De Moivre's Formula for the Pol. Rep. of Powers of Complex Numbers

# De Moivre's Formula for the Polar Representation of Powers of Complex Numbers

Recall from the Polar Representation with Multiplication of Complex Numbers page that if $z, w \in \mathbb{C}$ and if $r_z = \mid z \mid$, $r_w = \mid w \mid$, $\theta_z = \arg(z)$, and $\theta_w = \arg (w)$ then:

(1)\begin{align} \quad z \cdot w = r_zr_w (\cos (\theta_z + \theta_w) + i \sin(\theta_z + \theta_w)) \end{align}

This can be summarized by the following two statements:

**1)**$\mid z \cdot w \mid = \mid z \mid \mid w \mid$.**2)**$\arg (z \cdot w) = \arg(z) + \arg(w)$.

We will now use this theorem to prove a very useful formula called De Moivre's formula.

Theorem (De Moivre's Formula): Let $z \in \mathbb{C}$, $z \neq 0$ and let $r = \mid z \mid$ and $\theta = \arg (z)$. Then $z^n = r^n(\cos (n \theta) + i \sin (n \theta))$ for any $n \in \mathbb{N}$. |

**Proof:**We will carry this proof out by induction. Let $S(n)$ be the statement that $z^n = r^n (\cos (n \theta) + i \sin (n \theta))$.

- For the base step, $n = 1$, $S(1)$ is trivially true, and for $n = 2$ we have that:

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

- So $S(2)$ is true. Let $k \in \mathbb{N}$, $k > 2$ and assume that $S(k)$ is true, that is, assume that:

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

- Then multiplying $z^k$ by $z$ gives us:

\begin{align} \quad z^{k+1} &= z \cdot z^k \\ \quad &= r^{k+1}( \cos (\theta + k \theta) + i \sin (\theta + k \theta)) \\ \quad &= r^{k+1} (\cos ((k+1)\theta) + i \sin ((k+1) \theta)) \end{align}

- So $S(k+1)$ is true. So by the principle of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$ which proves De Moivre's theorem. $\blacksquare$