Question.(source) If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that
$$\frac{1}{1-z}= \frac{1}{2}\left( 1+ i\cot\frac{\theta}{2}\right)$$
I have been trying to rearrange the right-hand-side in order to make it look like the required form, but have been struggling to do so and want to see if anyone is able to show me what steps need to be taken to solve this question.
Best Answer
Hint:
Rewrite the expression, using the exponential notation: \begin{align} \frac{1}{1-z}&= \frac{1}{1-\mathrm e^{i\theta}}= \frac{1-\mathrm e^{-i\theta}}{(1-\mathrm e^{i\theta})(1-\mathrm e^{-i\theta})}\\ &= \frac{1-\mathrm e^{-i\theta}}{(2-\mathrm e^{i\theta}-\mathrm e^{-i\theta})} == \frac{1-\mathrm e^{-i\theta}}{2(1-\cos\theta)}\\ &=\frac{1-\cos\theta}{2(1-\cos\theta)}+i\,\frac{\sin\theta}{2(1-\cos\theta)} \end{align} continue with the duplication formulæ.