Problem involving complex numbers and modules

Question

During my complex numbers course, I came into the following problem: Let $z_k$ be complex numbers such that $z_k= \cos \frac {5k-4}{9}\pi + i \sin \frac {5k-4}{9}\pi$, where $k \in \{ 1,2,3 \}.$ Determine $z \in \Bbb C$ such that $$2z^3= \frac {|z|+z_1}{1+z_1+z_1z_2}+\frac {|z|+z_2}{1+z_2+z_2z_3}+\frac {|z|+z_3}{1+z_3+z_3z_1}.$$

Answer 1

Brute force \begin{align*} z_1 z_2 z_3 &= \operatorname{cis} \frac{(1+6+11)\pi}{9} \\ &= 1 \\ \frac{|z|+z_{1}}{1+z_1+z_1 z_2} \color{red}{\times \frac{z_3}{z_3}} &= \frac{z_3(|z|+z_1)}{z_3+z_3 z_1+1} \\ \frac{|z|+z_1}{1+z_1+z_1 z_2}+ \frac{|z|+z_3}{1+z_3+z_3 z_1} &= \frac{(1+z_3)|z|+z_3(1+z_1)}{1+z_3+z_3 z_1} \color{red}{\times \frac{z_2}{z_2}} \\ &= \frac{z_2(1+z_3)|z|+z_2 z_3+1} {z_2+z_2 z_3+1} \\ 2z^3 &= \frac{|z|+z_2}{1+z_2+z_2 z_3}+ \frac{z_2(1+z_3)|z|+z_2 z_3+1} {z_2+z_2 z_3+1} \\ &= |z|+1 \\ 2|z|^3 &= |z|+1 \\ (|z|-1)(2|z|^2+2|z|+1) &= 0 \\ |z| &= 1 \\ z^{3} &= 1 \\ z &= \operatorname{cis} \frac{2\pi k}{3} \end{align*}