2015/090) Find $\frac{1+a}{1-a}$ if $a = \cos\beta + i\sin\beta$.

$a = \cos\beta + i\sin\beta$
so
 $1+ a = 1+ \cos\beta + i\sin\beta= 2\cos^2\frac{\beta}{2}+2i\cos\frac{\beta}{2}\sin\frac{\beta}{2}= 2 \cos\frac{\beta}{2}(\cos\frac{\beta}{2} +i\sin\frac{\beta}{2})$

further
$1 - a = 1 - \cos\beta -i\sin\beta = ( 2 \sin ^2 \frac{\beta}{2} - 2 i \cos\frac{\beta}{2} \sin\frac{\beta}{2}) = 2 \sin \frac{\beta}{2}(\sin \frac{\beta}{2}- i \cos\frac{\beta}{2})$
= $- 2i \sin \frac{\beta}{2}(\cos\frac{\beta}{2} + i \sin \frac{\beta}{2})$

so $\frac{1+a}{1-a}= \frac{\cos\frac{\beta}{2}}{- i sin \frac{\beta}{2}} = i\cot \frac{\beta}{2}$