Find the number of roots of the equation $ z^4=\omega\overline{z} $ , for $ z \in \mathbb{C} $

Question

Let $\omega=(\frac{3+4i}{3-4i})^5 $

Find the number of roots of the equation $ z^4=\omega\overline{z} $ , for $ z \in \mathbb{C} $

I am curious how to approach this type of problem, because after I computed this on Wolfram, I realised the solutions can't be really found that easily :

http://www.wolframalpha.com/input/?i=%7B(24i-7)%2F25%7D%5E5%3D(a%2Bb+i+)%5E4%2F(a-b+i+)

Answer 1

HINT: $z=0$ is an obvious solution. For nonzero $z$ let us take $z=re^{i\varphi}$. Then $r^4e^{i4\varphi}=\omega re^{-i\varphi}$ or $r^3e^{5i\varphi}=\omega$.