Solve the equation $z^3=z+\overline{z}$ I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.
 A: You have that $z^3$ is real, since it equals $z+\bar{z}$. So $z$ is on one of six regularly spaced rays pointing from the origin. In four cases out of six, $z^3$ and $z+\bar{z}$ would have opposing real parts and thus cannot be equal. This only leaves the two rays where $z$ itself is real. And your equation is reduced to $$z^3=2z\implies z(z-\sqrt{2})(z+\sqrt{2})=0$$
A: Note that if $z=a+ib$ then $z+\overline{z}=2a$ and
$$
z^3=2a
$$
The cube roots of $2a$ are
\begin{align}
\sqrt[3]{2a}\cdot\omega^{0}
=
&
\sqrt[3]{2a}
\\
\sqrt[3]{2a}\cdot\omega^{1}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)
\\
\sqrt[3]{2a}\cdot\omega^{2}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big)
\end{align}
where $\omega=\frac{1}{2}+i\frac{\sqrt[3]{3}}{2}$ is any cubic root of the unit.The possible values ​​of $ a $ and $ b $ are obtained equaling $ z = a + ib$ to the roots of  given above $2a$.
\begin{align}
\sqrt[3]{2a}\cdot\omega^{0}
=
&
\sqrt[3]{2a}
\\
\sqrt[3]{2a}\cdot\omega^{1}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{1}
\\
\sqrt[3]{2a}\cdot\omega^{2}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big)
\end{align}
More explicitly
\begin{align}
a+ib
=
&
\sqrt[3]{2a}
\\
a+ib
=
&
\frac{1}{2}\sqrt[3]{2a}+i\frac{\sqrt{3}}{2}\sqrt[3]{2a}
\\
a+ib
=
&
\frac{1}{2}\sqrt[3]{2a}-i\frac{\sqrt{3}}{2}\sqrt[3]{2a}
\end{align}
A: You have $(a+bi)^3=2a$ Now expand the cube and equate real and imaginary parts
A: $$z^3=z+\bar{z}=2\,{\rm Re}(z)\implies z^3\in{\bf R}\implies z=\omega r ~~(\omega^3=1,r\in{\bf R})$$
$$r^3=(\omega+\overline{\omega})r\implies\begin{cases} r=0 \\ r^2=2\,{\rm Re}(\omega) \implies\begin{cases} r=\pm\sqrt{2} & \omega=1 \\ r^2<0 & \omega\ne1\end{cases}\end{cases} $$
Thus the solutions are $0$ and $\pm\sqrt{2}$.
A: Writing out real and imaginary parts of $z$ and separating real and imaginary parts yields
$$
x^3+3ix^2y-3xy^2-iy^3=2x
$$
therefore,
$$
x^3-3xy^2=2x\implies x=0\quad\text{or}\quad x^2-3y^2=2
$$
and
$$
3x^2y-y^3=0\implies y=0\quad\text{or}\quad3x^2=y^2
$$
If $y=0$, then $x^3=2x\implies x\in\{0,\sqrt2,-\sqrt2\}$.
If $x=0$, then $y=0$.
If $x^2-3y^2=2$ and $3x^2=y^2$, then $-8x^2=2$.
Thus, $(x,y)\in\{(0,0),(\sqrt2,0),(-\sqrt2,0)\}$; that is, $z\in\{0,\sqrt2,-\sqrt2\}$.
A: Conjugating you get
$$\bar{z}^3=\bar{z}+z,$$
so
$$z^3=\bar{z}^3.$$ Now use the initial equation to substitute $\bar{z}$, after some algebra you get
$$\begin{align*}
z^3 &= \left( z^3-z \right)^3 \\
0   &= z^3- \left( z^3-z \right)^3\\
    &= z^3\left( 2-z^2 \right)\left( z+i\left( z^2-1 \right) \right)\left( z- i\left( z^2-1 \right) \right).
\end{align*}$$ 
Now you have to solve two quadratic equations which you know how to.
Edit. As pointed in the comments you have to to keep only the real solutions because $z+\bar{z}=2\Re(z)$.
A: This is akin particularly to the methods described by Elias , Sujaan Kunalan and nbubis :
The sum of a complex number and its conjugate produce a pure real number,
$$ z \ + \ \overline{z} \ = \ 2a \ = \ 2a \ \cdot \ cis (0) = \ 2a \ \cdot \ cis (2 \pi)  \ , $$
$ \ cis( \theta ) \ $ being the abbreivated form of $ \ \cos \theta \ + \ i \sin \theta \ \ . $  DeMoivre's Theorem then gives us
$$ z^3 \ = \ r^3  \cdot \ cis(3 \theta) \ = \ 2a \ \cdot \ cis (2 \pi) , $$
with $ \ r \ $ being the modulus of $ \ z \ $ and $ \ \theta \ $ its argument.  We thus obtain  $ \ r^3 = 2a \ $  and the "trigonometric" equation  $ \ cis(3 \theta) \ = \ cis (2 \pi) , $
which is solved (within the "principal circle", $ \ 0 \ \le \ \theta \ < \ 2 \pi \ $ ) by  $ \ \theta \ = \ \frac{0 + 2k \pi}{3} \ , \ \text{for} \ k = 0,1,2 , $
which is described by what is also called DeMoivre's Theorem for roots. 
EDIT:  Oops, need to take this a bit further.
So there are three apparent solutions,
$$z \ = \ (2a)^{1/3} \ , \ (2a)^{1/3} \cdot cis(\frac{2\pi}{3}) \ , \ \text{and} \ (2a)^{1/3} \cdot cis(\frac{4\pi}{3}) \ . $$
For all three, however, we have $ \ z \ + \ \overline{z} \ = \ 2 \cdot (2a)^{1/3} \ , $ which must also equal $ \ 2a \ , $ which is real.   Hence, $ \ (2a)^{1/3} \ = \ a \ \ \Rightarrow \ \ a^{1/3} \ \cdot \ (a^{2/3} - 2^{1/3}) = 0 \ ,  $ which is solved by $ \ a = 0 \ , \ \sqrt{2} \ , \ \text{and} \ -\sqrt{2} \ , $
as found by other posters.  So there are three solutions are pure real numbers.
FURTHER EDIT:  Hmmm, not quite done, on thinking about this a little more.  We've established moduli for solutions, but we should also look at the results for   $ \ (2a)^{1/3} \cdot cis(\frac{2\pi}{3}) \ , \ \text{and} \ (2a)^{1/3} \cdot cis(\frac{4\pi}{3}) \ . $
For $ \ a = \sqrt{2} \ , \ z \ =  \ \sqrt{2} \cdot cis(\frac{2\pi}{3}) \  \ \text{and} \ \sqrt{2} \cdot cis(\frac{4\pi}{3}) \ $, with $ \ \overline{z} \ =  \ \sqrt{2} \cdot cis(\frac{4\pi}{3}) \  \ \text{and} \ \sqrt{2} \cdot cis(\frac{2\pi}{3}) \ , $ respectively, so
$ \ z + \overline{z} \ = \ -2 \sqrt{2} \  , $    but $ \ z^3 \ = \ ( \sqrt{2})^3 \ \cdot \ cis(3 \ \cdot \frac{2 \pi}{3}) \ =  \ 2 \sqrt{2} \ , \ $ and likewise for the argument $ \frac{4 \pi}{3} \ . $
The same thing happens for $ \ a = -\sqrt{2} \ , $  since we have $ \ z \ =  \ -\sqrt{2} \cdot cis(\frac{2\pi}{3}) \ =  \ \sqrt{2} \cdot cis(\frac{5\pi}{3}) \ ,  $ for which $ \ \overline{z} \ =  \ \sqrt{2} \cdot cis(\frac{\pi}{3}) \  $   so,
$ \ z + \overline{z} \ = \ 2 \sqrt{2} \  \   , $ while
$$z^3 \ = \ ( -\sqrt{2})^3 \ \cdot \ cis(3 \ \cdot \frac{2 \pi}{3}) \ = \ (-2 \sqrt{2}) \cdot cis(2 \pi) \ = \ (-2 \sqrt{2}) \cdot (+1) \ = \ -2 \sqrt{2} \ , $$ 
and similarly for   $ \ z \ =  \ -\sqrt{2} \cdot cis(\frac{4\pi}{3}) \ =  \ \sqrt{2} \cdot cis(\frac{\pi}{3}) \ .  $
So it would seem that all of the complex values really are extraneous (as alex.jordan points out).
A: Hint: $z^3=z+\overline{z} \implies z^3=a+bi+a-bi=2a$
A: One more way to think about it would be in polar coordinates:
$$z+z^*=re^{i\theta}+re^{-i\theta}= 2r\cos(\theta)=r^3 e^{3i\theta}$$
One obvious solution is $z=r=0$. If $r>0$, then since $e^{3i\theta}$ must be real, $\theta =n\pi/3$, which gives six  values for $\theta$, but not all are really possible. For instance, $\theta=\pi/3$ gives:
$$r=-r^3$$
Ultimately, only $\theta = 0,\pi$ are solutions for $r>0$, so that:
$$2r=r^3 \ \to \ z= \pm \sqrt{2}$$ 
