Question

If $a=\cos \theta +i\sin \theta$, find the value of $\frac{1+a}{1-a}$.

Answer 1

$$\begin{gathered} \frac{1+a}{1-a}=\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta } \\ =\frac{(1+cos\theta )+isin\theta }{(1-cos\theta )-isin\theta }\times \frac{(1-cos\theta )+isin\theta }{(1-cos\theta )+isin\theta } \\ =\frac{1-\cos \theta +i\sin \theta +\cos \theta -{\cos }^2\theta +i\cos \theta \sin \theta +i\sin \theta -i\sin \theta \cos \theta +i^2{\sin }^2\theta }{(1-\cos \theta {)}^2-i^2{\sin }^2\theta } \\ =\frac{1-{\cos }^2\theta -{\sin }^2\theta +2i\sin \theta }{1+{\cos }^2\theta -2i\cos \theta +{\sin }^2\theta }[\because i^2=-1] \\ =\frac{{\sin }^2\theta -{\sin }^2\theta +2i\sin \theta }{2-2i\cos \theta }[\because {\cos }^2\theta +{\sin }^2\theta =1] \\ =\frac{i\sin \theta }{1-\cos \theta } \\ =\frac{2i\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}{2{\sin }^2{\displaystyle \frac{\theta }{2}}} \\ =2i\cot \frac{\theta }{2} \end{gathered}$$

Thus, $\frac{1+a}{1-a}=2i\cot \frac{\theta }{2}$.