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?
$\endgroup$ 33 Answers
$\begingroup$$$z^2 = (a+bi)^2 = a^2 - b^2 +2iab $$ $$|z^2| = \sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4+2a^2b^2+b^4}=\sqrt{(a^2+b^2)^2}$$
and
$$|z| = \sqrt {a^2+b^2}$$
$$\implies |z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$$
If you use $z=r\cdot e^{i\theta}$, the proof is easy.
$|z^2|=|r^2\cdot e^{2i\theta}|=r^2$ and $|z|^2=r^2$.
$\endgroup$ $\begingroup$HINT: Assuming you know what complex conjugate means, use $$|z|^2=z\bar z$$ for any $z\in\mathbb{C}$ and $$\overline{uv}=\bar u \bar v$$ for any $u,v\in\mathbb{C}$.
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