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
 A: Note that\begin{align}\frac{z-1}{z+1}&=\frac{(z-1)\left(\overline z+1\right)}{(z+1)\left(\overline z+1\right)}\\&=\frac{|z|^2+z-\overline z-1}{|z+1|^2}\\&=\frac{|z|^2-1+2i\operatorname{Im}(z)}{|z+1|^2}\end{align}and that therefore $\frac{z-1}{z+1}$ is purely imaginary if and only $|z|^2-1=0$, which is the same thing as asserting that $|z|=1$.
A: Recall that $w\in \mathbb{C}$ is purely imaginary $\iff w=-\bar w$, then
$$\frac{z-1}{z+1}=-\overline{\left(\frac{z-1}{z+1}\right)} =-\frac{\bar z-1}{\bar z+1} \iff (z-1)(\bar z+1)=-(\bar z-1)(z+1)$$
$$z\bar z+z-\bar z-1=-z\bar z+z-\bar z+1$$
$$2z\bar z=2 \iff|z|^2=1$$
A: In the last step of your proof, before you use the assumption that $(a^2+b^2)=1$, you've deduced
$$\frac{z-1}{z+1}=\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}$$
Now, by assumption for the other direction, you have
$$\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}=xi$$
as $\frac{z-1}{z+1}$ is assumed to be imaginary, i.e. $\frac{z-1}{z+1}=xi$ for some $x$. Thus,
$$\frac{(a^2+b^2)-1}{(a^2+b^2)+1 +2a}+\frac{2bi}{(a^2+b^2)+1 +2a}=xi$$
and therefore 
$$\frac{(a^2+b^2)-1}{(a^2+b^2)+1 +2a}=0$$
i.e. $(a^2+b^2)-1=0$, and therefore$\sqrt{a^2+b^2}=1$.
A: Assume $z\ne\pm1$. Then ${z-1\over z+1}\in i{\mathbb R}$ means that, viewed from $z$, the two points $\pm1$ are seen under a right angle. By Thales' theorem this is the case iff $z$ is lying on a circle with diameter $[{-1},1]$.
A: Another approach
$${\bf Re}\dfrac{z-1}{z+1}={\bf Re}\dfrac{(z-1)(\bar{z}+1)}{|z+1|^2}=\dfrac{|z|^2-1}{|z+1|^2}=0 \iff |z|=1$$
