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I'm having a little trouble with this problem.

It's asking to find all $z\in\mathbb C$ that satisfy $z^3 = -2(1+i\sqrt{3})\overline z$, and to keep the answers in standard form.

I tried expanding and simplifying both sides by using $z=x+yi$ where $x,y\in\mathbb R$, then equating the imaginary and real parts separately to get:

real parts: $x^3-3xy^2 = -2x + 2\sqrt{3}y$.

imaginary parts: $4x^2y-y^3=-2\sqrt{3}x-2y$

But I don't really see how I can get very far with these two expressions because of the degree of the variables.

Am I using the wrong technique?

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Try converting to polar, i.e. write $z=re^{\theta}$ and $-2(1+i\sqrt{3})=-4e^{i\pi/3}$. Then solve for $r$ and $\theta$. And then convert back to standard form.

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  • $\begingroup$ But how does that account for the $\overline z$? $\endgroup$ – E 4 6 Nov 21 '14 at 0:30
  • $\begingroup$ Take the absolute value of both sides. Then you get $r^3= 4r$. $\endgroup$ – Avi Steiner Nov 21 '14 at 13:08

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