How to solve $z^4 + 4i\bar{z} = 0$ (efficiently)?
I managed to compute the radius of $z$:
Denote $z = rcis(\theta)$
Rearranged the equation to $z^4 = -4i\bar{z}$
Taken absolute value on both sides $r^4 = 16r$
And found $r=0$ or $r = \sqrt[3]{16}$
How can we calculate $\theta$ from here?
Or is there a better way to solve this?
Actually, if $z^4=-4i\overline z$, then $r^4=4r$, and therefore $r=\sqrt[3]4$.
If $r=0$, the $z=0$, which is a solution.
And if $r=\sqrt[3]4$, then, if $z=re^{i\theta}$, we have $z^4=4\sqrt[3]4e^{4i\theta}$ and $-4i\overline z=4ie^{-\pi i/2-i\theta}$. So,$$4\theta=-\frac\pi2-\theta+2k\pi,$$for some $k\in\Bbb Z$. Can you take it from here?
Note that $z=0$ is a solution. We can then proceed with the assumption that $z=re^{i\theta}$ is nonzero.
Your equation becomes $r^4e^{4i\theta}+4ire^{-i\theta}=0$ and therefore $$ r^3e^{5i\theta}=-4i $$ By uniqueness of the polar representation, you obtain $$ r^3=4,\qquad e^{5i\theta}=e^{-i\pi/2} $$ and so $$ r=\sqrt[3]{4},\qquad \theta=-\frac{\pi}{10}+\frac{2k\pi}{5},\ k=1,2,3,4,5 $$ if you want $\theta$ in the $[0,2\pi)$ interval.
By taking the modulus we get $|z|^4=4|z|\iff |z|=0\text{ or }|z|=\sqrt[3]{4}$
Let set aside the trivial solution $z=0$ then by multiplying by $z$ we get
$z^5=-4iz\bar z=-4i|z|^2=-8i\sqrt[3]{2}=-i\,2^\frac{10}{3}$
Therefore $z=\Big(-i\, 2^{\frac 23}\Big)\omega_k\ $ for $\, k=0,1,2,3,4\ $ and $\omega^5=1$ is a fifth root of unity.
