Describe the set whose points satisfy the relation $|{z-1\over z+1}|=1$ 
Describe the set whose points satisfy the relations $$|{z-1\over z+1}|=1$$ for any $z=a+ib\in\mathbb{C}$.

Solution: 
$$
\\ 1=|{z-1\over z+1}|=|{(z-1)(z^*+1)\over (z+1)(z^*+1)}| \Rightarrow
\\|z+1|^2=|(a^2+b^2-1)+i(2b)|$$
But here I get a complex equation of degree 4 which I don't know to solve.
 A: Remember that for all numbers $z,w\ne 0$ we have $$\Big|{z\over w}\Big| ={|z|\over |w|}$$
So we have $$|z-1| = |z+1|$$ and thus $z$ is at equal distance from $1$ and $-1$ and thus it is perpendicular bisector for segment between $1$ and $-1$, so it is an $y$-axis i.e. $z=bi$ where $b$ is real.  
A: Hint
Think geometrically, $|z-1|=|z+1|$ means all those points which are equidistant from both $1$ and $-1$. So it will be all the points lying on the perpendicular bisector of...
A: Just to show your approach using $z=a+bi$ could work:
If $$\left| \frac{z-1}{z+1} \right| = 1 $$ then $|z-1|^2=|z+1|^2, $ so $(a-1)^2+b^2=(a+1)^2+b^2,$ so $(a-1)^2=(a+1)^2.$
Clearly we can't have $a-1=a+1$, so we must have $a-1=-(a+1),$ i.e., $a=0, $
and there are no restrictions on $b$, so the solutions are $a=0+bi$ for all $b \in \mathbb R.$
A: $$\left| \frac{z-1}{z+1} \right| = 1 \quad \Leftrightarrow \quad \exists \theta \in \mathbb{R}, \text{} \frac{z-1}{z+1} =e^{i\theta}$$
You deduce that $z(1-e^{i\theta})=e^{i\theta}+1$, i.e. because $e^{i\theta}=1$ can't be a solution, 
$$z= \frac{1+e^{i\theta}}{1-e^{i\theta}}= i \mathrm{cotan} \left( \frac{\theta}{2} \right)$$
