#
**Complex Number**

Algebra of a Complex Number
Properties
Polar Form of Complex Number
De Moivre's theorem
Cube Roots of Unity
Solved Examples

## Introduction

Consider a simple quadratic equation x

^{2} + 1 = 0. There is no real number which satisfies
this equation. So there was a need to find a system which could answer to this problem.

** Euler** used the symbol

**'i'** to denote √-1 to solve the above equation.

Complex number system consists of the set of all ordered pairs of real numbers (a, b) denoted by a + ib,
where i = √-1

## Powers of i

i = √-1

i

^{2} = (√-1)

^{2}
i

^{3} = i

^{2} × i = – 1 × i = – i

i

^{4} = (i

^{2})

^{2} = (– 1)

^{2} = 1, etc.

## Conjugate complex numbers

If the complex numbers differ only in the sign of their imaginary parts, then they are called

**Conjugate Complex Numbers** .
Thus, a + ib and a – ib are conjugate complex numbers.
The conjugate of a complex number z is denoted by

"The sum, difference, product and quotient of two complex numbers is a complex number."

Let two complex numbers z

_{1} and z

_{2} be such that;

z

_{1} = a + ib and z

_{2} = c + id

**(a)** **Sum**
z

_{1} + z

_{2} = (a + ib) + (c + id)

= (a + c) + i (b + d)

Which is a complex number.

**(b)** **Difference**
z

_{1} – z

_{2} = (a + ib) – (c + id)

= (a – c) + i (b – d)

Which is a complex number.

**(c)** **Product**
z

_{1} × z

_{2} = (a + ib) (c + id)

= ac + iad + ibc + i

^{2}
= ac + i (ad + bc) – bd [ i

^{2} = – 1]

= (ac – bd) + i (ad + bc).

Which is a complex number.

**(d)** **Quotient**
C denotes is the set of complex numbers .

Let z

_{1}, z

_{2}, z

_{3}, be three complex numbers.

Addition is

**closed** in C = z

_{1} + z

_{2} is a complex number.

Addition is

**commutative** in C

i.e., z

_{1} + z

_{2} = z

_{2} + z

_{1}
Addition is

**associative** in C

i.e., z

_{1} + (z

_{2} + z

_{3}) = (z

_{1} + z

_{2}) + z

_{3}
## Properties of Multiplication in C

Multiplication is

**closed** in C

i.e., z

_{1} × z

_{2} is a complex number.

Multiplication is

**commutative** in C

i.e., z

_{1} × z

_{2} = z

_{2} × z

_{1}
Multiplication is

**associative** in C

i.e., z

_{1}(z

_{2}z

_{3}) = (z

_{1}z

_{2})z

_{3} .

## Division in C

The absolute value of quotient of two complex numbers is the quotient of absolute value of the
complex numbers. i.e

.

.

Let z = x + iy denote the point P(x,y) in the xy plane.

z = r{cosθ + isinθ} is called the polar form of the complex number.

is called the modulus of the complex number z denoted by |z| and θ = tan

^{-1} y/x is called
the

**amplitude** or argument of the complex number z denoted by amp(z) or arg(z).

The value of θ is such that -p < q ≤ p, is called the

**principal value** of the amplitude.

If 'n' is any integer (+ve, –ve, or zero)

Then,

**(cosθ + i sinθ )**^{n} = cosθ^{n} + i sinθ^{n} , where r = 1

Cube roots of unity are :

The second root is denoted by 'ω' ('ω' means omega )

The third root is denoted by ' ω

^{2 } '

##
Properties of Cube Roots of Unity

**(1)** The

**sum** of the cube roots of unity is zero

i.e.

**1 + ω + ω**^{2 } = 0
**(2)** The

**product** of the cube roots of unity is one

i.e.

**1 × ω × ω**^{2 } = 1
**
**

Example : Express

in the form a + ib .

**Solution** :

on rationalizing becomes

**
**

Example : Show that (1 – i)

^{2} = – 2i

**Solution** : Simplifying the L.H.S. of the above equation

(1 – i)

^{2} we get,

(1 – i)

^{2} = (1)

^{2} – 2(1)(i) + (i)

^{2}, [Since (a – b)

^{2} = a

^{2} – 2ab + b

^{2} ]

= 1 – 2i – 1

= – 2i

= RHS

**
Example** : Express the complex number 1 - i√3 in the polar form.

**Solution** : Let 1 - i√3 = z = r(cosθ + isinθ)

on comparing we get, rcosθ = 1 ....

**(1)** and rsinθ = -√3 ...

**(2)**
Squaring and adding (1) and (2) ,we get

r

^{2}cos

^{2}θ + r

^{2}sin

^{2}θ = 1

^{2} + -√3

^{2}
r

^{2} (cos

^{2}θ + sin

^{2}θ) = 1 + 3

r

^{2}(1) = r

^{2} = 4

r = 2

Therefore, 2cosθ = 1 and 2sinθ = -√3

cosθ = ½ and sinθ = -√3/2

If cosθ = ½, θ = 60

^{0} = π/3 or 5π/3

If sinθ = -√3/2, θ = 5π/3

Therefore, 1 - i√3 = z = r(cosθ + isinθ)

**
**

Example : Prove for any real θ , (cosθ + i sinθ) (cosθ – i sinθ) = 1

**Solution** : Consider the L.H.S. of the above equation,

(cosθ + i sinθ) (cosθ – i sinθ)

(a + b) (a – b) = a

^{2} – b

^{2} , we have

(cosθ + i sinθ) (cosθ – i sinθ) = cos

^{2}θ – i

^{2}sin

^{2}θ

= cos

^{2}θ + sin

^{2}θ [ i

^{2} = –1]

= 1

= R.H.S.

**
**

Example : Solve: (cosθ + i sinθ)

^{5} (cosθ – i sinθ)

^{3} .

**Solution** : (cosθ + i sinθ)

^{5} (cosθ – i sinθ)

^{3} = (cosθ + i sinθ)

^{5} (cosθ – i sinθ)

^{-3
} = (cosθ + i sinθ)

^{5-3} = (cosθ + i sinθ)

^{2
}
= cos 2θ + i sin 2θ [By De Moivre’s Theorem]

**
**

Example : Show that: (1 + 5ω

^{2 } + ω ) (1 + 5ω + ω

^{2 } ) (5 + ω + ω

^{2 } ) = 64 .

**Solution** :

LHS = (1 + 5ω

^{2 } + ω ) (1 + 5ω + ω

^{2 } ) (5 + ω + ω

^{2 } )
= (1 + ω + 5ω

^{2 } ) (1 + ω

^{2 } + 5ω) (5 + ω + ω

^{2 } )

since 1 + ω + ω

^{2 } = 0 ;
1 + ω

^{2 } = –ω ;

ω + ω

^{2 } = –1 and 1 + ω = -ω

^{2 }
Thus, (-ω + 5ω

^{2 } ) (–ω + 5) (5+(– 1)) = (4ω

^{2 } ) (4ω) (4)

= 64ω

^{3} = 64 (1) = 64 [ω

^{3} = 1 ]

= RHS