# induction proof of de moivres formula

• Nov 9th 2012, 02:58 AM
Petrus
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 :)
• Nov 9th 2012, 05:06 AM
HallsofIvy
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}\$.