Complex Numbers

What is Complex Numbers, Imaginary unit, Conjugate Complex Number, Modulus and Argument, Square root of Complex Numbers, Cube root of Complex Numbers:

Imaginary number or imaginary unit = √-1 = i

∴ i² = -1,   i³= i² × i = -1 × i = -i,   i⁴ = 1,   i⁵ = i etc.

i^{4n + 1} = i,   i^{4n + 2} = -1,   i^{4n + 3} = -i,   i^{4n + 1} = i etc. where, n ϵ Z     [in general i⁴ⁿ = 1]

Note:-

        √-a × √-b = √a × √-1 × √b × √-1 = √ab × √(-1²) = -√ab

But, √-a × √-b = √(-a × -b) = √ab is not valid.

Complex Number: A complex number is a number that can be expressed in the form a + bi, where ‘a’ and ‘b’ are real numbers and ‘i’ is the imaginary unit, that satisfies the equation i² = -1. In this expression, ‘a’ is the real part and ‘b’ is the imaginary part of the complex number and ‘i’ is the positive square root of -1.

Complex Number a + ib is denoted by Z

∴ a + ib = Z

Where, a = Real part, b = imaginary part

If b = 0, complex number is real.

And if a = 0, b ≠ 0, complex number is purely imaginary.

Note:-

Two complex numbers a + ib and c + id are equal if only a = c and b = d.

Conjugate Complex Number:

Complex numbers a + ib and a - ib are said to be conjugate to each other,

Denoted by Z = a + ib and Z̅ = a - ib

Properties of conjugate complex number: If Z₁ and Z₂ be any two complex numbers, then

Example: 01

Reduce (1 + i)³/(2 + 3i) to the form A = iB

Solution:

(1 + i)³/(2 + 3i) = (1 + 3i +3i² + i³)/2 + 3i) = (-2 + 2i)/(2 + 3i)

= (-2 + 2i) (2 - 3i) / (2 + 3i) (2 - 3i) = (2 + 10i) / (4 + 9) = (2 + 10i)/13

=> 2/13 + i(10/13) = A + iB form

Where, A = 2/13 and B = 10/13.

Some Important Properties:

If Z₁ = a + ib and Z₂ = c + id

Then, (i) Z₁ + Z₂ = (a + c) + i(b + d)

           (ii) Z₁ + Z₂ = (a - c) + i(b - d)

           (iii) Z₁. Z₂ = (ac - bd) + (ad + bc)

           (iv) Z₁/Z₂ = a + ib)/(c + id) = (a + ib) (c - id) / (c + id) (c - id)   [Since, (a + b) (a - b) = a² - b²]

Modulus and Argument (Amplitude):

If Z = x + iy, then modulus of Z, |Z| = √(x² + y²)

And = x + iy = r (cosθ + i sinθ)    [here, θ is called argument]

∴ |Z| = √(x² + y²) = r

And, arg. Z or amp. Z = θ

And, tan θ = y/x

Note:-

1. |1| = 1, arg.(1) = 0

2. |-1| = 1, arg. (-1) = π

3. |i| = 1, arg.(i) = π/2

4. |-i| = 1, arg.(-i) = -π/2

Example: 02

Find mod. and, arg. of

(A) -1 + i

(B) -1 - √3 i

Solution: A

Let, (-1 + i) = r (cosθ + i sinθ)

Then, r cosθ = -1…………..eq.(i)

           r sinθ = 1………..…..eq.(ii)

∴ Squaring both the eq. and adding,

      r² = 2   [since cos²θ + sin²θ = 1]

=> r = √2

On dividing eq.(ii) by eq.(i),

     tanθ = -1

=> θ = tan⁻ -1 = 3π/4

∴ |-1 + i| = r = √2 and, arg.(-1 + i) = θ = 3π/4

Solution: B

Let, (-1 -√3 i) = r (cosθ + i sinθ)

Then, r cosθ = -1……………eq.(i)

           r sinθ = -√3………….eq.(ii)

∴ Squaring both the eq. and adding,

     r² = 4   [since, cos²θ + sin²θ = 1]

=> r = 2

On dividing eq.(ii) by eq.(i),

     tanθ = √

=>   θ = -2π/3

Hence, |-1 - √3 i| = r = 2 and, arg. (-1 - √3 i) = θ = -2π/3

Note: If Z₁ and Z₂ are two complex numbers, then |Z₁ + Z₂| ≤ |Z₁| + |Z₂|

Example: 03

Find the modulus of (3 - 4i) (1 + 7i) / (1 + i) (12 + 5i)

Solution:

   |(3 - 4i) (1 + 7i) / (1 + i) (12 + 5i)|

= |(3 - 4i) (1 + 7i)| / |(1 + i) (12 + 5i)|

= |(3 - 4i)|.| (1 + 7i)| / |(1 + i)|.|(12 + 5i)|

= √(9 + 16). √(1 + 49) / √(1 + 1). √(144 + 25)

= 5. 5√2 / √2. 13

= 25/13

Example: 04

Find the square root of 7 - 30√-2

Solution:

Let, √(7 - 30√-2) = x - iy

Then, (x - iy)² = 7 - 30√-2

=> (x² - y²) - 2ixy = 7 - 30√2i   [since, √-2 = √2i]

=> (x² - y²) = 7…….…eq.(i)

     2xy = 30√2………..…..eq.(ii)

∴ eq.(i)² + eq.(ii)² = (x² - y²)² + (2xy)² = [(x² - y²)² + 4x²y²] = 49 + 1800

=>    (x²)² + (y²)² - 2x²y² + 4x²y² = (x² + y²)² = 49 + 1800

=>    (x² + y²) = √(49 + 1800)

=>    (x² + y²) = 43……………………eq.(iii)

∴ eq.(i) + eq.(iii) => 2x² = 50     => x = ± 5

And, eq.(iii) - eq.(i) = 2y² = 36    => y = ± 3√2

∴√(7 - 30√-2) = x - iy = ± (5 - 3√2)

Hence, the two roots are, (5 - 3√2) and, -(5 - 3√2)

Alternate Method:

(7 - 30√-2) = 25 - 18 - 2. 5. 3√2i   [since, 25 - 18 = 7, and, 2 × 5 × 3√2 = 30√2]

                    = (5 - 3√2i)²

∴ The square root of (7 - 30√-2) = ± (5 - 3√2)

Cube root of Complex Numbers:

Cube roots of unity are, 1, -½ + i √3/2 and -½ - i √3/2 of which, 1 is real and the other are complex (conjugate to each other)

If one of them is denoted by ω and the other is denoted by ω²

∴ -½ + i √3/2 = ω and, -½ - i √3/2 = ω²

Note:-

1. As the sum of cube roots of unity is zero, so, we have, 1 + ω + ω² = 0

2. The product of complex cube roots being unity, we have ω. ω² = 1 i.e. ω³ = 1

3. As a consequence, of which one complex root is the reciprocal of the other,

I.e. ω = 1/ ω²,    ω² = 1/ ω

As, ω³ = 1,   ω³ⁿ = 1,   ω^{3n + 1} = ω,    ω^{3n + 2} = ω² etc….

Example: 04

Find, [(√3 + i)/2]⁶ + [(i - √3)/2]⁶

Solution:

(√3 + i)/2 = (i √3 + i²)/2i = -i [(-1 + √3)/2] = -iω

(i - √3)/2 = (i² - √3 i)/2i= -i [(-1 - √3 i)/2] = -iω²

∴ [(√3 + i)/2]⁶ + [(i - √3)/2]⁶  = (-iω)⁶ + (-iω²)⁶  

                                                   = i⁶ω⁶ + i⁶ω¹²

                                                   = (-1) × (1) + (-1) × (1) = -2