de Moivre's Theorem

de Moivre's Theorem. Given any complex number $\cos\theta+i\sin\theta$ (which always has modulus $1$) and any integer $n$,

Corollaries (obtained by substituting $-\theta$ for $\theta$, and/or $-n$ for $n$, into the theorem):

1. $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
2. $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
3. $(\cos\theta-i\sin\theta)^{-n}=\cos n\theta+i\sin n\theta$
4. $\sin\theta+i\cos\theta=i(\cos\theta-i\sin\theta)$

How the theorem is applied to any complex number $z$ (not just one already in the form $\cos\theta+i\sin\theta$): first convert $z$ to polar form $z=r(\cos\theta+i\sin\theta)$ (find $r=|z|$ and the correct-quadrant argument $\theta$); then, since scalar factors pull straight out of a power, $z^n=r^n(\cos\theta+i\sin\theta)^n=r^n(\cos n\theta+i\sin n\theta)$ by de Moivre's theorem; finally reduce $n\theta$ modulo $2\pi$ to a convenient range before converting back to rectangular form if a numeric answer is required.

A recurring algebraic pattern. If $z=\cos\theta+i\sin\theta$, de Moivre's theorem gives $z^n=\cos n\theta+i\sin n\theta$ and $z^{-n}=\cos n\theta-i\sin n\theta$, so

$z^n+\frac1{z^n}=2\cos n\theta,\qquad z^n-\frac1{z^n}=2i\sin n\theta.$ …