# Help with a complex analysis exercise regarding an inequality

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?

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!