How to Use De Moivre’s Theorem to simplify (-2 + 2i)8?

Complex numbers are the combination of real and imaginary values. They are expressed in the form of x + iy where x and y are real numbers and i is the imaginary part also called iota. It is often represented by z. The value ‘x’ is called the real part, denoted by Re(z) and the value ‘y’ is called the imaginary part which is denoted by Im(z). The complex numbers are plotted on a plane called the Argand plane or a Complex plane where the x-axis is the real axis and y is the imaginary axis. 

Real and Imaginary numbers

Real Numbers are those whose square gives a positive result. They can be positive, negative, etc. It is represented by Re(). Imaginary Numbers are those numbers whose square gives a negative value. They are denoted by Im(). The imaginary numbers are of the form ‘bi’ where i is the iota and b is the real number. 

Example: z = 1 + 2i. Here in the above example, it is of the form a + ib where a = 1 and b = 2 which are real numbers.

• Re(z) = 1
• Im(z) = 2

More About Iota

An imaginary number is denoted by iota ‘i’. It is used to find the square root of negative numbers. Value of i = √(-1). If the square operation of i is performed,

i² = i × i = -1

i⁴ = 1

Operations on Complex numbers

In complex numbers, can perform addition, subtraction, multiplication, division, and conjugation. The operations are done separately for real values and imaginary values. 

• Addition: Addition of complex numbers is performed by adding the real parts and imaginary parts separately. Let’s assume two complex numbers x + ib and c + id. Therefore we get

x + ib + c + id = (x + c) + i(b + d)

• Subtraction: Subtraction of complex numbers is performed by subtracting the real and imaginary parts separately. Let’s assume two complex numbers a + ib and c + id. 

a + ib – (c + id ) = (a – c) + i(b – d)

• Multiplication: When two complex numbers say z₁ and z₂ are multiplied, the real part of z₁ is multiplied with both the real and imaginary parts of z₂ and similarly, it is done for the imaginary part as well. Let’s assume two complex numbers a + ib and c + id. 

(a + ib) × (c + id) = (ac – bd) + i(ad + bc)

• Conjugate: Let’s take a complex number z. The conjugate is found by changing the sign of the imaginary part of the complex number that means changing of + to – and – to +. Let’s assume one complex number a + ib.

conjugate(a + ib) = (a – ib)

• Division: When the division of two complex numbers z₁ and z₂ are performed we multiply the denominator z₂ by its conjugate and perform the division. Let’s assume two complex numbers a + ib and c + id. 

(a + ib)/(c + id) = {(a + ib) (c – id)}/(c² + d²) 

Modulus and Argument of A Complex Number

Modulus of a complex number is represented by |z| or ‘r’ which is the distance of the point z from the origin in the Argand plane or the complex plane. The numerical value of z = x + yi is given by . The argument of z often represented by arg(z) is the angle that the line joining z and the origin makes with the positive direction of the real axis. 

Arg(z) = Arg(x + iy)  = tan^-1(y/x) 

For example, To find modulus and argument of 1 + 2i

Let z = 1 + 2i

Modulus of z = = √5

Here, y = 2, x = 1

Arg(z) = tan^-1(2/1) 

Property of argument is Arg(zⁿ) = n Arg(z) 

Expressing Complex number in polar form

Complex numbers can be expressed in polar form and therefore they can be plotted on a complex plane with the x-axis as the real axis and the y-axis as the imaginary axis. 

Let x + iy be complex number. Therefore x = r cosθ, y = rsinθ,  r = 

z = r(cosθ + isinθ) 

De Moivre’s theorem

This theorem is one of the most useful theorems as it helps to establish a relationship between trigonometry and complex numbers. It helps to calculate the value of complex numbers in the polar form up to n times. The theorem states that for any real number x,

(cosx + isinx)ⁿ = cos(nx) + isin(nx) 

Where n is a positive integer and i is the imaginary part. 

How to Use De Moivre’s Theorem to simplify (-2 + 2i)⁸?

Solution: 

Let z = -2 + 2i

Arg(z) = tan^-1 (2/-2) = 3π/4

Absolute value =  = 2√2

Applying De Moivre’s Theorem,

z⁸ = [2√2{cos(3π/4) + isin(3π/4)}]⁸

= (2√2)⁸ [cos(24π/4) + isin(24π/4)]

= 4096(cos 6π + i sin 6π)

= 4096 (1 + 0i)

= 4096 (1)

= 4096

Similar Problems

Question 1: Write the formula for zⁿ where z = r(cos x + isin x) 

Solution: 

zⁿ = [r(cos x + isin x)]ⁿ

Using De Moivre’s formula,

zⁿ = rⁿ[cos nx + isin nx] 

Question 2: Let z = 2[cos (π/6) + isin (π/6)] find z³

Solution:

n = 3

Using De Moivre’s formula,

z³ = (2[ cos π/6 + isin π/6])³

= 2³[ cos 3π/6 + isin 3π/6]

= 8 [cos π/2 + isin π/2]

= 8(0 + i) 

= 8i

Question 3: Find value of (1 + i)² using De Moivre’s formula.

Solution:

Let z = 1 + i

Arg(z) = tan^-1 (1) = π/4

|z| =  = √2

So, z = √2(cos π/4 + i sin π/4) 

z² = (√2(cos π/4 + i sin π/4))²

= (√2)² [cos 2π/4 + i sin π/4]

= 2 [cos π/2 + i sin π/2]

= 2[0 + i]

= 2i

Question 4: Find the value of (√3 + i)⁴.

Solution: 

|z| =  = 2

Arg(z) = π/6

z⁴ = [2( cos π/6 + i sin π/6)]⁴

Using De Moivre’s formula,

z⁴ = 16(cos 4π/6 + i sin 4π/6)

= 16(cos 2π/3 + i sin 2π/3)

= 16(cos (π – π/3) + i sin (π – π/3))

= 16(-cos (π/3) + i sin (π/3))

= 16(-1/2 + i√3/2) 

= 8(-1 + i√3) 

Question 5: Write the formula of De Moivre’s theorem for z¹⁰⁰⁰ where z = r(cos θ + isin θ) 

Solution:

z¹⁰⁰⁰ = r¹⁰⁰⁰(cos θ + isin θ)¹⁰⁰⁰

= r¹⁰⁰⁰( cos 1000 θ + isin × 1000 × θ)