Given that $\displaystyle z=cos\theta + isin\theta$, show that
$\displaystyle
\frac {1}{1-z} = \frac {1}{2} + \frac {i}{2} cot \frac {\theta}{2}
$
$\displaystyle \frac{1}{(1 - \cos \theta - i \sin \theta)} \times \frac{(1 - \cos \theta + i \sin \theta)}{(1 - \cos \theta + i \sin \theta)} = \frac{1 - \cos \theta + i \sin \theta}{(1 - \cos \theta)^2 + \sin^2 \theta}$
$\displaystyle = \frac{1 - \cos \theta + i \sin \theta}{2(1 - \cos \theta)}$.
Seeing the real part of this is equal to 1/2 is trivial.
Your job is to now prove that $\displaystyle \frac{\sin \theta}{1 - \cos \theta}= \cot \frac {\theta}{2}$ ......