Solution of $(E):z^2-2mz+1=0$ in $\mathbb{C}$ Suppose the equation $(E):z^2-2mz+1=0 \quad / m\in \mathbb{C}\quad z\in\mathbb{C}$ and we suppose $z_{1}$ and $z_{2}$ are the two solution of this equation.
How can I prove that $|z_1|+|z_2| = |m-1|+|m+1|$? 

This is not very elegant but you can do it explicitly: Set $m=x+iy$ and choose a squareroot of $m^2-1$, $a+bi=\sqrt{m^2-1}$. Then $z_{1,2}= x+iy\pm (a+bi)$. We want to prove $|z_1|+|z_2|=|m-1|+|m+1|$ or equivalently (after squaring and remembering $z_1z_2=1$):
$$
|z_1|^2+2+|z_2|^2=|m-1|^2+|m+1|^2+2|m^2-1|
$$
The left hand side is
$$
(x+a)^2+(y+b)^2+2+(x-a)^2+(y-b)^2=2(x^2+y^2+1+a^2+b^2).
$$
The right hand side equals
$$
2(x^2+y^2+1+|m^2-1|),
$$
so it remains to show $a^2+b^2=|m^2-1|$. This is true since $a+bi=\sqrt{m^2-1}$ and so
$|m^2-1|=|\sqrt{m^2-1}|^2=a^2+b^2$.
