Solve $$z^5+4\overline z^3=0$$
This is what I did. Let $z=r(\cos\theta+i\sin\theta)$ so, the equation is: $$r^5(\cos5\theta+i\sin5\theta)+4r^3(\cos3\theta-i\sin3\theta)=0.$$ Suppose $r$ is not $0$ and divide both sides by $r^3$ to get: $$(r^2\cos5\theta+4\cos3\theta)+i(r^2\sin5\theta-4\sin3\theta)=0+0i$$ so we get the following system of equations: $$r^2\cos5\theta+4\cos3\theta=0$$ $$r^2\sin5\theta-4\sin3\theta=0$$ I'm not sure how to continue. I've tried subtract $4\cos3\theta$ from both sides in the first equation and in the second one add $4\sin3\theta$ to both sides, and then divide the equations, to get $\tan5\theta=-\tan3\theta$, but then i suppose im not dividing be zero and that's a problem. What can I do from here? Is there a simpler way to solve this problem? Thanks!