Solve $z^5 + 16\bar z = 0$ for $z\in \mathbb{C}$

Question

Solve $z^5 + 16\bar z = 0$ for $z\in \mathbb{C}$.

Need some help figuring out this problem.

Answer 1

We have $$z^5 + 16 \overline{z}=0 \iff z^5=-16\cdot \overline{z}$$ We use $z=|z|\cdot \exp(i\varphi)$ $$|z|^5 \exp(5i \varphi)= -16 |z|\exp(-i \varphi)$$ Now use that $$-1=\exp(i\pi)$$ Hence $$|z|^5 \exp(5 i \varphi) = 16 |z| \exp(i(\varphi+\pi))$$ Here we see that $z=0$ is a solution, now we search other solutions. $$|z|^4 \exp(5 i \varphi)= 16 \exp(i(\varphi+\pi))$$ Hence $|z|=2$ and we need to solve $$\exp(5 i \varphi) = \exp(i(\varphi+\pi))$$ As $\exp(z)\neq 0$ for all $z$ we have $$\exp(4 i \varphi)= \exp(i\pi)$$ We will be able to find $4$ more solutions here: we know that $$4 \varphi \equiv \pi \quad \operatorname{mod}\ 2\pi$$ The 4 solutions are $$\varphi_n = n\cdot \frac{2\pi}{4} +\pi$$ from $n=1\dots 4$. So the solutions are $$\left\{ \frac{3\pi}{2}, 0 , \frac{\pi}{2}, \pi\right\}$$ As the form $a+b i $ is easier for most, the solutions written in this form are $$\left\{ 0 , 2, 2i, -2,-2i\right\}$$

Answer 2

Here is a hint. $z^5 = -16 \overline{z}$. Hence $|z|^5 = 16 |z|$, from which we see that either $z=0$ or $|z| = 2$.

Then I would multiply across by $z$ to get $z^6 = -16 |z|^2 = -64$ (ignoring the zero solution, of course).

Answer 3

And this wouldn't qualify as a complex polynomial anyway, because of the $\bar z$.

I would rewrite the equation as $$ z^5 = -16\bar z. $$ From here you can take the modulus of both sides and deduce the possible values for $|z|$. Once you have that, you can find the possible arguments (angles) of $z$ as well.