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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?

  • 1
    Your first computation is wrong, how did you get $(a+bi)^2=a^2+b^2$? –  Oct 01 '16 at 15:07
  • 2
    @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! –  Oct 01 '16 at 15:10
  • $$|z^2|=\sqrt{z^2,\bar z^2}=\sqrt{|z|^4}=|z|^2$$ – Mark Viola Oct 01 '16 at 15:19

3 Answers3

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If you use $z=r\cdot e^{i\theta}$, the proof is easy.

$|z^2|=|r^2\cdot e^{2i\theta}|=r^2$ and $|z|^2=r^2$.

JMP
  • 21,771
3

$$z^2 = (a+bi)^2 = a^2 - b^2 +2iab $$ $$|z^2| = \sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4+2a^2b^2+b^4}=\sqrt{(a^2+b^2)^2}$$

and

$$|z| = \sqrt {a^2+b^2}$$
$$\implies |z|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$$

Aakash Kumar
  • 3,480
1

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 $u,v\in\mathbb{C}$.