DeMoivre's Theorem Proof

Silverfalcon

I just used DeMoivre's Theorem on one of old AHSME problem. But I wonder, can anyone prove this theorem so it makes easier to understand?

Thank you very much.

**Posted:** Sun Jan 02, 2005 12:58 pm

Myth

What theorem do you mean?
$(\cos t+i\sin t)^n=\cos nt+i\sin nt.$
Right?

**Posted:** Sun Jan 02, 2005 1:31 pm

_________________
Myth is out of here

Silverfalcon

I think that's the one.

It was one about complex numbers' relationships. Is there more than one DeMoivre's Theorem? Confused

**Posted:** Sun Jan 02, 2005 2:12 pm

kunny

Let $f(t)=\cos t+i\sin t$ ,then $f'(t)=-\sin t+i\cos t=i(\cos t+i\sin t)$ so we have $f'(t)=if(t)\Longleftrightarrow f(t)=Ce^{it}$ .
Since $f(0)=1$ ,we obtain $C=1$ .Therefore we obtain Euler's formula $e^{it}=\cos t+i\sin t$ .
Then $e^{int}=(e^{it})^n\Longleftrightarrow (\cos t+i\sin t)^n=\cos nt+i\sin nt\ (n=0,\pm1,\pm2,\cdots)$ .

ZetaX

Proof:
you have $cos(x)+i \cdot sin(x) = e^{ix}$ and from $e^{inx}=(e^{ix})^n$ follows $cos(nx)+i \cdot sin(nx)=(cos(x)+i \cdot sin(x))^n$

Edit: kunny was faster...

kunny

**Posted:** Sun Jan 02, 2005 2:32 pm