<s>[INST] How to prove that $|z^2| = |z|^2$ where $z = a+bi$? I just started my topic on complex numbers and I'm stuck on this question.
What I have managed to get (I go wrong here, I don't know where though):
$z^2 = (a+bi)^2 = a^2 + b^2$, so $|z^2| = \sqrt{(a^2 + b^2)^2 + 0^2} = \sqrt{a^4+2a^2b^2+b^4}$
and
$|z| = a^2+b^2$ so $|z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$
Suggestions?
 [/INST] $$z^2 = (a+bi)^2 = a^2 - b^2 +2iab $$
$$|z^2| = \sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4+2a^2b^2+b^4}=\sqrt{(a^2+b^2)^2}$$
and
$$|z| = \sqrt {a^2+b^2}$$
$$\implies |z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$$
</s>