Finding The Complex Roots Of $4z^5 + \overline z^3= 0$?

Question

Hello everyone how can I find all the complex roots of:

$4z^5 + \overline z^3= 0$?

I tried to mark $a+bi = z , a-bi = \overline z$ and

$4(a+bi)^5 +(a-bi)^3 = 0$

But I don't know how to continue.

Answer 1

Use the exponential form: if $z=r\mathrm e^{i\theta}\enspace (r\ge 0)$, the equation can be rewritten as $$4r^5\mathrm e^{5i\theta}+r^3\mathrm e^{-3i\theta}=0\iff 4r^5\mathrm e^{8i\theta}+r^3=0\iff\begin{cases} r=0\\[-1ex]\text{ or }\\r^2\mathrm e^{8i\theta}=-\frac 14= \frac14\mathrm e^{i\pi}\end{cases}$$ Can you proceed?

Some details: $r=0$ if of course equivalent to $z=0$. As to $r^2\mathrm e^{8i\theta}=\frac14\mathrm e^{i\pi}$, identify the positive factors of the complex exponentials, and the arguments of both sides (as elements of $\mathbf R/2\pi\mathbf Z$):

• $r^2=\frac 14\iff r=\frac12$ (as $r\ge 0$);
• $8\theta \equiv \pi\mod2\pi\iff\theta\equiv \frac\pi 8\mod \frac\pi 4$.

Answer 2

I would not use real and imaginary parts here. Instead, render

$4z^5=-\overline z^3$

and take absolute values getting $|z|\in\{0,1/2\}$. Thus $z=0$ or $\overline z=1/(4z)$. Substitute the latter into your original equation and you have an equation for $z$ alone which is solved for the nonzero roots.