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:
$$
  z=a+bi \\
  \bigg|\frac{(1+a)+bi}{(1-a)-bi}\bigg|=1 \\
  \mathrm{expression*conjugated} \\
  \bigg|\frac{1+2bi-(a^2)-(b^2)}{1+2a+a^2+b^2}\bigg|=1
$$

As you said, $z=a+bi$, and $|z+1|=|1-z|=|z-1|$ (with $z \not= 1$). Hence $(a+1)^{2} +b^{2}=(a-1)^{2}+b^{2}$.
Therefore $a=0$ and $$z=bi \,\,\,\,\,\,\,\text{for any} \,\,\,\,\,\,\,b\in\mathbb{R}.$$ 
