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 $z=\frac{1-\exp(-\pi i/15)}{1+\exp(-\pi i/15)}=i\tan\frac{\pi}{30}$.
Geometric reasoning for the first: $$\left\vert {1+z\over 1-z}\right\vert = 1 \iff |z - 1| = |z - (-1)|,$$ i.e., iff the distance from $z$ to $1$ is equal to the distance from $z$ to $-1$.
Hint for the second: $$2{(1+z)^5}-(1+{\sqrt 3}i){(1-z)}^5=0 \iff \left(\frac{1+z}{1-z}\right)^5 = \frac{1+\sqrt 3 i}2, $$ i.e., $(1+z)/(1-z)$ is a fifth root of $(1+\sqrt 3 i)/2$.
$$|1+z|^2=|1-z|^2$$ $$2Re(z)=-2Re(z)$$ $$Re(z)=0$$ 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.
