Solving $z^4 + 4i\bar{z} = 0$ in detail I asked this question here Solving $z^4 + 4i\bar{z} = 0$
Though I accepted the answer, later I found out that I and WA don't agree on the result.
What I did:
Find the absolute value of $z$ by solving $|z^4| = |-4i\bar{z}|$
Getting $|z|^4 = 4|z|$ and hence $|z| = 0$ or $|z| = 4^{\frac{1}{3}}$
Leaving the trivial solution $z=0$ aside, we have $|z| = 4^{\frac{1}{3}}$
Now back to the original equation, multiplying it by $z$ we get:
$z^5 = -4i|z|^2$ and hence $z^5 = -4i4^{\frac{2}{3}}$ hence $z^5 = 4^{\frac{5}{3}}\operatorname{cis}(-\frac{\pi}{2})$
Getting the roots $z_k = 4^{\frac{1}{3}}\operatorname{cis}(\frac{-\frac{\pi}{2} + 2\pi k}{5})$ where $k=0,1,2,3,4$
So we get the multiplication of the complex roots to be:
$z_0\cdot z_1\cdot z_2\cdot z_3\cdot z_4 = 4^{\frac{5}{3}}\operatorname{cis}(\frac{-\pi}{10} + \frac{3\pi}{10} + \frac{7\pi}{10} + \frac{11\pi}{10} + \frac{15\pi}{10}) = 4^{\frac{5}{3}}\operatorname{cis}(\frac{35\pi}{10}) = -4^{\frac{5}{3}}i=-2^{\frac{10}{3}}i$
But when calculating with WA I get $-2^{\frac{5}{2}}i$.
What is the correct result?

Wolframalpha returns the product of the roots of $\displaystyle\frac{z^4 + 4 i z^*}z = 0$ as
$$-10.0794 i,$$ which indeed actually equals the $$-2^{\frac{10}{3}}i$$ that you obtained.
(First dividing the equation by $z$ was to discard the trivial root $z=0,$ as desired.)
