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UNIVERSITY TEXT/CALCULUS I/COMPLEX NUMBERS
PROBLEMS
M
...
complex numbers problems
1
...

i) i5 + i15

ii) i2019 + i2021 + i2023 + i2024

2 + 3i
(3i − 2)i

iv)

v)

99
Y

i

vi)

n
Y

ik

iii) (2 + i)4

vii)

ik
...

i) i5 + i15 = i5 (1 + i10 ) = i · i4 (1 + (i4 )2 · i2 ) = i(1 − 1) = 0
ii) i2019 + i2021 + i2023 + i2024 = i2019 (1 + i2 + i4 + i5 ) = (i2 )
1009

= (−1)

1009

· i(1 − 1 + 1 + i)

2

i(1 + i) = −(i + i ) = 1 − i

iii) Using the Binomial expansion (a + b)n =
4

(i2 = −1 and i4 = 1)

v)

3

2

k=0
3

2

k

an−k bk we have,

(2 + i) = 2 + 4 × 2 ×i + 6 × 2 × i + 4×2×i + i4 = −7 + 24i
iv)

4

n  
X
n

2 + 3i
2 + 3i
(2 + 3i)(−3 + 2i)
−12 − 5i
1
=
=
=
= − (12 + 5i)
(3i − 2)i
(−3 − 2i)
(−3 − 2i)(−3 + 2i)
13
13
99
Y

99

}|
{
z
i = i × i × i ×
...
× i = i

...
Indeed, by division
theorem n = 8q + r, r = 0, 1, 2,
...
Therefore,
(8q + r)(8q + r + 1)
r(r + 1)
n(n + 1)
=
= 4q(8q + 2r + 1) +

...
, 7
...
+ in = i(1 + i +
...
+ in−1 )(1 − i)
1−i

2

M
...
Hence, in = ir = 1, i, −1, −i and
Pn
k
thereof k=1 i = 0, or i, or i − 1, or − 1
...
Solve the following equations
...
i) Using substitution z 2 = u, we have the quadratic equation u2 +u+1 = 0
...

2
2
The other two solutions to the equation are :
√
√
−1 − i 3
1+i 3
z=
and z =

...

Lemma 1
...
if z is a root of a real valued coefficient polynomial Pn (x) = an xn +
an−1 xn−1 +
...

ii) z 4 + 16 = 0
...

p
i(2kπ+π)
Arg(−16) = π
(−16)1/4 = 4 |−16|e 4
k = 0, 1, 2, 3
...

4
4
By Lemma 1
...

As an another approach one can easily find the latter two roots by direct calculation
...
We exert an absolute value to this equation
...
|z| = 0 implies z = 0
...


UNIVERSITY TEXT/CALCULUS I/COMPLEX NUMBERS

3

Hence, we should compute the forth roots of −1, and these are as follows:
(−1)1/4 = ei(
z1 = e

iπ
4

,

2kπ+π
)
4

z2 = e

i 3π
4

k = 0, 1, 2, 3
...


Therefore, the roots of the given equation are,
1
1
1
1
0, √ (1 + i), √ (−1 + i), √ (1 − i), √ (−1 − i)
...
Note that z  = |z| = z z¯
...
That is, z
is a real number
...

2
2
v) (1 + z)3 = (1 − z)3
...





(1 + z) − (1 − z) (1 + z)2 + (1 + z)(1 − z) + (1 − z)2 = 0
2z(z 2 + 3) = 0
2z = 0,
z = 0,

z 2 = −3
√
√
z = ± −3 = ±i 3

vi) z 4 + (3 − 6i)z 2 − 8 − 6i = 0
...

t2 + (3 − 6i)t − 8 − 6i = 0
p
−(3 − 6i) ± (3 − 6i)2 − 4(1)(−8 − 6i)
t=
2
√
−3 + 6i ± 5 − 12i
=
2p
−3 + 6i ± (3 − 2i)2
=
2
−3 + 6i ± (3 − 2i)
=
2
t1 = z 2 = −3 + 4i = (1 + 2i)2 and t2 = z 2 = 2i = (1 + i)2
z = ±(1 + 2i)

and z = ±(1 + i)

vii) z 6 = −1 − z 3
...
Then the given equation reduces to a quadratic
one
...

2
2

4

M
...

2

√
−1 − i 3

...


10π
i
4π
π
π

4π
= cos( ) + i sin( ) w2 = e 9 = − cos( ) + i sin( )
w1 = e
9
9
9
9
16π
i
2π
2π
w3 = e 9 = cos( ) − i sin( )
9
9
viii) z 4 + z 3 + z 2 + z + 1 = 0
...
+
z + 1), ∀z ∈ C
...


z4 + z3 + z2 + z + 1 =

All fifth roots of unity excluding 1 are solutions of the given equation
...


consequently,
11/5 = ei(
0

2kπ
5 )

z1 = e = 1,

k = 0, 1, 2, 3, 4
...
Indeed, using the
identities cos(3x) = 4 cos3 (3x) − 3 cos x and sin(2x) = 2 sin x cos x we have,
3π
π
π
3π
π 3π
π
) = 4 cos3 ( ) − 3 cos( ) and cos( ) = sin( −
) = sin( )
10
10
10
10
2
10
5
π
π
π
3 π
2 sin( ) cos( ) = 4 cos ( ) − 3 cos( )
(sin(2x) = 2 sin x cos x)
10
10
10
10
π
π
(cos2 x = 1 − sin2 x)
4 sin2 ( ) + 2 sin( ) − 1 = 0
10
10
√
√
π
−1 ± 5
π
−1 + 5
sin( ) =
⇒ sin( ) =
10
4
10p
4
√
r
10 + 2 5
π
π
cos( ) = 1 − sin2 ( ) =
10
10
4
p
√
√
π
π
π
−1 + 5
10 + 2 5
sin( ) = 2 sin( ) cos( ) = 2 ×
×
5 q
10
10
4
4
√
p
p
√
√
(−1 + 5)2
10 + 2 5
10 − 2 5
=2×
×
=
4
4
4
s
√
√
r
π
10 − 2 5
1+ 5
2 π
cos( ) = 1 − sin ( ) = 1 −
=
5
5
16
4
cos(

UNIVERSITY TEXT/CALCULUS I/COMPLEX NUMBERS

5

2π
2π
π 2π
π 2π
π
π
) + i sin( ) = sin( −
) + i cos( −
) = sin( ) + i cos( )
5q
5
2
5
2
5
10
10
√
√

1
− 1 + 5 + i 10 + 2 5
=
4
π
π
π
π
z3 = ei4π/5 = cos(π − ) + i sin(π − ) = − cos( ) + i sin( )
5
5
5
5
q
√
√

1
− 1 − 5 + i 10 − 2 5
=
4
q
√
√

π
1
π
z4 = ei6π/5 = − cos( ) − i sin( ) = − 1 + 5 + i 10 − 2 5
...

p
√
√
1 + i ± (1 + i)2 − 4 × 2(1 + i)
1 + i ± −8 − 6i
1 + i ± i 8 + 6i
z=
=
=
2
2
2
p
1 + i ± i (3 + i)2
1 + i ± i(3 + i)
=
=
2
2
1 + i − i(3 + i)
1 + i + i(3 + i)
z1 =
= 1 − i,
z2 =
= 2i
2
2
3
...
z = 1 − i satisfies the given equation
...
T hus,

b − 2a − 4 = 0,
4 − 2a = 0,

⇒ a = 2,

b = 8
...
Find real values a and b such that z = 1 + i is a solution of the equation
z 4 − 6z 3 + az 2 − 34z + b = 0
...

Solution
...
Thus,
(1 + i)4 − 6(1 + i)3 + a(1 + i)2 − 34(1 + i) + b = 0
(2i)2 − 6(2i)(1 + i) + a(2i) − 34(1 + i) + b = 0
b − 26 + i(2a − 46) = 0 ⇒ a = 23, b = 26
...

Hence, (z − (1 − i))(z − (1 + i)) = z 2 − 2z + 2 is a factor of the equation
...
So the other factor is
√
2±
−9
z 2 − 4z + 13
...
Thus, the roots of the equation are :
1
1 + i, 1 − i, 2 − 3i, 2 + 3i
...
If α and β are roots of the equation z 2 − 2z + 2 = 0, show that α2n + β 2n =
2
cos( nπ
2 )
...

z=

1±

√
1

−1

=1±i

6

M
...

= 2n+1 cos(
4
2
6
...
Prove that
α=1−i=

• i) z11 + z12 + z13 = 0
• ii)z12 + z22 + z32 = 0
Solution
...

|z1 |

ii) Squaring the equation z1 + z2 + z3 = 0, we get that,
z12 + z22 + z32 + 2(z1 z2 + z2 z3 + z3 z1 ) = 0
z12 + z22 + z32 + 2 × 0 = 0
z12
7
...
to i))

= 0
...

√
i
√
√
tan θ + √
tan θ + i cot θ
tan
θ = tan θ + i
√
i) √
=√
i
tan θ − i
tan θ − i cot θ
tan θ − √
tan θ
sin θ + i cos θ
i(cos θ − i sin θ)
=
=
sin θ − i cos θ
−i(cos θ + i sin θ)
ie−iθ
=
= −e−2iθ = −(cos(2θ) − i sin(2θ))
...
prove that
√
• i) 2 |z| ≥ |Rez| + |Imz|
• ii) |z1 − z2 | ≥ ||z1 | − |z2 ||


1
z
• iii) Re 1−z
≥ − , whenever |z| ≤ 1
 2 
 z1 
1|
• iv) ||z2|z
|−|z3 || ≥  z2 +z3 
• v) When does equality hold in triangle inequality?
Solution
...

Likewise, |z2 | − |z1 | ≤ |z2 − z1 | = |z1 − z2 |
Consequently,
− |z1 − z2 | ≤ |z1 | − |z2 | ≤ |z1 − z2 |
||z1 | − |z2 || ≤ |z1 − z2 |
...


iv) Referring to ii) we have,

||z2 | − |z3 || ≤ |z2 − z3 | = |z2 + z3 |
(z3 → −z3 )
1
1
≥
||z2 | − |z3 ||
|z2 + z3 |


 z1 
|z1 |
|z1 |


...
YEGAN

Re(z1 z¯2 ) = |z1 | |z2 | = |z1 z¯2 |

(z + z¯ = 2Rez)
2

Referring to the relations Rez = |z| and (Rez)2 + (Imz)2 = |z| we conclude
Imz = 0
...

x2
y2

Thus,
Re(z1 z¯2 ) = |z1 z2 | = x1 x2 + y1 y2 = k(x21 + y12 ) ≥ 0
and thereof k ≥ 0, that is, one of the numbers is a positive constant multiple of the
other
...

That is, equality holds
...

• i) Prove the some of all nth (n ≥ 2) roots of a complex number is zero
...

Find the sum z11 + z12 + z13 + z14 + z15
...
i) The nth roots of a complex number z are,
p
p
2kπ+θ
2kπ
θ
z 1/n = n |z|ei n = n |z|ei n ei n , k = 0, 1,
...

2kπ
2π
But ei n , k = 0, 1,
...
Putting w = ei n , then
1, w, w2 ,
...

n−1
X

p
n

|z|ei

2kπ+θ
n

θ

=

p
n

|z|ei n

=

p
n

θ
in

k=0

n−1
X

ei

2kπ
n

k=0

|z|e

(1 + w + w2 +
...
+ wn−1 )
1−w
p
θ
n
|z|ei n (1 − wn )
=
(wn = 1)
1−w
= 0
...
Hence, z11 , z12 ,
...
So by i) their sum is zero
...
If θ1 , θ2 and θ3 are the angles of a triangle, then what is the value of the
following expression ?
(sin θ1 − i cos θ1 )(sin θ2 − i cos θ2 )(sin θ3 − i cos θ3 )
i) −i

ii) −1

iii) i

iv) 1

UNIVERSITY TEXT/CALCULUS I/COMPLEX NUMBERS

9

Solution
...

(π/17)+i sin (π/17)
11
...

iπ
π
16π
cos (π/17) + i sin (π/17)
e 17
= − i16π = ei( 17 + 17 ) = eiπ = −1
...
Let 1, w, w2 , w3 , and w4 be the fifth roots of unity
...

(1 − w)(1 − w2 )(1 − w3 )(1 − w4 ) = (1 − w)(1 − w4 )(1 − w2 )(1 − w3 )
= (1 − w − w4 + w5 )(1 − w2 − w3 + w5 )
= (2 − w − w4 )(2 − w2 − w3 )

(w5 = 1)

= 4 − (w + w2 + w3 + w4 )
= 5 − (1 + w + w2 + w3 + w4 )
(1 + w + w2 + w3 + w4 )(1 − w)
1−w
1 − w5
=5−
=5−0=5
1−w
=5−

13
...


(1+i)15
(1−i)13

?
iv) 3
...

−i13π/4
2 e

14
...
z = z¯ ⇒ z  = z¯ 
3

2

2

|z| = |z| ⇒ |z| (|z| − 1) = 0
|z| = 0 ⇒ z = 0,
2
|z| = 1 ⇒ z 2 ·z 3 = z 2 ·¯
z 2 ⇒ z 5 = |z| = 1
...
YEGAN

The fifth roots of unity are : 1, w, w2 , w3 , w4 , where w = ei2π/5 = cos(2π/5) +
i sin(2π/5)
...

√
15
...
Arg(3 + i 3) = , and 3 + i 3 = 12
6
√
√

n √ n


12(cos(π/6) + i sin(π/6)) = 12 cos(nπ/6) + i sin(nπ/6)
(3 + i 3)n =
√ n
sin(nπ/6) = 0 implies (3 + i 3) ∈ R
...

Only 2022 is an integer multiple of 6
...

(3 + i 3)2022 = 12
16
...
What is the value of
1
1
i) −1
ii) √
iii) − √
iv) 1
2
2
Solution
...
Let z = (1+2i)(2+3i)(3+4i)(4+5i)
(4−3i)(5−4i)
i) 65
ii) 65 + 45i
iii) 45
iv) 45 + 65i
Solution
...
Suppose z = 3 + i 3, then Arg(¯
z 3 ) is :
π
π
i) −
ii) −π
iii) π
iv)
2
2
√
√


π
Solution
...
Thus, Arg(¯
19
...
eiθ ei3θ · · ·ei(2n−1)θ = ei(1+3+
...
Therefore θ = 2kπ
n2
20
...

ez = ex+iy = ex eiy = ex (cos y + i sin y) = 1 ⇒ ex cos y = 1

and ex sin y = 0
...
That is, z = 2miπ, m ∈ Z and the equation has countless solutions
...
com



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