I'm having some trouble understanding how to solve the following complex equation:
$$z^5-2\bar{z}|z|^2 = 0$$
I did find this other question Solve the complex equation $w^3=9\overline{w}$ that helped me realize that I could see mine as
$$z^5 =2\bar{z}|z|^2$$
And writing $z$ with the exponential form $e^{i\theta}$ and rewriting it
$$r^5e^{5i\theta}=2r^3e^{-i\theta}$$
This lead me to having $r=\sqrt{2}$ and $\theta=\frac{k\pi}{2}$. I think that from here I just need to solve with:
$$z=\sqrt{2} \left(\cos{\frac{k\pi}{2}}+i\sin{\frac{k\pi}{2}} \right)$$
However I'm not sure for which are the actual solutions, to which $z_k$ I should be solving and why
Best Answer
As you know already, absolute value of each solution is $\sqrt 2.$
For the argument:
The equation $e^{5i\theta}=e^{-i\theta}$ is equivalent to $$e^{6i\theta}=1,$$ which has six solutions ${\theta}_k=\frac{k\pi}{3}$ with $k\in\{0,1,2,3,4,5\}.$
The six solutions to the given equation in cis form are $$z_k=\sqrt{2}\left(\cos\frac{k\pi}{3}+i\sin\frac{k\pi}{3}\right),\quad \text{where}\quad k\in\{0,1,2,3,4,5\}.$$