induction proof of de moivres formula

Is this correct and enough proof

$\displaystyle (cosx+isinx)^n=cos(nx)+isinx$ is true so i put 1 and its true now ima prove for

$\displaystyle S(k+1) = (Cosx+isinx)^k+1= (Cosx+isinx)^k(cosx+isinx)$ now we can rewrite that ^k with a formula i cant remember as

$\displaystyle (coskx+isinkx)(cosx+isinx)$

and now i foctour out cos x and sin x

$\displaystyle Cos x ( k+1) + iSinx(k+1)$

is this enough :)

Re: induction proof of de moivres formula

You "can't remember"? You just wrote it two lines above!

For n a positive integer, yes, that **will** be a valid proof once you use trig identities to show how you got from $\displaystyle (cos(kx)+ isin(kx))(cos(x)+ i sin(x))$ to $\displaystyle cos((k+1)x)+ i sin((k+1)x)$.

DeMoivre's formula is true for n any number, of course, not just positive integers. It would think the simplest way to prove it is to use the fact that $\displaystyle cos(x)+ i sin(x)= e^{ix}$.