De Moivre's Formula Examples 1

Recall from the De Moivre's Formula for the Polar Representation of Powers of Complex Numbers page that if $z \in \mathbb{C}$ is a nonzero complex number with $z = r (\cos \theta + i \sin \theta)$ then for any $n \in \mathbb{N}$ we have that:

(1)

\begin{align} \quad z^n = r^n (\cos n \theta + i \sin n \theta) \end{align}

We will now look at some example problems involving De Moivre's formula.

Example 1

Use De Moivre's formula to find $z^7$ where $z = 1 + i$.

In polar coordinates we have that:

(2)

\begin{align} \quad z = (1 + i) = \sqrt{2} \left ( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right ) \end{align}

So by De Moivre's formula we have that:

(3)

\begin{align} \quad z^7 = (1 + i)^7 &= \sqrt{2}^7 \left ( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right ) \\ &=8\sqrt{2} \left ( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i \right ) \\ &= 8 - 8i \end{align}

Example 2

Use De Moivre's formula to find a formula for $\sin 2 \theta$.

Let $z = \cos \theta + i \sin \theta$. Then by De Moivre's formula, setting $n = 2$ yields:

(4)

\begin{align} \quad z^2 = (\cos 2 \theta + i \sin 2 \theta) \end{align}

In other words:

(5)

\begin{align} \quad (\cos \theta + i \sin \theta)^2 = \cos 2 \theta + i \sin 2 \theta \\ \quad \cos^2 \theta + 2i \cos \theta \sin \theta - \sin^2 \theta = \cos 2 \theta + i \sin 2 \theta \end{align}

We use the identity that $\cos^2 \theta - \sin^2 \theta = \cos 2 \theta$ to get:

(6)

\begin{align} \quad \cos 2 \theta + 2i \cos \theta \sin \theta = \cos 2 \theta + i \sin 2 \theta \\ \quad 2i \cos \theta \sin \theta = i \sin 2 \theta \end{align}

Therefore:

(7)

\begin{align} \quad \sin 2 \theta = 2 \cos \theta \sin \theta \end{align}