For convinience, $\tau = 2\pi$.

De Moivre’s Theorem

De Moivre’s theorem is that $(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta$ when $n$ is an integer and $\theta$ is a real number.

Noting that $e^{i \theta} = \cos \theta + i \sin \theta$, we can see that $(\cos \theta + i \sin \theta)^n=(e^{i \theta})^n=e^{i (n \theta)}=\cos n\theta + i \sin n\theta$.

Complex Exponentation

For our “proof” our De Moivre’s theorem, we have to show that $(e^{i \theta})^n=e^{i (n \theta)}$.

However, we have to be careful with complex exponention as it is not always true that $(e^a)^b=e^{ab}$ when $a,b$ are complex numbers. For example, in Clausen’s paradox, we have $e = (e^{1+ i\tau})^{1 + i\tau} \neq e^{(1 + i \tau) ^2} = e^{-\tau^2} \approx 7.157 \times 10^{-18}$. The reason for why this exponentiation rule does not work is that in the definition for complex exponentation is $a^b = \exp{(b \cdot \log a)}$ . Correctly expanding $(e^a)^b = \exp{(b \cdot \log e^a)}= \exp{(b \cdot (a \text{ mod }i \tau))}$.

At first glance, it does not seem obvious why $n$ must be an integer. However, consider when $\theta=\tau$ and $n=\frac{1}{2}$. We have $(\cos \theta + i \sin \theta)^n =1$ and $\cos n\theta + i \sin n\theta = -1$. Therefore, it is important that $n$ is constrained to only be an integer.

The reason why we can conclude that $(e^{i \theta})^n=e^{i (n \theta)}$ is because $(n \cdot (\theta \text{ mod } \tau)) \text{ mod } \tau = (n \theta) \text{ mod } \tau$ when $n$ is an integer. Therefore, we have $(e^{i \theta})^n = \exp{(n \cdot(i \theta \text{ mod } i \tau))} = \exp{((n \cdot (\theta \text{ mod } i \tau)) \text{ mod } i \tau)} = \exp{((i n \theta) \text{ mod } i \tau)} = e^{i (n \theta)}$. In fact, one can show that $(e^a)^b=e^{ab}$ holds when $a$ is a real number or $b$ is an integer.

Roots of Unity

We wish to solve the equation $z^n=1$ for integers $n$. Obviously $\lvert z \rvert=1$ so $z$ can be written in the form $e^{i \theta}$. Since $n$ is an integer, $(e^{i \theta})^n = e^{i (n \theta)} = 1$. So, $n\theta$ must be a multiple of $\tau$ which implies that $\theta = k \frac{\tau}{n}$ for integer $k$. To restrict $\theta$ to be in the range $[0,\tau)$, we restrict $k$ to be in the range $[0,n)$. Note that we have found $n$ different roots, which are all the roots since a polynomial of degree $n$ has exactly $n$ roots (including repeated roots). Geometrically, the roots are $n$ equally spaced points on the unit circle on the complex plane.

In the more general case where we wish to solve the equation $z^n = a \cdot e^{i b}$, where $a$ is a positive real and $0 \leq b < \tau$, the solutions are given by $a^{\frac{1}{n}} \cdot e^{i (k \frac{\tau}{n} + \frac{b}{n})}$ for integer $k$ in the range $[0,n)$.

Trigonometric Identities

We can obtain expression of $\cos n\theta$, $\sin n\theta$ and $\tan n\theta$ in terms of sum of powers of $\cos \theta$, $\sin \theta$ and $\tan \theta$ respectively by proper expansion.

  • $\cos n \theta = \Re((\cos \theta +i \sin \theta)^n)$
  • $\sin n \theta = \Im((\cos \theta +i \sin \theta)^n)$
  • $\tan n \theta = \frac{\Re((\cos \theta +i \sin \theta)^n)}{\Im((\cos \theta +i \sin \theta)^n)} = \frac{\Re((\cos \theta)^n (1 +i \tan \theta)^n)}{\Im((\cos \theta)^n (1 +i \tan \theta)^n)} = \frac{\Re((1 +i \tan \theta)^n)}{\Im((1 +i \tan \theta)^n)}$
\[\begin{align}\sin(5\theta) =& \Im((\cos \theta + i \sin \theta)^5) \\ =& 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta \\ =& 5 (1- 2 \sin^2 \theta + \sin^4 \theta) \sin \theta - 10 (1-\sin^2 \theta) \sin^3 \theta + \sin^5 \theta \\ =& 16 sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta \end{align}\]

We can also do the opposite, that is represent powers of $\cos \theta$ and $\sin \theta$ in terms of sums of $\cos n\theta$ and $\sin n\theta$ respectively by using the fact that $(e^{i \theta})^n=e^{i (n \theta)}$ when $n$ is an integer.

  • $\cos^n \theta = (\frac{e^{i \theta}+e^{-i\theta}}{2})^n$
  • $\sin^n \theta = (\frac{e^{i \theta}-e^{-i\theta}}{2i})^n$
\[\begin{align}\cos^4 \theta =& (\frac{e^{i \theta}+e^{-i\theta}}{2})^4 \\ =& \frac{1}{16}(e^{4i\theta} + 4 e^{2i\theta} + 6 e^{0i\theta} + 4 e^{-2i\theta} + e^{-4i\theta} ) \\ =& \frac{1}{16}(\cos4\theta + 4\cos2 \theta + 6 \cos0\theta + 4 \cos-2 \theta + \cos-4\theta) \\ =& \frac{1}{16} ( 2\cos 4\theta + 8 \cos 2\theta + 1) \end{align}\]

Summation of Series

Since we have $\cos(n \theta) = \Re((e^{i\theta})^n)$, we can turn some series into geometric series.

$\sum\limits_{k=0}^\infty \frac{1}{2^k} \cos(k \theta) = \Re(\sum\limits_{k=0}^\infty (\frac{e^{i \theta}}{2})^k) = \Re(\frac{2}{2-e^{i\theta}}) = \Re(\frac{2(2-e^{-i \theta})}{5 -2 e^{i \theta} - 2e^{-i \theta}}) = \frac{4-2 \cos \theta}{5 - 4 \cos \theta}$

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