Proof of $(1-e^{ix})^{-1}$

Question

In G.H. Hardy's book 'Divergent Series' there is a claim that $(1-e^{ix})^{-1} = \frac {1}{2} + \frac {1}{2} i \cot (\frac {1} {2} x) $ I, for the life of me, can't get past showing that $(1-e^{ix})^{-1} =$ $$(1+\cos(x)+i\sin(x))/(1-\cos^2(x)-2i\sin(x)\cos(x)-i^2\sin^2(x))$$ Any help is appreciated.

Answer 1

I would recommend starting from the RHS, and converting trig functions to exponentials:

$$\begin{align} {1\over2}+{1\over2}i\cot\left({1\over2}x\right)&={1\over2}\left(1+i{\cos(x/2)\over\sin(x/2)} \right)\\ &={1\over2}\left({\sin(x/2)+i\cos(x/2)\over\sin(x/2)} \right)\\ &={1\over2}\left({e^{ix/2}-e^{-ix/2}\over2i}+i{e^{ix/2}+e^{-ix/2}\over2}\over{e^{ix/2}-e^{-ix/2}\over2i} \right)\\ &={1\over2}\left({e^{ix/2}-e^{-ix/2}\over2i}-{e^{ix/2}+e^{-ix/2}\over2i}\over{e^{ix/2}-e^{-ix/2}\over2i} \right)\quad\text{(look closely for what changed here!)}\\ &={1\over2}\left(-2e^{-ix/2}\over{e^{ix/2}-e^{-ix/2}}\right)\\ &={1\over1-e^{ix}} \end{align}$$

Note: This could probably be more cleanly done by making the second line read

$$={1\over2}\left({2i\sin(x/2)-2\cos(x/2)\over2i\sin(x/2)} \right)$$

so that you can go straight to

$$={1\over2}\left({(e^{ix/2}-e^{-ix/2})-(e^{ix/2}+e^{-ix/2})\over e^{ix/2}-e^{-ix/2}} \right)$$

etc.

Answer 2

Let $y=\cos(x) + i\sin(x)$. This factors into $\frac{1+y}{1-y^2}=\frac{1}{1-y}=(1-y)^{-1}$, implying $e^{ix}=\cos(x)+i\sin(x)$.