$\arg(\frac{1+z}{1-z})$

Question

Let $z:=e^{i\theta}$ with $0 < \theta < \pi$. We want to show that $\arg(\frac{1+z}{1-z})=\frac\pi2$. Instead of a writing $z=\cos(\theta)+i\sin(\theta)$ and doing many tricky manipulations on $\arg(\frac{1+z}{1-z})$, is there any more efficient/elegant way to compute it ?

Answer 1

$$\frac{1+e^{i\theta}}{1-e^{i\theta}}=\frac{1+e^{i\theta}}{1-e^{i\theta}}\times\frac{1-e^{-i\theta}}{1-e^{-i\theta}}=\frac{1+2i\sin\theta-1}{2-2\cos\theta}=\frac{i\sin\theta}{1-\cos\theta}$$

This is wholly imaginary and has a positive imaginary part, so the argument is $\frac{\pi}{2}$.