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 $

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+)

$|z|^4 = |w||z| = |z|$ thus $z=0$ or $|z|=1$. Next write $z=e^{ia}$ and $w=e^{i5b}$, then $e^{i4a} = e^{i(5b-a)}$ which implies $4a = 5b-a$ mod $2\pi$. — Gabriel Romon, Apr 30 '18 at 08:45

Przemysław Scherwentke · Accepted Answer · 2018-04-30T13:42:55.620

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 $r^3e^{5i\varphi}=\omega$.

edited Apr 30 '18 at 13:42

answered Apr 30 '18 at 08:52

Przemysław Scherwentke

13,668
5
35
56

We know that $ r = 1 $ and $ z = 0 $ is a solution , does that mean that we get $ 5 $ solutions from that equation? because the correct answer is $ 6 $ – SADBOYS Apr 30 '18 at 13:29
Yes, because $(r_1)^5e^{5i\vartheta}=\omega$ has 5 solutions (the number of roots of degree 5 of nonzero complex number). $r_1$ is in this case equal to 1. – Przemysław Scherwentke Apr 30 '18 at 13:34
That makes sense, thanks. – SADBOYS Apr 30 '18 at 13:52

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

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