If a = cos θ + i sin θ,

According to the question,

We have,

a = cos θ + i sin θ

$\Rightarrow \frac{1+a}{1-a}=\frac{(1+\cos \theta)+i \sin \theta}{(1-\cos \theta)-i \sin \theta}$

$=\frac{2 \cos ^{2} \frac{\theta}{2}+12 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}-12 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$

$=\frac{2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)}{2 \sin \frac{\theta}{2}\left(\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}\right)}$

$=\frac{i \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)}{\sin \frac{\theta}{2}\left(i \sin \frac{\theta}{2}-i^{2} \cos \frac{\theta}{2}\right)}$

$=\frac{i \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)}{\sin \frac{\theta}{2}\left(i \sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)}$

$=i \cot \frac{\theta}{2}$