Solve for $z$ in $z^3=8i$ I only have a question about the end results. I answered the question fully but my professor knocked off 1 point for my $w_1^0$ result, but I don't know why. He circled the $i\pi /6$ in my answer but I can't figure out what I did wrong. Does anyone know what might be wrong here?

$$z^3=8i$$
$$z=(8i)^{1/3}$$
Converting to polar form and letting $k=0,1,2$ and $-\pi \leq \theta \leq \pi$, we get

$$w_1^0 =2e^{i\pi /6}$$
$$w_2^1=2e^{i5\pi /6}$$
$$w_3^2=2e^{9i \pi /6} = 2e^{-i \pi /2}$$

What's wrong with the $\frac{i\pi}{6}$ in $2e^{i\pi /6}$??
Perhaps your teacher expected a little more details, in particular on how you obtained that angle:

$$z^3=8i=8e^{\frac{\pi i}2+2k\pi i}=8e^{\frac{\pi i}2\left(4k+1\right)}\implies$$

$$z_k=(8i)^{1/3}=\sqrt[3]8e^{\frac{\pi i}6\left(4k+1\right)}=2e^{\frac{\pi i}6\left(4k+1\right)}\;,\;\;k=0,1,2\implies$$

$$\begin{cases}z_0=2e^{\frac{\pi i}6}=2\left(\frac{\sqrt3}2+\frac12i\right)=\sqrt3+i\\{}\\z_1=2e^{\frac{5\pi i}6}=2\left(-\frac{\sqrt3}2+\frac12i\right)=-\sqrt3+i\\{}\\z_2=2e^{\frac{3\pi i}2}=-2i\end{cases}$$
Hint use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ solve them and get the roots .