# 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 $2e^{i\pi /6}$??

• It could simply be a misreading of your handwriting. I don't see anything wrong with it. Mar 8 '16 at 4:15
• I don't either! But here's the weird part - he circled $i\pi /6$ 3 times in my homework as well! I'm going to ask him tomorrow but I was wondering if anyone know what I was missing here. Mar 8 '16 at 4:16
• Did your professor want you to convert back to $a+bi$? Or perhaps draw a diagram showing the three points forming a triangle? Mar 8 '16 at 4:20
• No way because otherwise he would have taken off points for the other 2 answers as well. Plus, in the homework he was also only taking off 1 point when I wrote $i\pi /6$. Mar 8 '16 at 4:21
• I guess the three circles meant "$\pi/6$ is an angle you were drilled in school to know the sine and cosine of, you should write it out." Mar 8 '16 at 5:25

$$z^3=8i=8e^{\frac{\pi i}2+2k\pi i}=8e^{\frac{\pi i}2\left(4k+1\right)}\implies$$
$$z_k=(8i)^{1/3}=\sqrt8e^{\frac{\pi i}6\left(4k+1\right)}=2e^{\frac{\pi i}6\left(4k+1\right)}\;,\;\;k=0,1,2\implies$$
$$\begin{cases}z_0=2e^{\frac{\pi i}6}=2\left(\frac{\sqrt3}2+\frac12i\right)=\sqrt3+i\\{}\\z_1=2e^{\frac{5\pi i}6}=2\left(-\frac{\sqrt3}2+\frac12i\right)=-\sqrt3+i\\{}\\z_2=2e^{\frac{3\pi i}2}=-2i\end{cases}$$
Hint use $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ solve them and get the roots .
• Maybe your professor wants $a+bi$ form rather than polar Mar 8 '16 at 4:58