# 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?

• Your first computation is wrong, how did you get $(a+bi)^2=a^2+b^2$?
– user223391
Commented Oct 1, 2016 at 15:07
• @ZacharySelk Ah thank you, for some reason I was thinking about $(a+bi)(a-bi) = a^2+b^2$. Thanks for making me realise my mistake!
– user373679
Commented Oct 1, 2016 at 15:10
• $$|z^2|=\sqrt{z^2\,\bar z^2}=\sqrt{|z|^4}=|z|^2$$ Commented Oct 1, 2016 at 15:19

$$z^2 = (a+bi)^2 = a^2 - b^2 +2iab$$ $$|z^2| = \sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4+2a^2b^2+b^4}=\sqrt{(a^2+b^2)^2}$$
$$|z| = \sqrt {a^2+b^2}$$
$$\implies |z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$$
If you use $z=r\cdot e^{i\theta}$, the proof is easy.
$|z^2|=|r^2\cdot e^{2i\theta}|=r^2$ and $|z|^2=r^2$.
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 $u,v\in\mathbb{C}$.