Finding the cubic root of a complex number given its relation to the sum of itself and its conjugate

I am trying to solve for $z$, given that $z^3=z+\bar{z}$.

I tried reducing this seemingly easy equation by rewriting to polar form, completing the square, and some trig manipulation but with no success. How do I tackle this problem?

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What is the sum of a complex number and its conjugate? – Kolmo Mar 14 '12 at 12:51
@Kolmo It's $2 \Re(z)$? I can't, however, figure out where to go from there. I made several attempts at it, like I said, with no luck. – Milosz Wielondek Mar 14 '12 at 12:56
Yes it is. And it is real so $z^3$. now consider Re(z)>0 than what is the argument of z. And if Re(z)<0? – Kolmo Mar 14 '12 at 13:01
@Kolmo $\arg{z}$ would be $0$ and $-\pi +2n\pi$. Still no epiphany. – Milosz Wielondek Mar 14 '12 at 13:05
If Re(z)>0 then it is a complex number with argument 0. So z is a complex with argument 0, 2/3 pi and 4/3 pi. But in the last two cases the cosine is negative so it remains only the first case that gives $z=\sqrt{2}$. – Kolmo Mar 14 '12 at 13:18

Let's denote : $z=a+bi$ , then :

$(a+bi)^2(a+bi)=2a \Rightarrow (a^3-3ab^2)+(3a^2b-b^3)i=2a$

So , you should solve following system of equations :

$\begin{cases} a^3-3ab^2=2a \\ 3a^2b-b^3=0 \end{cases}$

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I thought there was a way around this, so as to not to have to expand the $(a+bi)$ binomial (which I mistakenly thought would be a pain), however this seems to do the trick! – Milosz Wielondek Mar 14 '12 at 13:16

A hint: From your equation it follows that $z^3$ is real. Now write $z=r\,e^{i\phi}$ and draw your conclusions about possible $\phi$'s and $r$'s. There will be several cases.

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I did write it like that, but I missed the fact that you can figure out $\operatorname{Im} z$ from the argument. Thanks! – Milosz Wielondek Mar 14 '12 at 15:45