Find all the complex numbers $z$ satisfying 
Find all the complex numbers $z$ satisfying
  $$
  \bigg|\frac{1+z}{1-z}\bigg|=1
$$

So far I´ve done this:
$$
  z=a+bi \\
  \bigg|\frac{(1+a)+bi}{(1-a)-bi}\bigg|=1 \\
  \mathrm{expression*conjugated} \\
  \bigg|\frac{1+2bi-(a^2)-(b^2)}{1+2a+a^2+b^2}\bigg|=1
$$

$\bf{My\; Solution}::$ Given $$\displaystyle \left|\frac{1+z}{1-z}\right| = 1\Rightarrow \left|1+z\right| = |1-z| = |z-1|.$$
Now Put $z=x+iy\;,$ Then $$|x+iy+1|=|x+iy-1|$$
So we get $$\sqrt{(x+1)^2+y^2} = \sqrt{(x-1)^2+y^2}\Rightarrow (x+1)^2=(x-1)^2$$.
So we get $x=0$. So all Complex no. are in the form of $z=iy\;,$ where $y\in \mathbb{R}.$
