Chapter

COMPLEX NUMBERS AND

QUADRATIC EQUATIONS

W. R. Hamilton

(1805-1865)

vMathematics is the Queen of Sciences and Arithmetic is the Queen of

Mathematics. – GAUSS v

5.1 Introduction

In earlier classes, we have studied linear equations in one

and two variables and quadratic equations in one variable.

We have seen that the equation x

2

+ 1 = 0 has no real

solution as x

2

+ 1 = 0 gives x

2

= – 1 and square of every

real number is non-negative. So, we need to extend the

real number system to a larger system so that we can

find the solution of the equation x

2

= – 1. In fact, the main

objective is to solve the equation ax

2

+ bx + c = 0, where

D = b

2

– 4ac < 0, which is not possible in the system of

real numbers.

5.2 Complex Numbers

Let us denote

1

−

by the symbol i. Then, we have

2

1

i

= −

. This means that i is a

solution of the equation x

2

+ 1 = 0.

A number of the form a + ib, where a and b are real numbers, is defined to be a

complex number. For example, 2 + i3, (– 1) +

3

i

,

1

4

11

i

−

+

are complex numbers.

For the complex number z = a + ib, a is called the real part, denoted by Re z and

b is called the imaginary part denoted by Im z of the complex number z. For example,

if z = 2 + i5, then Re z = 2 and Im z = 5.

Two complex numbers z

1

= a + ib and z

2

= c + id are equal if a = c and b = d.

5

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98 MATHEMATICS

Example 1 If 4x +

i(3x – y) = 3 + i (– 6), where

x and y are real numbers, then find

the values of x and y.

Solution

We have

4x + i (3x – y) = 3 +

i (–6) ... (1)

Equating the real and the imaginary parts of (1), we get

4x = 3, 3

x – y = – 6,

which, on solving simultaneously, give

3

4

x

=

and

33

4

y =

.

5.3 Algebra of Complex Numbers

In this Section, we shall develop the algebra of complex numbers.

5.3.1 Addition of two complex numbers Let z

1

= a + ib and z

2

= c + id be any two

complex numbers. Then, the sum z

1

+ z

2

is defined as follows:

z

1

+ z

2

= (a + c) + i (b + d), which is again a complex number.

For example, (2 + i3) + (– 6 +i5) = (2 – 6) + i (3 + 5) = – 4 + i 8

The addition of complex numbers satisfy the following properties:

(i) The closure law The sum of two complex numbers is a complex

number, i.e., z

1

+ z

2

is a complex number for all complex numbers

z

1

and z

2

.

(ii) The commutative law For any two complex numbers z

1

and z

2

,

z

1

+ z

2

= z

2

+ z

1

(iii) The associative law For any three complex numbers z

1

, z

2

, z

3

,

(z

1

+ z

2

) + z

3

= z

1

+ (z

2

+ z

3

).

(iv) The existence of additive identity There exists the complex number

0 + i 0 (denoted as 0), called the additive identity or the zero complex

number, such that, for every complex number z, z + 0 = z.

(v) The existence of additive inverse To every complex number

z = a + ib, we have the complex number – a + i(– b) (denoted as –

z),

called the additive inverse or negative of z. We observe that z + (–z) = 0

(the additive identity).

5.3.2 Difference of two complex numbers Given any two complex numbers z

1

and

z

2

, the difference z

1

– z

2

is defined as follows:

z

1

– z

2

= z

1

+ (– z

2

).

For example, (6 + 3i) – (2 – i) = (6 + 3i) + (– 2 + i ) = 4 + 4

i

and (2 – i) – (6 + 3i) = (2 – i) + ( – 6 – 3i) = – 4 – 4i

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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99

5.3.3 Multiplication of two complex numbers Let z

1

= a +

ib and z

2

= c + id be any

two complex numbers. Then, the product z

1

z

2

is defined as follows:

z

1

z

2

= (ac

– bd) + i(ad + bc)

For example, (3 + i5) (2 + i6) = (3 × 2 – 5 × 6) + i(3 × 6 + 5 × 2) = – 24 + i28

The multiplication of complex numbers possesses the following properties, which

we state without proofs.

(i) The closure law The product of two complex numbers is a complex number

,

the product z

1

z

2

is a complex number for all complex numbers z

1

and z

2

.

(ii) The commutative law For any two complex numbers z

1

and z

2

,

z

1

z

2

= z

2

z

1

.

(iii) The associative law For any three complex numbers z

1

, z

2

, z

3

,

(z

1

z

2

) z

3

= z

1

(z

2

z

3

).

(iv) The existence of multiplicative identity There exists the complex number

1 +

i 0 (denoted as 1), called the multiplicative identity such that z.1 =

z,

for every complex number z.

(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ≠ 0, b ≠ 0), we have the complex number

2 2 2 2

a –b

i

a b a b

+

+ +

(denoted by

1

z

or z

–1

), called the multiplicative inverse

of z such that

1

1

z.

z

=

(the multiplicative identity).

(vi) The distributive law For any three complex numbers z

1

, z

2

, z

3

,

(a) z

1

(z

2

+ z

3

) = z

1

z

2

+ z

1

z

3

(b) (z

1

+ z

2

) z

3

= z

1

z

3

+ z

2

z

3

5.3.4 Division of two complex numbers Given any two complex numbers z

1

and z

2

,

where

2

0

z

≠

, the quotient

1

2

z

z

is defined by

1

1

2 2

1

z

z

z z

=

For example, let z

1

= 6 + 3i and z

2

= 2 – i

Then

1

2

1

(6 3 )

2

z

i

z i

= + ×

−

=

(

)

6 3

i

+

( )

( )

( )

2 2

2 2

1

2

2 1 2 1

i

− −

+

+ − + −

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100 MATHEMATICS

=

( )

2

6 3

5

i

i

+

+

=

( ) (

)

1 1

12 3 6 6 9 12

5 5

i i

− + + = +

5.3.5 Power of i we know that

(

)

3 2

1

i i i i i

= = − = −

,

(

)

( )

2

2

4 2

1 1

i i

= = − =

(

)

( )

2

2

5 2

1

i i i i i

= = − =

,

(

)

( )

3

3

6 2

1 1

i i

= = − = −

, etc.

Also, we have

1 2

2

1 1 1

, 1,

1 1

i i

i i i

i i

i

− −

= × = = − = = = −

− −

3 4

3 4

1 1 1 1

, 1

1 1

i i

i i i

i i

i i

− −

= = × = = = = =

−

In general, for any integer k, i

4k

= 1, i

4k + 1

= i, i

4k + 2

= –1, i

4k + 3

= – i

5.3.6 The square roots of a negative real number

Note that i

2

= –1 and ( – i)

2

= i

2

= – 1

Therefore, the square roots of – 1 are i, – i. However, by the symbol

1

−

, we would

mean i only.

Now, we can see that i and –i both are the solutions of the equation x

2

+ 1 = 0 or

x

2

= –1.

Similarly

(

)

(

)

2 2

3 3

i =

i

2

= 3 (– 1) = – 3

(

)

2

3

i

−

=

(

)

2

3

−

i

2

= – 3

Therefore, the square roots of –3 are

3

i

and

3

i

−

.

Again, the symbol

3

−

is meant to represent

3

i

only, i.e.,

3

−

=

3

i

.

Generally, if a is a positive real number,

a

−

=

1

a

−

=

a i

,

We already know that

a b

×

=

ab

for all positive real number a and b. This

result also holds true when either a > 0, b < 0 or a < 0, b > 0. What if a < 0,

b < 0?

Let us examine.

Note that

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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101

( ) ( )

2

1 1 1 1

i

= − − = − −

(by assuming

a b

×

=

ab

for all real numbers)

=

1

= 1, which is a contradiction to the fact that

= −

2

1

i

.

Therefore,

a b ab

× ≠

if both a and b are negative real numbers.

Further, if any of a and b is zero, then, clearly,

a b ab

× =

= 0.

5.3.7 Identities We prove the following identity

( )

2

2 2

1 2 1 2 1 2

2

z z z z z z

+ = + +

, for all complex numbers z

1

and z

2

.

Proof We have, (z

1

+ z

2

)

2

= (z

1

+ z

2

) (z

1

+

z

2

),

= (z

1

+ z

2

) z

1

+ (z

1

+ z

2

) z

2

(Distributive law)

=

2 2

1 2 1 1 2 2

z z z z z z

+ + +

(Distributive law)

=

2 2

1 1 2 1 2 2

z z z z z z

+ + +

(Commutative law of multiplication)

=

2 2

1 1 2 2

2

z z z z

+ +

Similarly, we can prove the following identities:

(i)

( )

2

2 2

1 2 1 1 2 2

2

z z z z z z

− = − +

(ii)

( )

3

3 2 2 3

1 2 1 1 2 1 2 2

3 3

z z z z z z z z

+ = + + +

(iii)

( )

3

3 2 2 3

1 2 1 1 2 1 2 2

3 3

z z z z z z z z

− = − + −

(iv)

(

)

(

)

2 2

1 2 1 2 1 2

z – z z z z – z

= +

In fact, many other identities which are true for all real numbers, can be proved

to be true for all complex numbers.

Example 2 Express the following in the form of a + bi:

(i)

( )

1

5

8

i i

−

(ii)

(

)

(

)

2

i i

−

3

1

8

i

−

Solution (i)

( )

1

5

8

i i

−

=

2

5

8

i

−

=

(

)

5

1

8

−

−

=

5

8

=

5

0

8

i

+

(ii)

( ) ( )

3

1

2

8

i i i

− −

=

5

1

2

8 8 8

i

× ×

× ×

=

( )

2

2

1

256

i

1

256

i i

=

.

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102 MATHEMATICS

Example 3 Express (5 – 3

i)

3

in the form a + ib.

Solution We have, (5 – 3i)

3

= 5

3

– 3 × 5

2

× (3i) + 3 × 5 (3i)

2

– (3i)

3

= 125 – 225

i – 135 + 27

i = – 10 – 198i.

Example 4

Express

(

)

(

)

3 2 2 3

i

− + − −

in the form of a + ib

Solution We have,

(

)

(

)

3 2 2 3

i

− + − −

=

(

)

(

)

3 2 2 3

i i

− + −

=

2

6 3 2 6 2

i i i

− + + −

=

(

)

(

)

6 2 3 1 2 2

i

− + + +

5.4 The Modulus and the Conjugate of a Complex Number

Let z = a + ib be a complex number. Then, the modulus of z, denoted by | z |, is defined

to be the non-negative real number

2 2

a b

+

, i.e., | z | =

2 2

a b

+

and the conjugate

of z, denoted as

z

, is the complex number a – ib, i.e.,

z

= a – ib.

For example,

2 2

3 3 1 10

i+ = + =

,

2 2

2 5 2 ( 5) 29

i− = + − =

,

and

3 3

i i

+ = −

,

2 5 2 5

i i

− = +

,

3 5

i

− −

= 3i – 5

Observe that the multiplicative inverse of the non-zero complex number z is

given by

z

–1

=

1

a ib

+

=

2 2 2 2

a b

i

a b a b

−

+

+ +

=

2 2

a ib

a b

−

+

=

2

z

z

or z

2

z z

=

Furthermore, the following results can easily be derived.

For any two compex numbers z

1

and z

2

, we have

(i)

1 2 1 2

z z z z

=

(ii)

1

1

2 2

z

z

z z

=

provided

2

0

z

≠

(iii)

1 2 1 2

z z z z

=

(iv)

1 2 1 2

z z z z

± = ±

(v)

1 1

2 2

z z

z z

=

provided z

2

≠ 0.

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COMPLEX NUMBERS AND QUADRATIC EQUATIONS 103

Example 5 Find the multiplicative inverse of 2 – 3i.

Solution Let z = 2 – 3

i

Then

z

= 2 + 3i and

2

2 2

2 ( 3) 13

z

= + − =

Therefore, the multiplicative inverse of

2 3

i

−

is given by

z

–1

2

2 3 2 3

13 13 13

z i

i

z

+

= = = +

The above working can be reproduced in the following manner also,

z

–1

=

1 2 3

2 3 (2 3 )(2 3 )

i

i i i

+

=

− − +

=

2 2

2 3 2 3 2 3

13 13 13

2 (3 )

i i

i

i

+ +

= = +

−

Example 6 Express the following in the form a + ib

(i)

5 2

1 2

i

i

+

−

(ii) i

–35

Solution (i) We have,

5 2 5 2 1 2

1 2 1 2 1 2

i i i

i i i

+ + +

= ×

− − +

( )

2

5 5 2 2 2

1 2

i i

i

+ + −

=

−

=

3 6 2 3(1 2 2 )

1 2 3

i i

+ +

=

+

=

1 2 2

i

+

.

(ii)

( )

35

35 17

2

1 1 1

i

i

i i

i

i i

−

= = = ×

−

=

2

i

i

i

=

−

EXERCISE 5.1

Express each of the complex number given in the Exercises 1 to 10 in the

form a + ib.

1.

( )

3

5

5

i i

−

2.

i

i

9

19

+

3.

i

−

39

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