Complex variable problem I would appreciate some help about the following problem.


  Given $z\in\mathbb{C}$ and
  $$\left\vert {1+z\over 1-z}\right\vert =1 $$ 
  show that $z$ is purely imaginary. 
  Deduce that the roots of the equation
   $$2{(1+z)^5}-(1+{\sqrt 3}i){(1-z)}^5=0 $$
  are purely imaginary, then write down the five distinct roots of the above equation in $z$ .
A slightly different approach is to notice that $f(z)=\frac{1-z}{1+z}$ is an involutive map: 
$$\forall z\in\mathbb{C}\setminus\{-1\},\quad f(f(z))=z $$
hence $f(z)$ gives a bijection between the punctured plane and itself. It is simple to notice that $f$ maps the imaginary axis into the unit circle minus the point $-1$ in a bijective way, hence the inverse image of the unit circle minus the point $-1$ is the imaginary axis.

The second equation is equivalent to
$$ \left(\frac{1-z}{1+z}\right)^{5} = \exp\left(-\frac{\pi i}{3}\right) $$
and every solution lies on the imaginary axis by the previous lemma. A solution, for instance, is given by $z=\frac{1-\exp(-\pi i/15)}{1+\exp(-\pi i/15)}=i\tan\frac{\pi}{30}$.
$$|1+z|^2=|1-z|^2$$
$$2Re(z)=-2Re(z)$$
$$Re(z)=0$$
Hence $z$ is purely imaginary.

You can also see this geometrically: If $z$ is equidistant from both $1$ and $-1$, then it has to lie on imaginary axis.