<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] 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$\sqrt{a^2+b^2}=1$.
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