# de Moivre identity, proof of

For the case $n=0$, observe that

 $\cos(0\theta)+i\sin(0\theta)=1+i0=\left(\cos(\theta)+i\sin(\theta)\right)^{0}.$

Assume that the identity holds for a certain value of $n$:

 $\cos(n\theta)+i\sin(n\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{n}.$

Multiply both sides of this identity by $\cos(\theta)+i\sin(\theta)$ and expand the left side to obtain

 $\displaystyle\cos(\theta)\cos(n\theta)-\sin(\theta)\sin(n\theta)+i\cos(\theta)% \sin(n\theta)+i\sin(\theta)\cos(n\theta)$ $\displaystyle=\left(\cos(\theta)+i\sin(\theta)\right)\left(\cos(n\theta)+i\sin% (n\theta)\right)$ $\displaystyle=\left(\cos(\theta)+i\sin(\theta)\right)^{n+1}.$

By the angle sum identities,

 $\displaystyle\cos(\theta)\cos(n\theta)-\sin(\theta)\sin(n\theta)$ $\displaystyle=\cos(n\theta+\theta)$ $\displaystyle\cos(\theta)\sin(n\theta)+\sin(\theta)\cos(n\theta)$ $\displaystyle=\sin(n\theta+\theta)$

Therefore,

 $\cos((n+1)\theta)+i\sin((n+1)\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{% n+1}.$

Hence by induction de Moivre’s identity holds for all natural $n$.

Now let $-n$ be any negative integer. Then using the fact that $\cos$ is an even and $\sin$ an odd function, we obtain that

 $\displaystyle\cos(-n\theta)+i\sin(-n\theta)$ $\displaystyle=\cos(n\theta)-i\sin(n\theta)$ $\displaystyle=\frac{\cos(n\theta)-i\sin(n\theta)}{\cos^{2}(n\theta)+\sin^{2}(n% \theta)}$ $\displaystyle=\frac{1}{\cos(n\theta)+i\sin(n\theta)}\cdot\frac{\cos(n\theta)-i% \sin(n\theta)}{\cos(n\theta)-i\sin(n\theta)}$ $\displaystyle=\frac{1}{\cos(n\theta)+i\sin(n\theta)},$

the denominator of which is $\left(\cos(n\theta)+i\sin(n\theta)\right)^{n}$. Hence

 $\cos(-n\theta)+i\sin(-n\theta)=\left(\cos(\theta)+i\sin(\theta)\right)^{-n}.$
Title de Moivre identity, proof of DeMoivreIdentityProofOf 2013-03-22 14:34:08 2013-03-22 14:34:08 rspuzio (6075) rspuzio (6075) 10 rspuzio (6075) Proof msc 12E10 proof of de Moivre’s formula   proof of de Moivre’s theorem AngleSumIdentity