<s>[INST] Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} $$
and therefore $\ \frac{z-1}{z+1}$ is imaginary
now let me assume $\ \frac{z-1}{z+1} $ is imaginary number, how could I conclude that $\ |z| =1 $ I really can't think of any direction..
Thanks
 [/INST] Recall that $w\in \mathbb{C}$ is purely imaginary $\iff w=-\bar w$, then
$$\frac{z-1}{z+1}=-\overline{\left(\frac{z-1}{z+1}\right)} =-\frac{\bar z-1}{\bar z+1} \iff (z-1)(\bar z+1)=-(\bar z-1)(z+1)$$
$$z\bar z+z-\bar z-1=-z\bar z+z-\bar z+1$$
$$2z\bar z=2 \iff|z|^2=1$$
</s>