I would need some help with this, also it's my first time using this website.

Question

Let $a\in(0,\pi)$

if $z=\frac{1-i\cot a}{1+i\cot a}$

Prove that $z^2=\cos 4a+i\sin4a$

Answer 1

Direct way is what @Dr.SonnhardGraubner said.

For another one suppose $w = 1-i\cot a$, so we have: $$z = \frac{1-i\cot a}{1+i\cot a} = \frac{w}{\bar{w}} \Longrightarrow \left\{\begin{array}{c} |z| = \left|\frac{w}{\bar{w}}\right| = \frac{|w|}{|\bar{w}|}=1 \\ \arg z = \arg w - \arg \bar{w}=2\arg w \end{array}\right\} \Longrightarrow$$ $$\Longrightarrow \left\{\begin{array}{c} |z^2| = |z|^2 = 1^2 = 1 \\ \arg z^2 = 2\arg z = 4\arg w \end{array}\right\}$$ Now suppose $\arg w = \theta$: $$\tan \theta = -\cot a \Longrightarrow \theta = a + \frac{\pi}{2} \Longrightarrow 4\theta = 4a+2\pi \Longrightarrow \arg z^2 = 4a$$ That completes the proof.