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So I can prove quite easily by induction that for any integer n>0, De Moivre's Theorem (below) holds.

If ##\DeclareMathOperator\cis{cis} z = r\cis\theta, z^n= r^n\cis(n\theta)##

My proof below:

Prove true for n = 1:

##z^1 = z = r\cis\theta = r^1\cis(1\theta)##

Assume true for n = k:

##z^k = r^k\cis(k\theta)##

Prove true for n = k + 1:

##z^{k+1} = z^k \times z\\

= r^k\cis(k\theta) * r\cis\theta\\

= r^{k+1}\cis(k\theta)\cis\theta\\

= r^{k+1}(\cos(k\theta) + i\sin(k\theta))((\cos\theta + i\sin\theta)\\

= r^{k+1}(\cos(k\theta)\cos\theta + \cos(k\theta)i\sin\theta + i\sin(k\theta)\cos\theta - \sin(k\theta)\sin\theta)##

then by compound angle identities

##= r^{k+1}(\cos(k\theta + \theta) + i\sin(k\theta + \theta))\\

= r^{k+1}\cis((k+1)\theta)##

However I struggle to do this with negatives. I tried the simple case of

##\frac{1}{r\cis\theta}## and from there, try to get to ##\frac{1}{r}\cis(-\theta)##, but I've only managed to do that by replacing the numerator of 1 with ##r\cis0##, which is equivalent, and then from there using De Moivre's Theorem to get an argument of ##-\theta##, which is... not very useful considering De Moivre's Theorem is what I'm trying to prove in the first place.

I just can't manipulate it into a form that would allow me to use compound angle identities like I could with positive numbers. Any ideas?