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Let $z=x+yi$, I get $y=xi$ finally, but what's the next step?

Winther
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Linda dadad
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    Hint: $x$ and $y$ are real. – Stahl Dec 01 '16 at 21:47
  • yes, x and y are real, so z=0? – Linda dadad Dec 01 '16 at 21:48
  • If $z = 5 + i 0 = 5$ then $z^2 = |z|^2$ so $z=0$ is not the only solution here. You have done a small mistake at a previous step: $y^2 = xyi$ does not imply $y = xi$ it implies $y =0$ or $y = xi$. – Winther Dec 01 '16 at 21:50
  • Something must be off, since $y=xi$ does not hold for a noticeable class of solutions to your equation. –  Dec 01 '16 at 21:50
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    @Lindadadad Realize that the right hand side is a real value? For what values of $a$ and $b$ is $(a+bi)^2$ a real output? – imranfat Dec 01 '16 at 21:52
  • Try to think of a complex number as a vector and its multiplication as both a rotation and a "distortion" in its absolute value. – Arthur Dec 01 '16 at 21:52

6 Answers6

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Note that $|z|^2 = z\bar z$. So, we have $$ z^2 = |z|^2\\ z^2 - z\bar z = 0\\ z(z - \bar z) = 0 $$ so: we must have $z = 0$, or $z = \bar z$ (which is to say that $Im(z) = 0$).

Paolo Leonetti
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Ben Grossmann
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Clearly $|z|^2$ is real and non-negative. So you need $z^2$ real and non-negative. That happens if $z$ is real, and in that case it's clear that $z^2=|z|^2$. Finally, observe that if $z$ is not real, then $z^2$ either is not real or is negative.

So the solution is: $z$ is real.

Michael Hardy
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There isn't really any need to break down $z$ into real parts or polar form or multiple cases. Firstly $|z|^2=z\bar z$, so you are looking at the equation $z\bar z = zz$.

Either $z=0$ or else you can cancel $z$ from both sides, whence $z=\bar z$, and in either case that means $z$ is real.

So, only real numbers will work, and it's clear they all work.

rschwieb
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Of course $z=0$ is a solution. If $z\ne0$, write $z=|z|u$, where $|u|=1$. Then the equation becomes $$ |z|^2u^2=|z|^2 $$ Can you say what $u$ should be and finish?

egreg
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Since $z^2=|z|^2 e^{2i\theta}$, for some $\theta$, then the equation is $e^{2i\theta}=1$. Hence $\theta \in \pi\mathbf{Z}$, i.e., $z \in \mathbf{R}$.

Paolo Leonetti
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In polar coordinates you get $(r(\cos \theta + \sin \theta))^2 = r^2(\cos 2\theta + \sin 2\theta) = r^2 = |r(\cos \theta + \sin \theta)|^2$ so $\theta \in \pi \Bbb Z$ and therefore $z \in \Bbb R$

cronos2
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