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 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
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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

.

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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.
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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
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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
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