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:

\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. $\blacksquare$
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