Prove: if |$z|=1$ then $|iz-\overline{z}|\leq 2$ 
Prove: if |$z|=1$ then $|iz-\overline{z}|\leq 2$

$$|iz-\overline{z}|=|i(a+bi)-(a-bi)|=|-(a+b)+(a+b)i|=\sqrt{(a+b)^2+(a+b)^2}=\sqrt{2(a+b)^2}$$
On the other hand $|z|=1\iff \sqrt{a^2+b^2}=1\iff a^2+b^2=1$
So it seems that $\sqrt{2(a+b)^2}\leq \sqrt{2}$ maybe it is $|iz-\overline{z}|\leq \sqrt{2}$

One way to proceed from your point is by using the inequality $2ab\le a^2+b^2$ $$0\le (a+b)^2=1+2ab\le1+a^2+b^2= 2$$ And thus your quantity ends up being in the interval $\left[\sqrt0,\sqrt4\,\right]$.
