Help with a complex analysis exercise regarding an inequality

Question

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?

Answer 1

Assume $z \ne 0$ as $z=0, u=v=0$ works and write $z=re^{i\theta}, |\theta|\le \pi$, so $|z|-\Re z=2r\sin^2\theta/2$

Hence if $z-\Re z \le 1/2$ one gets $r\sin^2\theta/2 \le 1/4$ so $|\sqrt re^{i\theta/2}-\sqrt re^{-i\theta/2}|^2=4r\sin^2\theta/2 \le 1$ so one can take $u=v=\sqrt re^{i\theta/2}$ and obtain the required result.

Conversely if $z=uv, |u-\bar v| \le 1$ by squaring $|u-\bar v| \le 1$ we get $|u|^2+|v|^2-2\Re (uv) \le 1$ and by the mean inequality we get $2|uv|-2\Re (uv) \le 1$ which is precisely $|z|-\Re z \le 1/2$ so we are done this way too!