is fixed
...

It follows that
rn = R,
and we can take
nθ = α + 2kπ,

where k = 0, 1,
...
, n − 1
...
It follows that
z=

√
n



α + 2kπ
α + 2kπ
R cos
+ i sin
,
n
n

where k = 0, 1,
...
On the other hand, it follows from (15) and de Moivre’s theorem that
each of the n numbers in (15) satisfies z n = c
...

PROPOSITION 1F
...
Then the solutions
of the equation z n = c are given by (15)
...
7
...
The 7-th roots of 1 − i can be calculated as follows
...
It follows from Proposition 1F that the
7-th roots of 1 − i are given by





√
π 2kπ
π 2kπ
14
z=
2 cos
+
+ i sin
+
, where k = 0, 1, 2, 3, 4, 5, 6
...
7
...
The case c = 1 is particularly important, as we get the n-th roots of 1
...
It follows that the n-th roots of unity are given by
z = cos

2kπ
2kπ
+ i sin
,
n
n

where k = 0, 1,
...


Example 1
...
3
...

Furthermore,
p(z) = (z − 1)(z 2 + 2),
so that two other solutions are given by the roots of the equation z 2 = −2
...
It follows that the two roots of z 2 = −2 are given by
z=

√ 
π √
π
= 2i
2 cos + i sin
2
2

and

z=

√



√
3π
3π
+ i sin
2 cos
= − 2i
...
7
...
Consider the polynomial p(z) = z 6 − 2z 3 + 4 = 0
...

2
To find the roots of p(z), we have to find all the roots of
z3 = 1 +

√

3i,

(16)

3i
...
It follows from Proposition 1F that the






π 2kπ
π 2kπ
2 cos
+
+ i sin
+
,
9
3
9
3

where k = 0, 1, 2;

in other words,

√
π
π
3
,
z1 = 2 cos + i sin
9
9
To study (17), note that 1 −
roots of (17) are given by
z=

√
3

√

z2 =

√
3



7π
7π
2 cos
+ i sin
,
9
9

z3 =

√
3



13π
13π
2 cos
+ i sin
9
9



...
It follows from Proposition 1F that the






5π 2kπ
5π 2kπ
2 cos
+
+ i sin
+
,
9
3
9
3

where k = 0, 1, 2;

in other words,
z4 =

√
3



5π
5π
2 cos
+ i sin
9
9


,

z5 =

√
3



11π
11π
2 cos
+ i sin
9
9


,

z6 =

√
3



17π
17π
2 cos
+ i sin
9
9



...
8
...
If
we write z = x + iy, then such an equation can be equally well described as a relation between z and
z
...
It follows that to obtain a genuine locus, these two equations should be essentially the same
...

Here, we shall restrict our discussion to three examples
...

Example 1
...
1
...


(18)

To see this, suppose that z = x + iy and c = a + ib, where x, y, a, b ∈ R
...
Note that the equation (18) can also be written
in the form
(z − c)(z − c) = r2
...
Next, we consider the inequality |z − c| < r
...
This represents the region on the xy-plane inside the
circle (x − a)2 + (y − b)2 = r2
...

Chapter 1 : The Number System

page 15 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

Example 1
...
2
...
For writing z = x + iy, where x, y ∈ R, equation (20) becomes
|(x − 1) + iy| = |(x + 1) + iy|,
so that squaring both sides, we obtain
(x − 1)2 + y 2 = (x + 1)2 + y 2
...
Interpreted geometrically, note that |z − 1| represents the distance
between the points z and 1 on the Argand plane, while |z + 1| represents the distance between the points
z and −1 on the Argand plane
...

To achieve this, z must lie on the y-axis; in other words, we must have x = 0
...
This is the region on the complex plane containing all points z such that the
distance of z from 1 is smaller than the distance of z from −1
...

Example 1
...
3
...

We now place the parallelogram on the Argand plane so that the vertex O is precisely at the point 0
...
Then
the vertex B is represented by the complex number z + w
...
This proves that the two diagonals of a parallelogram bisect each other
...
Suppose that a, b, c, d are positive real numbers satisfying a < b and c < d
...

[Hint: Use the Field axioms and the Order axioms only
...
Find x, y ∈ R such that x < y and x−1 < y −1
...
Suppose that x, y, z ∈ R
...

4
...

[Hint: Use the Field axioms and the Order axioms only
...
Prove that 13 + 23 + 33 +
...

6
...

7
...

8
...
Such numbers can
be represented on the xy-plane by points of the form (a, b)
...
Draw a
picture of the xy-plane, clearly indicating the origin (0, 0) and the point (3, 4)
...
What is the length of this line segment?
c) We now multiply the complex number 3 + 4i by the complex number i to obtain the product
i(3 + 4i)
...

d) Draw a line segment joining (0, 0) and the point in part (c)
...

Draw a line segment joining (0, 0) and this point
...
Repeat steps (a)–(e) with a + bi in place of 3 + 4i
...
Let z1 = 2 + 4i and z2 = 12 (1 − 5i)
...
Let z1 = 5, z2 = 3 + 4i and z3 = 1 − 3i
...
Suppose that z = x + iy, where x, y ∈ R and i2 = −1
...
Express each of the following numbers in the form x + yi, where x, y ∈ R:
a) (1 + 3i)3
b) (3 − 2i)2 − (3 + 2i)2
3 + 4i
c) (1 + i + i2 + i3 )100
d)
5 + 6i
Chapter 1 : The Number System

page 17 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

13
...

1−i
3 + 4i
i
14
...
Show that a = c and b = d
...
For each of the following complex numbers z, find real numbers x and y such that z = x + iy, then
show the positions of z and z on the Argand diagram, and determine the modulus and the principal
argument of z:
3 + 4i
a) z = (1 + i)4
b) z =
1 − 2i
16
...

a) Write down |z|2 and (Rez)2
...

c) For what values of z does equality hold?
17
...

a) Find z, |z| and z −1
...

18
...
Solve the equation z 2 + z + 1 = 0
...

20
...

a) Solve the equation
...

21
...

b) By considering the real and imaginary parts of your result in (a), solve for x and y
...
Consider the equation z 3 − 3z 2 + 4z − 2 = 0
...

b) Find also the other solutions
...
You are given that z = 1 is a solution of the cubic equation z 3 − 5z 2 + 9z − 5 = 0
...

24
...
Find the other
two solutions
...
You are given that z = −1 is a solution of the equation z 3 + 3z 2 + 6z + 4 = 0
...
Then indicate the positions of the three solutions in the Argand diagram
...
Suppose that a non-zero complex number z has modulus r and argument θ
...
Express each of the√
following in polar form:
a) −7 + 7i
b) 3 + 3i
c) −i
Chapter 1 : The Number System

d) 1 +

√

3i

e) 1 −

√

3i

f) −2 − 2i
page 18 of 20

c


First Year Calculus

28
...
a) On the Argand diagram, choose a point z with positive real and imaginary parts and satisfying
|z| = 2
...

b) Explain in simple English how you come to your conclusions
...
Suppose that the complex number z satisfies |z| = 1
...

31
...
Explain why 0, z −1 and z lie in a straight line on the Argand
plane
...
Suppose that the complex number z1 is a cube root of unity and the complex number z2 is a 4-th
root of unity
...
Show that z is a 12-th of unity
...
Use de Moivre’s theorem to show that for every real number θ, we have cos 2θ = cos2 θ − sin2 θ and
sin 2θ = 2 sin θ cos θ
...
Consider the equation z 6 = −64
...

b) Convert your answers in part (a) to rectangular form
...
Use instead the well known fact that
√
3
π
cos =
6
2

and

sin

π
1
=
...

35
...
Let z = −1 − i
...

b) Find the modulus |z|
...

d) Express z 3 in polar form and then indicate its position in the Argand diagram you have drawn
in part (a)
...
Consider the equation z 4 = −16
...

b) Draw an Argand diagram clearly indicating the positions of the four roots
...

38
...

a) Show that z = ±1 are solutions of the equation
...

c) Draw an Argand diagram clearly showing all six solutions of the equation
...
Find in polar form the cube roots of −2 − 2i
...

40
...
What is the equation of this line?
Chapter 1 : The Number System

page 19 of 20

c


First Year Calculus

W W L Chen, 1982, 2008

41
...
What is it?
42
...


Harder Problems for Chapter 1
43
...
Find sup A and inf A
...
For each of the following sets, determine whether the set is bounded above, bounded below, both
or neither
...
Suppose that a, b ∈ R and a < b + n−1 for every n ∈ N
...

[Hint: Suppose on the contrary that a > b
...
]
46
...

a) Show that sup(A ∪ B) = max{sup A, sup B}
...
a) Suppose that x ≤ a for every x ∈ E
...

b) Show that the corresponding statement with ≤ replaced by < does not hold
...
Suppose that A and B are two non-empty bounded sets of real numbers
...
Show that sup E = sup A+sup B
and sup F = sup A − inf B
...
a) Suppose that A is a non-empty bounded set of real numbers
...
Show that inf A ≤ inf B ≤ sup B ≤ sup A
...
Show that if C = {bx : x ∈ A}, then sup C = b sup A
...
Show that if
C = {xy : x ∈ A and y ∈ B}, then sup C = (sup A)(sup B)

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