A complex number is made up of both real and imaginary components. It can be represented by an expression of the form **(a+bi)**, where **a** and **b** are real numbers and **i** is imaginary. When defining **i** we say that **i = √(-1)**. Along with being able to be represented as a point (a,b) on a graph, a complex number **z = a+bi** can also be represented in polar form as written below:

*z = r (*cos* θ + i *sin*θ)
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where *r = [z] = √(a*^{2} + b^{2})

and

*θ = *tan^{-1}(b/a) or *θ = *arctan*(b/a)*

and we also have: *a = r *cos*θ* and *b = r* sin*θ*

**Statement of DeMoivre's Theorem
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Let 'n' be any rational number, positive or negative, then

[cos* **θ** + i *sin* **θ* ]^{n} = cos *n**θ** + i *sin* n**θ***
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Basically, in order to find the nth power of a complex number we need to take the nth power of the absolute value or length and multiply the argument by n.

Let *z = r (*cos* θ + i *sin*θ) *and n be a positive integer. Then z has n distinct nth roots given by:

, where k = 0, 1, 2, ... , n-1

De Moivre's Theorem states that for any complex number as given below:

z = r ∙ *cos*θ + i ∙ r ∙ *sin*θ

the following statement is true:

z^{n} = r^{n} (*cos*θ + i ∙ *sin*(nθ)), where n is an integer.

If the imaginary part of the complex number is equal to zero or i = 0, we have:

z = r ∙ *cos*θ and z^{n} = r^{n} (*cos*θ)

**Exponential form of complex number:**

*z = re*^{i}^{θ} where *r* is the modulus of *z* and* θ* is its argument.

**Example 1:** Compute the three cube roots of -8.

**Solution:** Since -8 has the polar form 8 (cos π + i sin π), its three cube roots have the form

^{3}√8 {cos[(π + 2πm)/3] + i sin[(π + 2πm)/3]} for m=0, 1, and 2.

Thus the roots are

2 (cos π/3 + i sin π/3) = 1 + √3 i,

2 (cos π + i sin π) = -2, and

2 (cos 5π/3 + i sin 5π/3) = 1 - √3 i.

**Example 2: **Use De Moivre's Theorem to compute (1 + i)^{12}.

__Solution__**: ** The polar form of 1 + i is √2 (cos π/4 + isin π/4). Thus, by De Moivre's Theorem, we have:

(1 + i)^{12} = [√2 (cos π/4 + i sin π/4)]^{12}

= (√2)^{12}(cos π/4 + i sin π/4)^{12}

= 2^{6} (cos 3π + i sin 3π)

= 64(cos π + i sin π)

= 64(-1) = -64.

**Example 3: **Use De Moivre's Theorem to compute (√3 + i)^{5}.

__Solution__**: **It is straightforward to show that the polar form of √3 + i is 2(cos π/6 + i sin π/6). Thus we have:

(√3 + i)^{5} = [2(cos π/6 + i sin π/6)]^{5}

= 2^{5}(cos π/6 + i sin π/6)^{5}

= 32(cos 5π/6 + i sin 5π/6)

= 32(-√3/2 + 1/2 i)

= -16√3 + 16 i.

**Example 4:** Find the sixth roots of *√3+i*

**Solution: **The modulus of *√3**+i* is 2 and the argument is π/6.

The sixth roots are therefore