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I have to find $z$.

$|z|^2 = z\cdot \bar{z}$, but I don't know if this helps in this situation.

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up vote 9 down vote accepted

You can prove first that $z^2$ is real, and then use the fact that $|z|^2=|z^2|$.

Can you find $z^2$ then?

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How do I prove that $z^2$ is real? – Daniel Nov 3 '11 at 15:46
can you see any reason why the second and third term would be real? (taking the conjugate also helps, but this is simpler) – N. S. Nov 3 '11 at 15:48
So finally it will be $4z^2+8z^2-3=0$? Thank you! – Daniel Nov 3 '11 at 15:56
Carefull, $z$ is complex, so $z^2$ can be also negative.... – N. S. Nov 3 '11 at 16:00
@nikita2 The last two are $\pm i \frac{\sqrt{3}}{2}$. – N. S. Apr 5 '13 at 14:27

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