Prove $1 \leq |z^2-5| \leq 9$ if $|z| \leq 2$, $z = a + bi$ (complex number)

Question

Prove $1 \leq |z^2-5| \leq 9$ if $|z| \leq 2$, $z = a + bi$ (complex number)
I guess this is really easy question, but I still having troubles with it. Here is my try:
$$z^2 = (a+bi)^2 = (a^2 - b^2) + 2abi$$ $$|z| = \sqrt{a^2 + b^2} \leq 2 \implies a^2 + b^2 \leq 4$$
Since $a^2 \geq 0$ and $b^2 \geq 0$, $a^2 + b^2 \geq 0$

We need to evaluate $a^2 - b^2$.
$b^2$ can be equal $4$, so in this case $a^2$ must be equal $0$ and $a^2 - b^2 \leq a^2 + b^2$. Hence
$$-4 \leq a^2 - b^2 \leq 4$$ In order to evaluate $2ab$: $$a^2 + b^2 = (a - b)^2 + 2ab; (a - b)^2 \geq 0 \implies 2ab \leq a^2 + b^2 \leq 4$$ Unfortunately, I don't know how to prove that $-4$ is lower bound of $2ab$, so just let it be.

Then: $$|z^2 - 5| = |(a^2 - b^2) + 2abi - 5| \leq |4 + 4i - 5| = |-1 + 4i| = \sqrt{1 + 16} < 9$$ $$|z^2 - 5| \geq |-4 -4i - 5| = |-9 - 4i| = \sqrt{81 + 16} > 9 > 1$$

So there is some mistake. Multiple actually. And I strongly suspect that my way of "solving" this problem is at least inefficient (wrong).

Answer 1

If $|z|\leqslant2$, then $|z^2|=|z|^2\leqslant4$. Therefore:

• $|z^2-5|\leqslant|z|^2+5\leqslant 9$;
• $|z^2-5|\geqslant\bigl||z|^2-5\bigr|=5-|z|^2\geqslant1$.

This is one of the many cases in which it is better not to write $z$ as $a+bi$, with $a,b\in\Bbb R$.