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

Need some help figuring out this problem.
 A: 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 
$$\left\{ 0 , 2, 2i, -2,-2i\right\}$$
A: 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).
A: 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.
