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