Complex Number

Consider a simple quadratic equation x² + 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
i² = (√-1)²
i³ = i² × i = – 1 × i = – i
i⁴ = (i²)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

Algebra of a Complex Number

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

Let two complex numbers z₁ and z₂ be such that;
z₁ = a + ib and z₂ = c + id

(a) Sum
z₁ + z₂ = (a + ib) + (c + id)
= (a + c) + i (b + d)
Which is a complex number.

(b) Difference
z₁ – z₂ = (a + ib) – (c + id)
= (a – c) + i (b – d)
Which is a complex number.

(c) Product
z₁ × z₂ = (a + ib) (c + id)
= ac + iad + ibc + i²
= ac + i (ad + bc) – bd [ i² = – 1]
= (ac – bd) + i (ad + bc).
Which is a complex number.

(d) Quotient

Top

Properties of addition in C

C denotes is the set of complex numbers .

Let z₁, z₂, z₃, be three complex numbers.
Addition is closed in C = z₁ + z₂ is a complex number.

Addition is commutative in C
i.e., z₁ + z₂ =  z₂ + z₁

Addition is associative in C
i.e., z₁ + (z₂ + z₃) = (z₁ + z₂) + z₃

Properties of Multiplication in C

Multiplication is closed in C
i.e., z₁ × z₂ is a complex number.

Multiplication is commutative in C
i.e., z₁ × z₂ = z₂ × z₁

Multiplication is associative in C
i.e.,  z₁(z₂z₃) = (z₁z₂)z₃ .

Division in C

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

.

Polar Form of Complex Number

.










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.

Top

De Moivre's theorem

If 'n' is any integer (+ve, –ve, or zero)
Then, (cosθ + i sinθ )ⁿ = cosθⁿ + i sinθⁿ , where r = 1

Cube Roots of Unity

Cube roots of unity are :




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



The third root is denoted by ' ω² '   

Properties of Cube Roots of Unity

(1) The sum of the cube roots of unity is zero
i.e. 1 + ω + ω² = 0

(2) The product of the cube roots of unity is one
i.e. 1 × ω × ω² = 1

Top

Solved Examples

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 = a² – 2ab + b² ]
= 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
r²cos²θ + r²sin²θ = 1² + -√3²
r² (cos²θ + sin²θ) = 1 + 3
r²(1) = r² = 4
r = 2

Therefore, 2cosθ = 1 and 2sinθ = -√3
cosθ = ½ and sinθ = -√3/2
If cosθ = ½, θ = 60⁰ = 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) = a² – b² , we have
(cosθ + i sinθ) (cosθ – i sinθ) = cos²θ – i²sin²θ

= cos²θ + sin²θ [ i² = –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 + 5ω² + ω ) (1 + 5ω + ω² ) (5 + ω + ω² ) = 64 .

Solution :
LHS = (1 + 5ω² + ω ) (1 + 5ω + ω² ) (5 + ω + ω² ) = (1 + ω + 5ω² ) (1 + ω² + 5ω) (5 + ω + ω² )

since 1 + ω + ω² = 0 ; 1 + ω² = –ω ;
ω + ω² = –1 and 1 + ω = -ω²

Thus, (-ω + 5ω² ) (–ω + 5) (5+(– 1)) = (4ω² ) (4ω) (4)
        = 64ω³ = 64 (1) = 64 [ω³ = 1 ]
        = RHS

Top