What Is de Moivre's Formula and How Do You Use It?

Complex numbers can be used to solve problems that seem at first glance to only deal with real numbers. An important tool you can use in these cases is de Moivre’s formula.

Formula

De Moivre’s Formula

For all natural numbers n , you have

= (cos 𝜃 + i sin 𝜃)n = cos (n𝜃) + i sin (n𝜃).

(cos 𝜃 + i sin 𝜃)n = cos (n𝜃) + i sin (n𝜃).

Calculations are often simplified by moving the exponent as in de Moivre’s formula. This is demonstrated in Example 1.

De Moivre’s formula can be proved using Euler’s formula and simple power rules:

= (cos 𝜃 + i sin 𝜃)n = (ei𝜃) n = ein𝜃 = cos (n𝜃) + i sin (n𝜃).

(cos 𝜃 + i sin 𝜃)n = (ei𝜃) n = ein𝜃 = cos (n𝜃) + i sin (n𝜃).

Q.E.D

Example 1

Prove the following trigonometric identities:

cos (2𝜃) = cos2𝜃 sin2𝜃,

and

sin (2𝜃) = 2cos𝜃sin𝜃,

using de Moivre’s formula

The expressions include cos (2𝜃) and sin (2𝜃), so you can use n = 2 in de Moivre’s formula:

= cos (2𝜃) + i sin (2𝜃) = (cos 𝜃 + i sin 𝜃)2 = cos 2𝜃 + 2i sin 𝜃 cos 𝜃i = + (i sin 𝜃)2 = cos 2𝜃 + 2i sin 𝜃 cos 𝜃i = sin 2𝜃 = cos 2𝜃 sin 2𝜃 = + 2i sin 𝜃 cos 𝜃.

cos (2𝜃) + i sin (2𝜃) = (cos 𝜃 + i sin 𝜃)2 = cos 2𝜃 + 2i sin 𝜃 cos 𝜃i + (i sin 𝜃)2 = cos 2𝜃 + 2i sin 𝜃 cos 𝜃i sin 2𝜃 = cos 2𝜃 sin 2𝜃 + 2i sin 𝜃 cos 𝜃.

For this equation to be valid, the real parts on both sides of the equal sign must be the same, and the imaginary parts on both sides must also be the same. This yields the identities that you set out to prove:

cos (2𝜃) = cos 2𝜃 sin 2𝜃, sin (2𝜃) = 2 sin 𝜃 cos 𝜃.

Q.E.D

Using Euler’s formula, you can derive a relationship between the exponential function and the trigonometric functions:

rei𝜃 = r (cos 𝜃 + i sin 𝜃).

You can therefore define cosine and sine using complex numbers via Euler’s formula.

Theory

Cosine and Sine Using Complex Numbers

For all complex numbers z , the following holds:

cos z = eiz + eiz 2 , sin z = eiz eiz 2i .

The definition can be justified for real numbers 𝜃 using Euler’s formula like this:

ei𝜃 + ei𝜃 2 = cos 𝜃 + i sin 𝜃 + cos (𝜃) + i sin (𝜃) 2 = cos 𝜃 + i sin 𝜃 + cos 𝜃 i sin 𝜃 2 = 2 cos 𝜃 2 = cos 𝜃,

and

ei𝜃 ei𝜃 2i = cos 𝜃 + i sin 𝜃 (cos (𝜃) + i sin (𝜃)) 2i = cos 𝜃 + i sin 𝜃 cos 𝜃 + i sin 𝜃 2i = 2i sin 𝜃 2i = sin 𝜃.

Q.E.D

The relationship between the exponential function and the trigonometric functions is useful in a variety of situations. It is often easier to work with the exponential function than the trigonometric functions. So when you’re working with trigonometric functions, it can be a good idea to reformulate the problem using the exponential function.

Example 2

Rediscover the differentiation rules for sine and cosine,

(sinx) = cosx, (cosx) = sinx,

using the exponential function

First, you write sine in exponential form:

sin x = eix eix 2i .

Then you find the derivative of both sides of the expression with respect to x. Remember the differentiation rules for the exponential function. The imaginary unit i is derived in the same manner as for any other number:

(sin x) = (eix eix 2i ) = ieix + ieix 2i = i (eix + eix) 2i = eix + eix 2 = cos x.

You can do the same with cosine by first writing it in exponential form:

(cos x) = (eix + eix 2 ) = ieix ieix 2 = i (eix eix) 2 = i2 (eix eix) 2i = (eix + eix) 2i = sin x.

(cos x) = (eix + eix 2 ) = ieix ieix 2 = i (eix eix) 2 Expand the fraction by i = i2 (eix eix) 2i Use the fact that i2 = 1 = (eix + eix) 2i = sin x.

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