Simplify $\frac{1}{1-z}$, given $z = \cos \vartheta+ i\sin \vartheta$

Question

I'm stuck on this question and was hoping for a hint:

Given: $z = \cos(\vartheta) + i\sin(\vartheta),$ show: $$\frac{1}{1-z}=\frac{1}{2} + \frac{i\cot\left({\frac{\vartheta}{2}}\right)}{2}$$

I tried half angle and $\tan$ trig identity, but got nowhere!

Thanks.

Answer 1

Note,

$$\frac{1}{1-z}= \frac{1}{1-e^{i\theta}} = \frac{e^{-i\frac{\theta}2}}{e^{-i\frac{\theta}2}-e^{i\frac{\theta}2}} =\frac i2 \frac{\cos\frac{\theta}2-i \sin\frac{\theta}2 }{\sin\frac{\theta}2} =\frac{1}{2} + \frac{i}{2} \cot{\frac{\theta}{2}} $$

Answer 2

\begin{align*} \frac{1}{1-z} & = \frac{1}{1-\cos(\theta) - i\sin(\theta)}\\\\ & = \frac{1}{2\sin^{2}(\theta/2) - 2i\sin(\theta/2)\cos(\theta/2)}\\\\ & = \frac{1}{2\sin(\theta/2)}\frac{1}{\sin(\theta/2) - i\cos(\theta/2)}\\\\ & = \frac{1}{2}\times\csc(\theta/2)\times(\sin(\theta/2) + i\cos(\theta/2))\\\\ & = \frac{1}{2} + \frac{i\cot(\theta/2)}{2} \end{align*}

Answer 3

$$\frac{1}{1-z}=\frac{1}{1-(\cos(\theta) + i\sin(\theta))}\\ \frac{1}{(1-\cos(\theta)) - i\sin(\theta)}=\\ \frac{1}{(1-\cos(\theta)) - i\sin(\theta)}\times \frac{(1-\cos(\theta)) + i\sin(\theta)}{(1-\cos(\theta)) + i\sin(\theta)}$$ Next step is simplifying.