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What $z \in \mathbb{C}$ numbers solve $ \overline{z} = z^2$?

It seems obvious that $\left|z\right|$ can only be $1$, otherwise $\left|z^2\right| \ne \left|z\right|$.

Since $\varphi$ arguments add up upon square, I suppose solutions for $\varphi + \varphi = \varphi$, when $\left|z\right|=1$ solve the original problem.

So I think $z=1+0i$ solves the equation, because then $\varphi = 0°$ and $\varphi +\varphi = 0°$ and mirroring to $x$ does nothing to $\varphi$ as well.

But I'm not totally sure, that are there any other solutions.

kdani
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3 Answers3

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HINT:

Let $z=a+ib$

So,we have $a-ib=(a+ib)^2=a^2-b^2+2ab i$

Equating the imaginary parts, $-b=2ab$

Can you take it home from here?

lab bhattacharjee
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  • Well then if $a=1$ then $-b=2b$ so $b$ can only be $0$. As @bof said the solutions are cube roots of unity. $1+0i$ seems to be one cube. So would the solutions be $1\angle (k*120°)$ where $k\in \mathbb{Z}$? EDIT: Also $0i$ seems to work. – kdani Jan 20 '14 at 16:13
  • @kdani, if $b=0,$ we have $a=a^2\iff a=1,0$ Else $2a+1=0\iff a=-\frac12$ Set this value in the real part $$a=a^2-b^2$$ – lab bhattacharjee Jan 20 '14 at 16:27
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Multiply both sides of the equation $z^2=\overline z$ by $z$.

So $z^3=z\overline z=|z|^2=1$.

I.e., the solutions of your equation are cube roots of unity.

bof
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Put $z=re^{i\theta}$. Then we have $re^{-i\theta} = r^2e^{2i\theta}$.
So $r = 0$ or $1$.
If $r = 1$, then $e^{-i\theta} = e^{2i\theta}$, which means that $e^{-3i\theta}=1$, so $3\theta$ is a multiple of $2\pi$.

Thus there are four solutions in total.

TonyK
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