So, today in my complex analysis course we were given the following exercise:
Prove there exists a complex number $z$, such that $\vert z\vert-Re(z)\leq\frac{1}{2}$ if and only if there exists $u,v$ also complex numbers such that $z=uv$ and $\vert u-\overline{v}\vert\leq 1$.
What I have done so far is the following:
We know that $Re(z)=\frac{z+\overline{z}}{z}$ and $\vert z\vert=\sqrt{z\overline{z}}$.
Therefore:
$|z|-Re(z)=\sqrt{z\overline{z}}-\frac{z+\overline{z}}{z}=-\frac{1}{2}\left(\sqrt z-\overline{\sqrt z}\right)^2$
But I do not know how to prove the inequality after that, can you help me?