Question:
Prove that $\frac{e^Z}{e^W} = e^{z-w}$
My Attempt:
The given equation on left-hand side can be re-written as $e^z.e^{-w}$.
Let $z = x + iy$ and $w = u + iv$.
$exp(w) = e^u(cosv+isinv)$. This can be re-written by expanding the brackets and yields $exp(w) = e^ucosv + (e^usinv)i$
After calculation, $exp(-w) = \frac{cosv}{e^u} - \frac{sinv}{e^u}i$
Now, calculating $e^z.e^{-w}$ yields the following:
$$\frac{e^x}{e^u}[(cosy.cosv + siny.sinv) + i(cosv.siny - sinv.cosy)]$$
Problem:
The answer is all correct expect that the signs are wrong within the brackets. Where did I go wrong ?
Update:
Looking at the notes given to me by Prof. Robert Israel (who is also a community member), I got this:
and I guess my answer is indeed right, I did not mess up the signs :)