Session 1

Further Algebra of Complex Numbers

Contents
Introduction, p3
1.1 Historical examples, p3
1.2 Basic definitions, p6
1.3 Modulus of a complex number, p9
1.4 The polar form of a complex number, p10
1.5 Argument and principle argument of a complex number, p10
1.6 Theorems, p13
Summary, p25
Objectives, p26

Introduction

The student who has followed the level 2 has learned the basis about the
complex numbers. In this session we discuss about the complex numbers in
detail.

This session contains basic definitions of complex number system. Some

problems are solved using algebraic methods.

1.1 Historical examples

Example 1.1.1

One of the first problems in which the square root of a negative

number occurred was posed by the algebraist Diophantus of
Alexandria.

He was trying to find the dimensions of a right angled triangle which

would have the area 7 units and the perimeter 12 units. If x , y are
the lengths of the two sides adjacent to the right angle then the
hypotenuse has length x 2 + y2

3
 2 2
x+y
x
area = 7
perimeter =
12

Figure 1.1.1.1

Solution

2 2
x+y
x

A B
y

Figure 1.1.1.2
1
7 = xy
2
14
\y =
x

Given that

x 2 + y 2 + x + y = 12
x 2 + y2 = 12 - x - y
12x + 12y - xy - 72 = 0 ------- (1)
14
12x + 12. - 14 - 72 = 0
x
6x - 43x + 84 = 0
2

43 ± -167
\x =
12

The quadratic equation has no real solutions.

There is no right-angled triangle such that the area and perimeter are
7 and 12. The solutions of the above equation are imaginary
numbers.

4
Example 1.1.2

Find x , y numbers such that the sum and product 10 and 40. This
problem was tried to solve by the Italian mathematician and doctor
Gerolamo Cardano in 1545.
Solution

xy = 40 ------ (1)
x + y = 10 ---- (2)
\ x satisfy x 2 - 10x + 40 = 0
( x - 5) = -15
2

x = 5 ± -15
x = 5 ± -15 and y = 5 ± -15
x + y = +5 + -15 + 5 - -15
x + y = 10

( )
2
xy = 52 - -15
xy = 40 ----------- (2)

Introducing i
The symbol i to represent -1 was first introduced by the famous
mathematician Euler.
-15 = ( -1)15
= -1 15
= i 15
-15 = ( -1) 15 = i 15

1.1.1 Note
i 2 = -1 , i3 = -i
i4 = 1
when n Î z
\ i 4n = 1
i 4n +1 = i
i 4n + 2 = -1
i 4n +3 = -i

5
S.A.Q: 1.1

Solve the following equations. Express roots of the form x + iy when

x , yÎR
1. z 2 + 2z + 2 = 0 7. 5z 2 + 4z + 1 = 0
2. z2 - z + 1 = 0 8. 5z 2 + 36 = 0
3. z = 16
4
9. z2 + z + 1 = 0
5. z2 - z - 2 = 0 10. z 4 + 13z 2 + 36 = 0

1.2 Basic definitions

Complex number
When x , y Î R the number x + iy where i 2 = -1 is defined as a complex
number.
If z = x + iy
x is called the real part of z and it is denoted by Re(z).
y is called the imaginary part of z and it is denoted by Im(z).

Complex conjugate
If z = x + iy where x , y Î R the conjugate of z is defined as x – iy. It is
denoted by z or z *
Let z1 = x1 + iy1 , z 2 = x 2 + iy 2 where i 2 = -1
x1 , x 2 , y1 , y2 Î R
Equality of two complex numbers
z1 = z 2 Û x1 = x 2 , y1 = y2
That is Im ( z1 ) = Im ( z 2 ) , Re ( z1 ) = Re ( z 2 )
Addition of two complex numbers
z1 + z 2 = ( x1 + x 2 ) + i ( y1 + y 2 )

Example 1.2.1

(i) ( 4 + 5i ) + ( 3 - i ) = ( 4 + 3) + ( 5 - 1) i
= 7 + 4i

(ii) ( 4 + 5i ) - ( 3 - i ) = éë 4 + ( -3)ùû + ( 5 + 1) i
= 1 + 6i
(iii) ( 5 + 7i ) (
+ 3 - 4i ) - ( 6 - 3i )
= (5 + 3 – 6) + (7 – 4 + 3)i
= 2 + 6i

6
Multiplication of two complex numbers

z1z 2 = ( x1 + iy1 )( x 2 + iy 2 )
= (x x
1 2 + ix 2 y1 + ix1 y2 + i 2 y1 y2 )
z1z 2 = ( x1x 2 - y1y 2 ) + i ( x 2 y1 + x1y 2 )

Example 1.2.2

(i) (3 + 4i )( 4 - 3i )
= 12 - 12i2 - 9i + 16i
= 12 – 12(-1) + 7i
= 24 + 7i

(ii) (4 – i)(1 + i)
= 4 - i 2 - i + 4i
= 4 –(-1) + 3i
= 5 + 3i

(1 + i ) = (1 + i )(1 + i )
2
(iii)
= 1 + i 2 + 2i
= 1 – 1 + 2i
= 2i

Division of two complex numbers

When z 2 ¹ 0
z1 x1 + iy1
=
z 2 x 2 + iy2

=
( x1 + iy1 )( x 2 - iy 2 )
( x 2 + iy2 )( x 2 - iy2 )

x1x 2 - i 2 y1y 2 + ( x 2 y1 - x1 y 2 ) i
=
x 22 - i 2 y 22

x1x 2 - ( -1) y1y 2 + ( x 2 y1 - x1 y 2 ) i

=
x 22 - ( -1) y 22

z1 ( x1x 2 + y1y 2 ) i ( x 2 y1 - x1 y 2 )
= +
z2 x12 + y 22 x 22 + y 22

7
Example 1.2.3

1 + 2i
1. Find
1- i

1 + 2i (1 + 2i )(1 + i )
=
1- i (1 - i )(1 + i )

1 + 2i 2 + i + 2i
=
1 - (i)
2

1 + 2 ( -1) + 3i
=
1 - ( -1)
-1 + 3i
=
2
1 + 2i 1 3
=- + i
1- i 2 2

4 + 3i ( 4 + 3i )( 4 + 3i )
2. =
4 - 3i ( 4 - 3i )( 4 + 3i )

16 + 9i2 + 24i
=
16 - 9i 2

16 + 9 ( -1) + 24i
=
16 - 9(-1)

7 24
= + i
25 25

S.A.Q: 1.2

Express the following numbers of the form x + iy where x , y Î R

1. (5 – 9i) + (2 – 4i) 2. 3(2 – i) – 4(5 + 2i)

3. i(5 + 8i) 4. i(4 + i) + 3i(i + 3)
æ 1 3 öæ 2 5 ö
5. (2 + 3i)(4 + i) 6. ç - i ÷ç + i ÷
è 2 2 øè 3 3 ø
i i
7. 3i + 8.
2-i 1+ i

8
 2-i 10 - 5i
9. 10.
3 + 5i 8 + 2i
11.
4 - 2i
12.
(1 + i )(1 - 2i )
3 + 5i ( 2 + i )( 4 - 3i )
i (1 - i )( 2 - i )( 3 - i ) (1 + i ) (1 - i )
2 3
13. 14.
2-i ö
2

( 2 + 3i ) æç
1
15. 3 + 6i + 16. ÷
2-i è 1 + 2i ø
(1 + i ) (1 - i ) (1 + 2i ) (1 - 3i )
2 2 2

17. 18.
i (2 + i) (4 + i)
12 - 5i 1- i
19. 20
4 + 3i 3 + 2i

The complex number system is denoted by C.

{
C = x + iy : x , y Î R and i 2 = 1 }

1.3 Modulus (absolute value) of a complex number

Definition

If z = x + iy where x , y Î R x 2 + y2 is defined as the modulus of z

It is denoted by z

Example 1.3.1

Find the modulus of the following complex numbers.

(a) 4+3i
z = 4 2 + 32 = 5

(b) 4 – 5i
z = 42 + ( -5 ) = 41
2

(c) -1 – 3i
z = ( -1) + ( -3) = 10
2 2

9
 1.4 The polar form of a complex number
If z = x + iy where x, y Î R and x ¹ 0 or y ¹ 0
z = x + iy
é x y ù
= x 2 + y2 ê + iú
êë x 2 + y 2 x 2 + y 2 úû
y x
Take sin q = , cos q =
x +y
2 2
x + y2
2

Then cos 2 q + sin 2q = 1 and x 2 + y2 = r > 0

\ z = r ( cos q + i sin q ) and z = r

This form is called the polar form of z.

1.5 Argument and principle argument of a complex

number

If z = r ( cos q + i sin q ) . q is defined as the argument of z.

The argument of z is denoted by Arg(z).
Principle argument of a complex number .
If z = r ( cos q + i sin q ) the value of q such that -p < q £ p ( or 0 £ q<2p ) is
defined as the principle argument of z. The principle argument of z is
denoted by arg z.
If z = x + iy x , y Î R

æyö
tan -1 ç ÷ if x > 0 , y > 0
èxø
æyö
p + tan -1 ç ÷ if x < 0, y > 0
èxø
æyö
tan -1 ç ÷ if x > 0 , y < 0
èxø
æ yö
-p + tan -1 ç ÷ if x < 0 , y < 0
arg(z) = èxø
p
if x = 0 , y > 0
2
p
- if x = 0 , y < 0
2
0 if x > 0 , y = 0
p if x < 0 , y = 0
Undefined if x = 0 , y = 0

10
Example 1.5.1

Find the modulus, argument and principle argument of the following

complex numbers.

æ 1 1 ö
1. z = 1+ i = 2 ç +i ÷
è 2 2ø
æ p pö
= 2 ç cos + i sin ÷
è 4 4ø
p
\ arg ( z ) = z = 2
4
p
\ A rg ( z ) = + 2kp where k Î z
4

æ p pö
2. z = i = ç cos + i sin ÷
è 2 2ø
p
\ arg z = z =1
2
p
A rg z = + 2kp where k Î z
2
æ1 3 ö
3. z = 1 - 3i = 2 çç - i ÷÷
è2 2 ø
é p pù
= 2 ê cos - i sin ú
ë 3 3û
é æ pö æ p öù
= 2 ê cos ç - ÷ + i sin ç - ÷ ú
ë è 3ø è 3 øû
p
\ arg z = - z =2
3
p
\ A rg z = 2kp - where k Î z
3

æ4 3 ö
4. 4 + 3i = 5 ç + i ÷
è5 5 ø
æ3ö
Take q = tan -1 ç ÷
è4ø
= 5 ( cos q + i sin q ) 5
4
æ3ö
\ arg ( z ) = tan ç ÷
-1
z =5 q
è4ø 3
æ3ö
\ A rg ( z ) = 2kp + tan ç ÷ where k Î z
-1

è4ø

11
5. z = -4i
æ p pö
= 4 ç cos - i sin ÷
è 2 2ø
é æ pö æ p öù
z = 4 ê cos ç - ÷ + i sin ç - ÷ ú
ë è 2ø è 2 øû
p
arg ( z ) = - z =4
2
p
A rg ( z ) = 2kp - Where k Î z
2

6. z = -2 + 3i
é 2 3 ù
= 13 ê - + iú 13
ë 13 13 û 3
æ3ö q
Take q = tan -1 ç ÷ 2
è2ø
= 13 [ - cos q + i sin q]
= 13 éëcos ( p - q ) + i sin ( p - q ) ùû
æ3ö
\ arg ( z ) = p - tan -1 ç ÷ z = 13
è2ø
æ3ö
A rg ( z ) = 2kp - tan -1 ç ÷ where k Î z
è2ø

7 z = -3 - 4i
æ 3 4 ö
= 5ç - - i ÷
è 5 5 ø
4
Take q = tan -1
3 5
= 5 ( - cos q - i sin q ) 4
q
= 5 éëcos ( p - q ) - i sin ( p - q ) ùû
3
= 5 éëcos ( q - p ) + i sin ( q - p ) ùû
æ4ö
\ arg ( z ) = tan -1 ç ÷ - p z =5
è3ø
æ4ö
A rg ( z ) = 2kp - p + tan -1 ç ÷ where k Î z
è3ø

8. z = -1 = 1( cos p + i sin p )
\ arg z = p z =1
A rg ( z ) = 2kp + p where k Î z

12
9. z = 4 = 4 ( cos q + i sin q )
\ arg ( z ) = 0 z =4
A rg ( z ) = 2kp, where k Î z

S.A.Q: 1.5

Find the modulus, argument and principle argument of the following

complex numbers.

(1) -5 (2) 3 (3) 4 + 9i (4) 4i (5) -5 + 12i (6) -3 – 4i

(7) - 5i (8) 4 – 9i (9) 4 – 4i (10) – 3 - 4i

1.6 Theorems
Theorem 1.6.1

Let z1 , z 2 , z Î C

1. (i) Re ( z ) =
1
2
(z+z ) (ii) Im ( z ) =
1
(
2i
z-z )
2. z1 + z 2 = z1 + z 2 3. z1z 2 = z1 z 2

æ z1 ö z1
4. ç ÷= where z 2 ¹ 0 5. z = z Û z ÎR
è z2 ø z2

6. (z) = z
Proofs

Let z = x + iy where x , y Î R

1. (i) Im ( z ) = y , Re ( z ) = x
z = x - iy
z + z = x + iy + x - iy
2x = z + z
z+z
x=
2
z+z
Re ( z ) =
2

13
 (ii) z - z = ( x + iy ) - ( x - iy )
= 2iy
\y =
1
2i
(z-z )
Im ( z ) =
1
2i
(
z-z )
2. z1 = x1 + iy1 , z 2 = x 2 + iy 2 . Where x1 , x 2 , y1 , y 2 Î R
z1 = x1 - iy1 , z 2 = x 2 - iy 2
z1 + z 2 = ( x1 + x 2 ) + i ( y1 + y2 )
= ( x1 + x 2 ) - i ( y1 + y2 )
= ( x1 - iy1 ) + ( x 2 - iy2 )
\ z1 + z 2 = z1 + z 2

3. z1z 2 = ( x1 + iy1 )( x 2 + iy 2 )
= ( x1x 2 - y1y 2 ) + i ( x1y 2 + x 2 y1 )
= ( x1x 2 - y1y 2 ) - i ( x1 y2 + x 2 y1 )
= x1x 2 - ix1 y 2 + i 2 y1 y 2 - ix 2 y1
= x1 ( x 2 - iy 2 ) - iy1 ( x 2 - iy 2 )
= ( x1 - iy1 )( x 2 - iy2 )
z1z 2 = z1 z 2

4. Proof for reader as an exercise

Deduction from (3)
Since z 2 ¹ 0
z
z1 = 1 z 2
z2
Taking conjugates both sides
z
z1 = 1 z 2
z2
æz ö
z1 = ç 1 ÷ z 2 (using 3)
è z2 ø
Since z 2 ¹ 0 z 2 ¹ 0
z1 æ z1 ö æz ö z
=ç ÷ \ç 1 ÷ = 1
z2 è z2 ø è z2 ø z2

14
5. z = z Û x + iy = x - iy
2iy = 0
Ûy=0
Û z ÎR

6. ( z ) = x - iy
= x – (- iy)
= x + iy
(z) = z
Theorem 1.6.2

Let z , z1 , z 2 Î C

2
1. zz = z 2. z = -z = z = -z
z1 z
3. z1z 2 = z1 z 2 4. = 1 where z 2 ¹ 0
z2 z2

Proofs

Let z , z1 , z 2 Î C

1. z = x + iy where x , y Î R
z = x - iy
zz = ( x + iy )( x - iy )
= x 2 - i2 y2
= x 2 + y2
2
zz = z

2. z = x + iy , z = x - iy , - z = - x - iy , - z = - x + iy

z = x 2 + y2

z = x2 + ( -y) = x 2 + y2
2

( -x ) + ( -y )
2 2
-z = = x 2 + y2

( -x )
2
-z = + y2 = x 2 + y2

\ z = -z = z = -z

15
3. z1z 2 = z1 z 2
Proof using the basic definitions for the reader
Deduction
z1z 2 = ( z1z 2 )( z1z 2 )
2

= ( z1z 2 ) (z1 z 2 )
= z1z 2 z1 z 2
= ( z z )( z z )
1 1 2 2
2 2
= z1 z 2
= ( z1 z 2 )
2 2
z1z 2
z1z 2 = z1 z 2

z1 z
4. = 1 where z2 ¹ 0
z2 z2

Proof using the basic definitions for the reader

Deduction
Since z 2 ¹ 0
z
z 2 1 = z1
z2

Taking modulus of both sides

z1
z2 = z1
z2
z1
z2 = z1
z2
Since z2 ¹ 0 z2 ¹ 0
z1 z
= 1
z2 z2

Theorem 1.6.3

1. ( Re ( z ) £ z and Im ( z ) £ z
2. z1 + z 2 £ z1 + z 2
3. z1 - z 2 ³ z1 - z 2

16
Proof
Let z , z1 , z 2 Î C

1. z = x + iy where x , y Î R
Im ( z ) = y , Re ( z ) = x
x £ x 2 + y2 , y £ x 2 + y2
Re ( z ) £ z , Im ( z ) £ z

z1 + z 2 = ( z1 + z 2 )( z1 + z 2 )
2
2.
= ( z1 + z 2 ) ( z1 + z 2 )
= z1 z1 + z1 z 2 + z1z 2 + z 2 z 2
2 2
= z1 + z1 z 2 + z1 z 2 + z 2
2 2
= z1 + z1 z 2 + z1 z 2 + z 2
=
2 2
z1 + z 2 + 2 Re z1 z 2 ( )
+ 2 Re ( z z )
2 2
£ z1 + z 2 1 2

2 2
£ z1 + z 2 + 2 z1 z 2
2 2
£ z1 + z 2 + 2 z1 z 2
2 2
£ z1 + z 2 + 2 z1 z 2
z1 + z 2 £ ( z1 + z 2 )
2 2

\ z1 + z 2 £ z1 + z 2

3. z1 , z 2 Î C
z1 - z 2 , z 2 Î C
z1 - z 2 + z 2 £ z1 - z 2 + z 2
z1 £ z1 - z 2 + z 2
z1 - z 2 ³ z1 - z 2 ----------- (1)
z 2 - z1 , z1 Î C
z 2 - z1 + z1 £ z 2 - z1 + z1
z 2 £ z1 - z 2 + z1
\ z1 - z 2 ³ z 2 - z1 = - ( z1 - z 2 )
z1 - z 2 ³ - ( z1 - z 2 ) ----------- (2)
\ from (1) & (2)
z1 - z 2 ³ z1 - z 2

17
Theorem 1.6 .4

zi Î C for i = 1, 2, …n

1. z1 + z 2 + .... + z n = z1 + z 2 + .... + z n
2. z1z 2 ....z n = z1 z 2 .....z n
3. z1z 2 ....z n = z1 z 2 ........ z n
4. Re ( z1 + z 2 + ... + z n ) = Re ( z1 ) + Re ( z 2 ) + ... + Re ( z n )
5. Im ( z1 + z 2 + .... + z n ) = Im ( z1 ) + Im ( z 2 ) + ... + Im ( z n )
6. z1 + z 2 + ... + z n £ z1 + z 2 + ... + z n

Proof

Proof for the theorem 1.6.4 for the reader

Hint ; Use the mathematical induction principle to prove.

1.1.2 Note

z is purely imaginary
Û Re ( z ) = 0
p
Û arg ( z ) = ±
2
Û z+z =0

z is purely real
Û Im ( z ) = 0
Û arg ( z ) = 0 or p
Û z-z =0

Example 1.6.1

1. Find the modulus of (3 + 4i)(5 + 12i)

( 3 + 4i )( 5 + 12i ) = 3 + 4i 5 + 12i
= 32 + 4 2 52 + 122
= 5.13
= 65

5 - 12i
2. Find the modulus of
4 - 3i

18
 5 - 12i 5 - 12i
=
4 - 3i 4 - 3i

52 + ( -12 )
2

=
4 2 + ( -3 )
2

13
=
5
1 1 1
3. If = + a , b , c Î R prove that
z a b + ic
b2 + c 2
z = a
(a + b) + c2
2

1 1 1
= +
z a b + ci
1 b + ci + a
=
z a ( b + ci )
a ( b + ci )
\z =
( a + b ) + ci
a ( b + ci )
z =
( a + b ) + ci
a b + ci
=
a + b + ic
b2 + c 2
= a
(a + b)
2
+ c2
b2 + c 2
= a
(a + b)
2
+ c2

1 1
4. z = 2 prove that £
z - 5z + 1 5
4

z 4 - 5z + 1 = z 4 - ( 5z + 1)
³ z 4 - 5z + 1

³ z - ( 5 z + 1)
4

³ 24 - ( 5.2 + 1) = 5
z 4 - 5z + 1 ³ 5

19
 1 1
£
z - 5z + 1 5
4

1 1
£
z - 5z + 1 5
4

1 + it
5. If t Î R prove that =1
1 - it
1 + it
Let z =
1 - it
1 + it
z =
1 - it
1+ t2
=
1 + ( -t )
2

z =1
1 + it
\ =1
1 - it

6. Prove that z + w
2
- z-w
2
( )
= 4 Re zw

z+w - z-w
2 2

( ) ( )
= (z + w ) z + w - (z - w ) z - w
= (z + w )(z + w ) - (z - w )(z - w )
= zz + zw + zw + ww - zz - w w + zw + zw
(
= 2 zw + zw )
(
= 2 zw + z w )
= 2 ( zw + zw )

(
= 2 2 Re zw ( ))
= 4 Re ( zw )

1 1 1
7. If = + prove that
z a ib
ab
z = where a , b Î R
a + b2
2

20
 1 1 1
= +
z a ib
1 a + ib
=
z abi
abi
z=
a + ib
abi ab i
z = =
a + ib a + ib
ab
=
a 2 + b2

z-a
8. If z , a Î C and = 1 prove that a = 1 or z = 1
1 - az

z-a
=1
1 - az
z - a = 1 - az
2 2
z-a = 1 - az

( )
( z - a ) ( z - a ) = (1 - az ) 1 - az
( z - a ) ( z - a ) = (1 - az ) (1 - az )
( z - a ) ( z - a ) = (1 - az ) (1 - az )
( z - a ) ( z - a ) = (1 - az )(1 - az )
zz - a z - az + aa = 1 - a z - az + a zza
z + a = 1 + aazz
2 2

2 2 2 2
0 = 1- z - a + a z
0 = 1- z - a
2 2
(1 - z ) 2

(
0 = 1- z
2
)(1 - a )
2

\ z = 1 or a = 1
2 2

z = 1 or a =1

9. Prove that z + w
2
+ z-w
2
=2 w + z ( 2 2
)
2 2
z+w + z-w

21
 ( )
= (z + w ) z + w + (z - w ) z - w ( )
= (z + w )(z + w ) + (z - w )(z - w )
= zz + zw + zw + ww + zz - zw - zw + ww
2 2 2 2
= z + w + z + w
= 2 z + w ( 2 2
)
Theorem 1.6.5

If a is a root of a polynomial equation with real coefficients then a is a root

of the polynomial equation.

Proof

Let p ( z ) = a 0 + a1z + a 2 z 2 + ... + a n z n = 0 be a polynomial equations where

a i Î R for i = 0, 1, 2, …n

If a is a root of p(z) = 0. ihen p (a ) = 0

a 0 + a1a + a 2a 2 + .... + a n a n = 0
a 0 + a1a + a 2a 2 + .... + a n a n = 0
a 0 + a1 a + a 2 a 2 + ... + a n a n = 0
Since a i Î R a i = a i for i = 0 , 1 , 2 , 3 , ........n

( )
m
am = a

+ a ( a ) + a (a ) ( )
2 n
a0 1 2 + ... + a n a =0

\ p (a) = 0
\a is a root of p(z) = 0

Example 1.6.2

Prove that -1 + i is a root of z 4 + 6z3 + 15z 2 + 18z + 10 = 0 Hence find the

other roots of the above equation.

Let p ( z ) = z 4 + 6z 3 + 15z 2 + 18z + 10 = 0

p ( -1 + i ) = ( -1 + i ) + 6 ( -1 + i ) + 15 ( -1 + i ) + 18 ( -1 + i ) + 10
4 3 2

( ) ( )
= 1 + i 2 - 2i + 6 ( -1 + i ) 1 + i 2 - 2i +15 1 + i 2 - 2i + 18 ( -1 + i ) + 10 ( )
2

( -2i ) + 6 ( -1 + i )( -2i ) + 15 ( -2i ) + 18 ( -1 + i ) + 10

2
=
= 4i + 12i - 12i 2 - 30i - 18 + 18i + 10
2

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 = -4 + 12 – 18 + 10
= 0
\ -1 + i is a root of p(z) = 0
Since the coefficients of p(z) = 0 are real -1 + i is also a root is a root of
p(z) = 0 i.e. -1 – i
\ ( z + 1 - i )( z + 1 + i ) is a factor of p(z)
= ( z + 1) - i 2
2

= z 2 + 2z + 2
\ p ( z ) = ( z 2 + 2z + 2 )( z 2 + Az + 5 ) where A is a constant
Equating the coefficients of z3
6=A+2 \A=4
p(z) = 0
( z2 + 2z + 2 )( z2 + 4z + 5 ) = 0
é( z + 1)2 + 1ù é( z + 2 )2 + 1ù = 0
ë ûë û
( z + 1 - i )( z + 1 + i )( z + 2 - i )( z + 2 + i ) = 0
\ The roots of p(z) = 0 are

-1 + i, -1 – i, -2 + i, -2 – i

Example 1.6.3

Prove that 2 + 3i is a root of 2z 4 - 11z 3 + 39z 2 - 43z + 13 = 0 .

Find the other roots of the equation.
Solution

Let p ( z ) = 2z 4 - 11z3 + 39z 2 - 43z + 13 = 0

p ( 2 + 3i ) = 2 ( 2 + 3i ) - 11( 2 + 3i ) + 39 ( 2 + 3i ) - 43 ( 2 + 3i ) + 13
4 3 2

( ) (
= 2 4 + 9i 2 + 12i - 11( 2 + 3i ) 4 + 9i 2 + 12i )
2

( )
+39 4 + 9i + 12i - 43 ( 2 + 3i ) + 13
2

= 2(-5 + 12i) - 11(2 + 3i)( -5 + 12i ) + 39(-5 + 12i)

2

-86 - 129i + 13
( ) (
= 2 25 + 144i - 120i - 11 -10 + 9i + 36i2
2
)
-195 + 468i - 73 - 129i
= 2 ( -119 - 120i ) - 11( -46 + 9i ) - 268 + 339i
= -238 - 240i + 506 - 99i - 268 + 339i
= 506 – 339i + 506 + 339i
p ( 2 + 3i ) = 0
\ 2 + 3i is a root of p(z) = 0

Since p(z) = 0 has real coefficients

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2 + 3i is a root of p(z) = 0
2 + 3i = 2 - 3i
\ ( z - 2 + 3i )( z - 2 - 3i ) is a factor of p(z)
( z - 2 + 3i )( z - 2 - 3i ) = ( z - 2 )
2
+9
= z - 4z + 13
2

p ( z ) = ( z - 4z + 13)( 2z 2 + Az + 1)
2

Equating the coefficients of z3

-11 = -8+A \ A = -3
2z 2 - 3z + 1 = 0
( 2z - 1)( z - 1) = 0
\ The roots are
1
, 1 , 2 + 3i , 2 - 3i
2

S.A.Q: 1.6

1. Using the algebraic methods prove that

2 2
z + w = z 2 + w + 2 Re zw ( )
Deduce that
z+w+u
2 2 2
= z + w + u + 2 Re zw + wu + uz
2
( )
2. (i) If a, b Î R and z, w Î C prove that
2
az + bw + bz - aw
2
(
= a 2 + b2 )( z 2
+ w
2
)
2 2 2 z1
(ii) If z1 + z 2 = z1 + z 2 prove that is purely imaginary.
z2

3. If A Î C , B Î R and z 2 + Re ( zA ) + B = 0 has a solution prove

2
that A ³ 4B and solutions lie on a circle or a point.

4. (i) (
Prove that Re z1 z 2 = Re z1 z 2 ) ( )
(ii) Prove that z - a < 1 - az Û a < 1 and z < 1

5. az 2 + bz + c = 0 has real roots if and only if b 2 - 4ac ³ 0. whether

the above statement is true or false justifies your answer.

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6. Prove that 2 + 3i is a root of 2z 3 - 9z 2 + 30z - 13 = 0 . Hence solve
the equation.

7. Prove that 2i is a root of z 4 + 3z 3 + 12z - 16 = 0 . Find the other roots

of the equation.

8. Prove that one root of the equation z 4 - 11z 3 + 27z 2 - 25z + 7 = 0 is

2 - i 3 . Find the other roots of the equation.

9. Given that 4 + i is a root of the equation z3 + az 2 + 33z + b = 0 .

Where a, b Î R . Find the values of a , b and solve the equation.

10. Given that 1 – 2i is a root of the equation 2z 3 + az 2 + bz - 5 = 0. If

a, b Î R find the values of a,b and solve the equation.

Summary

Complex number system

{
C = x + iy : x , y Î R , i 2 = -1 }
Let z , z1 , z 2 Î C
z1 = x1 + iy1 , z 2 = x 2 + iy2 , z = x + iy
Where x1 , x 2 , y1 , y 2 , x, y Î R

Equality
z1 = z 2 Û x1 = x 2 , y1 = y2

Addition
z1 + z 2 = ( x1 + x 2 ) + i ( y1 + y 2 )

Multiplication

z1z 2 = ( x1x 2 - y1y 2 ) + i ( x1y 2 + x 2 y1 )

Division
z1 ( x1x 2 + y1y 2 ) i ( x 2 y1 - x1 y 2 )
= +
z2 x 22 + y 22 x 22 + y 22

Modulus
z = x 2 + y2

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 z1 z
z1z 2 = z1 z 2 , = 1
z2 z2

Conjugate
z = x - iy z=z

z1 + z 2 = z1 + z 2 z1z 2 = z1 z 2

æ z1 ö z1
ç ÷=
è z2 ø z2

Argument and principle argument

z = r ( cos q + i sin q ) r > 0

q = Arg ( z )
-p < q £ p then arg z = q

Theorems

Objectives

At the end of the session the student should have a thorough understanding
the algebra of complex numbers.

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