from a handpicked tutor in LIVE 1-to-1 classes

# De Moivre's Formula

De Moivre's formula (or) De Moivre's theorem is related to complex numbers. We can expand the power of a complex number just like how we expand the power of any binomial. But De Moivre's formula simplifies the process of finding the power of a complex number much simple. To apply De Moivre's formula, the complex number first needs to be converted into polar form. Let us learn more about this formula in the upcoming sections.

## What Is De Moivre's Formula?

To understand De Moivre's formula, let us consider a complex number in polar form z = r (cos θ + i sin θ). Let us raise this to powers 2 and see what happens.

\(\begin{align} z^{2} &=[r(\cos \theta+i \sin \theta)]^{2} \\ &=r^{2}(\cos \theta+i \sin \theta)(\cos \theta+i \sin \theta) \\ &=r^{2}\left(\cos \theta \cos \theta+i \sin \theta \cos \theta+i \sin \theta \cos \theta+i^{2} \sin \theta \sin \theta\right) \\ &=r^{2}[(\cos \theta \cos \theta-\sin \theta \sin \theta)+i(\sin \theta \cos \theta+\sin \theta \cos \theta)] \\ &=r^{2}(\cos 2 \theta+i \sin 2 \theta) \end{align}\)

Similarly, we can see that (r (cos θ + i sin θ))^{3} = r^{3} (cos 3θ + i sin 3θ) by expanding it manually. This is the basic idea of De Moivre formula. If a complex number (in polar form) r (cos θ + i sin θ) is raised to some power 'n' (where n is an integer), then the modulus of the result is r^{n} and the argument of the result is nθ. Thus, De Moivre Formula is:

(r (cos θ + i sin θ))^{n} = r^{n} (cos nθ + i sin nθ), where n ∈ Z

**Proof of De Moivre's Formula**

Let us prove De Moivre's theorem by the principle of mathematical induction.

Let us assume that S(n) : (r (cos θ + i sin θ))^{n} = r^{n} (cos nθ + i sin nθ).

**Step 1: **To prove S(n) for n = 1.

LHS = (r (cos θ + i sin θ))^{1} = r (cos θ + i sin θ)

RHS = r^{1} (cos (1)θ + i sin (1)θ) = r (cos θ + i sin θ)

Thus, S(n) is true for n = 1.

**Step 2: **Assume that S(n) is true for some natural number n = k. Then

(r (cos θ + i sin θ))^{k} = r^{k} (cos kθ + i sin kθ)

**Step 3: **To prove S(n) for n = k + 1.

LHS = (r (cos θ + i sin θ))^{k + 1}

= (r (cos θ + i sin θ))^{k} • (r (cos θ + i sin θ))

= r^{k} (cos kθ + i sin kθ) • (r (cos θ + i sin θ)) (By Step 2)

= r^{k + 1 }[(cos kθ cos θ - sin kθ sin θ) + i (cos kθ sin θ + sin kθ cos θ)]

= r^{k + 1 }[ cos (kθ + θ) + i sin (kθ + θ) ]

= r^{k + 1 }[ cos (k + 1)θ + i sin (k + 1)θ ]

= RHS

So S(n) is true for n = k + 1.

Thus, by the principle of mathematical induction, S(n) is true for all the values of n.

We know that cos θ + i sin θ can be written as cis θ. Thus, the De Moivre formula can also be written as:

(r cis θ)^{n} = r^{n} cis nθ, where n ∈ Z

Let us have a look at a few solved examples to understand De Moivre's formula better.

## Solved Examples Using De Moivre's Formula

**Example 1:** Find the value of (1 - √3 i)^{5} using the De Moivre formula.

**Solution:**

Let z = 1 - √3 i = a + ib.

Its modulus is, r = √(a^{2} + b^{2}) = √(1+3) = 2.

α = tan^{-1} |b/a| = tan^{-1} √3 = π/3.

Since a > 0 and b < 0, θ is in the 4^{th} quadrant. So

θ = 2π - π/3 = 5π/3

Thus, z = r (cos θ + i sin θ) = 2 (cos 5π/3 + i sin 5π/3)

Now, z^{5} = (2 (cos 5π/3 + i sin 5π/3))^{5}

By De Moivre formula,

z^{5} = 2^{5} (cos 25π/3 + i sin 25π/3)

= 32 (1/2 + √3/2 i)

= 16 + 16 √3 i

**Answer: **(1 - √3 i)^{5} = 16 + 16 √3 i.

**Example 2:** Find the cube roots of unity using De Moivre's formula.

**Solution:**

We have to find \(\sqrt[3]1\).

1 in polar form is, 1 = 1 (cos 0 + i sin 0) = cos 2nπ + i sin 2nπ.

Here n = 0, 1, 2 (We have taken 3 numbers because we have to find the cube roots).

\(\sqrt[3]1\) = 1^{1/3}

= (cos 2nπ + i sin 2nπ)^{1/3}

= cos 2nπ/3 + i sin 2nπ/3 (By De Moivre formula)

When n = 0, \(\sqrt[3]1\) = cos 0 + i sin 0 = 1.

When n = 1, \(\sqrt[3]1\) = cos cos 2π/3 + i sin 2π/3 = -1/2 + (√3/2) i.

When n = 2, \(\sqrt[3]1\) = cos cos 4π/3 + i sin 4π/3 = -1/2 - (√3/2) i.

**Answer: **\(\sqrt[3]1\) = 1, -1/2 + (√3/2) i, and -1/2 - (√3/2) i.

visual curriculum