Let $z$ be a complex number satisfying $|z|=1$. If $\text{arg}(z)=\theta$, then $\text{arg}\left(\frac{1+z}{1-z}\right)$ is equal to which of the following?

Solution

1. Let $$\begin{array}{rl}z& =e^{i\theta }\\ & =\cos \theta +i\sin \theta \end{array}$$.
2. Then $$\begin{array}{rl}\frac{1+z}{1-z}& =\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta }\\ & =\frac{2{\cos }^2(\theta /2)+2i\sin (\theta /2)\cos (\theta /2)}{2{\sin }^2(\theta /2)-2i\sin (\theta /2)\cos (\theta /2)}\end{array}$$.
3. Factoring out $2\cos (\theta /2)$ in the numerator and $2i\sin (\theta /2)$ in the denominator gives $\frac{\cos (\theta /2)+i\sin (\theta /2)}{i(\sin (\theta /2)-i\cos (\theta /2))}$.
4. This simplifies to $$\begin{array}{rl}\frac{e^{i\theta /2}}{i{e}^{-i\theta /2}}& =\frac{1}{i}{e}^{i\theta }\\ & =-i{e}^{i\theta }\\ & =e^{-i\pi /2}{e}^{i\theta }\\ & =e^{i(\theta -\pi /2)}\end{array}$$.
5. Hence the answer is (B).