Geometric Proof of DeMoivre's Formula I want to know if there were a geometric proof of DeMoivre's formula.
My attempt was starting with an easy complex number and observing patterns, then generalizing that pattern.
If you start with $(1+i)^n$ and graph it for $n=0,1,2,3,\dots$ the result should lead directly to a realization of DeMoivre's formula, and if you write down the exact absolute values and angles, you should get the desired result.
Then you could try to see if it works for fractional $n$ using the new formula, and make some way to prove that it works for $n\in\mathbb{R}$?
Are there any flaws in this method, and is there a better geometric proof?
 A: To write a proof by induction show that for any compex number $z = x + i y$ you can translate it to polar form
$z = \rho (\cos\theta + i \sin \theta)$
The base case of the proof by induction take pretty much takes care of itself.
$z^1 = \rho^1 (\cos1*\theta + i \sin 1*\theta)$
assuming $z^n = \rho^n (\cos n\theta + i \sin n\theta)$
if $z^{n+1} = \rho^{n+1} (\cos (n+1)\theta + i \sin (n+1)\theta)$
then it is true for all natrual numbers.  i.e. you showed that is true when n = 1 and when it is true for n=1 it will be true for n=2, and when it is true for n=2 it is true for n=2+1...
$z^{n+1} = z z^n = \rho (\cos \theta + i \sin \theta)*\rho^n (\cos n\theta + i \sin n\theta)$ from the inductive hypothesis.
Now, crank through the algebra to show what you need to show.
Again, this is algebra.  It doesn't have the elegance of a geometric proof.  But the key insight is that to multiply complex numbers you multiply the lengths and add the angles.  I suggest working with complex numbers of length 1 to see that it works.
