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I found the equation $z^2 = z^*$, where * means the complex conjugate. I managed to solve this equation: $z:=(a+bi)$, then $a^2-b^2+2abi=a-bi$, which means $$a^2-b^2=a \land 2ab=-b$$ The roots of this system $(a;b)$ are $(0;0),(1;0)$ and $(-\frac12;\pm\frac{\sqrt3}2)$. It is obvious that the nonzero roots are the cube roots of unity. Is there a way to turn $z^2 = z^*$ into $z^4=z$, which has the same roots? If it's not possible, I'm also interested in whether we can solve the original equation with a different, purely complex method?

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  • $\begingroup$ You have $z^2=z^*$. Taking the conjugate we get $((z)^2)^*=(z^*)^2=(z^*)^*=z$, and now from the original equation $z^4=(z^2)^2=(z^*)^2=z$ $\endgroup$ – Kelenner Jul 9 '17 at 16:19
  • $\begingroup$ Wonderful, thanks $\endgroup$ – Lambert Jul 9 '17 at 16:22
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Multiply both the sides by $z$.(Considering $z\neq 0$)

$$z^2=\bar z \implies z^3= \bar z \cdot z$$

It's a standard result that $z \cdot \bar z=|z|^2$. Using this we get -

$$z^3=|z|^2$$

Taking modulus both the sides we get $|z^3|=|z|^2\implies |z|=1$.

Thus we get $$z^3=|z|^2=1$$

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HINT:

Taking modulus in both sides, $$|z|^2=|z^*|=|z|$$

$$\implies|z|(|z|-1)=0$$

If $|z|\ne0,|z|=1$

WLOG $z=e^{it}$ where $t$ is real.

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