1. ## Some trig help please

Given that $z=cos\theta + isin\theta$, show that

$
\frac {1}{1-z} = \frac {1}{2} + \frac {i}{2} cot \frac {\theta}{2}
$

2. Originally Posted by rednest
Given that $z=cos\theta + isin\theta$, show that

$
\frac {1}{1-z} = \frac {1}{2} + \frac {i}{2} cot \frac {\theta}{2}
$
Sub in c+si for z
make the denominator real
simplify

You'll need a trig identity...

$\frac{\sin\theta}{1-\cos\theta}=\cot\left(\frac{\theta}{2}\right)$

3. Originally Posted by rednest
Given that $z=cos\theta + isin\theta$, show that

$
\frac {1}{1-z} = \frac {1}{2} + \frac {i}{2} cot \frac {\theta}{2}
$
$\frac{1}{(1 - \cos \theta - i \sin \theta)} \times \frac{(1 - \cos \theta + i \sin \theta)}{(1 - \cos \theta + i \sin \theta)} = \frac{1 - \cos \theta + i \sin \theta}{(1 - \cos \theta)^2 + \sin^2 \theta}$

$= \frac{1 - \cos \theta + i \sin \theta}{2(1 - \cos \theta)}$.

Seeing the real part of this is equal to 1/2 is trivial.

Your job is to now prove that $\frac{\sin \theta}{1 - \cos \theta}= \cot \frac {\theta}{2}$ ......