Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find  $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these

The correct answer is : $2, 2 e^{\pi/3}, 2 e^{5\pi/3}$
My calculation is here:
$r=\sqrt{-8^{2}}=8$
Then,
$= 2\ cis\ \frac{2\pi k}{n}$
If k is $0$,
$= 2\ cis\ 0=2$
If k is $1$,
$= 2\ cis\ \frac{\pi }{3}$
If k is $2$,
$= 2\ cis\ \frac{4\pi }{3}$
Therefore, the results are $2$, $= 2\ cis\ \frac{\pi }{3}$, and $= 2\ cis\ \frac{4\pi }{3}$.
So, why the correct answer is $2$, $2\ cis (\frac{\pi}{3})$, $2\ cis (\frac{5\pi}{3})$?
 A: We normally do exponents before multiplication and the minus sign here is a multiplication.  I think the problem wants
$$-(8^{1/3}) \mbox{ non } (-8)^{1/3}.$$
So work out  $8^{1/3}$ to get
$$2, 2 \mbox{ cis } \left(\frac{2\pi}{3}\right), 2 \mbox{ cis } \left(\frac{4\pi}{3}\right).$$
Now multiply by $-1$
$$-2, -2 \mbox{ cis } \left(\frac{2\pi}{3}\right), -2 \mbox{ cis } \left(\frac{4\pi}{3}\right)$$
And recall that $-\mbox{ cis } \left(\frac{2\pi}{3}\right)= \mbox{ cis } \left(\frac{5\pi}{3}\right)$ and $-\mbox{ cis } \left(\frac{4\pi}{3}\right) = \mbox{ cis } \left(\frac{\pi}{3}\right).$  so you have
$$-2, 2 \mbox{ cis } \left(\frac{5\pi}{3}\right), 2 \mbox{ cis } \left(\frac{\pi}{3}\right).$$
I believe the right answer has $-2$ rather than $2$.
A: Since $-8=8e^{i({\pi+2n\pi})}$ we have $(-8)^{\frac{1}{3}}=2e^{i\frac{\pi+2n\pi}{3}}$ for $n\in\mathbb Z$ (and you want the cases when $n=0,1$ and $2$).
A: We can rewrite,
$$-8^{\frac13}=2.(-1)^{\frac13}=2\cdot \text{CiS}\left(\frac{2k\pi + \pi}{3}\right)$$
And on substituting,
$k=0,1 $ and $2$ we get our desired answer
A: $-8 = 8(-1) = 8\operatorname{cis} (\pi) = 8\operatorname{cis}( (2k+1)\pi)$
So $(-8)^{\frac 13} = \sqrt[3]{8}\operatorname{cis}(\frac {2k+1}3\pi)=$.
$2\operatorname{cis}(\frac \pi 3 + \frac {2k\pi}3); k=0,1,2$
$2\operatorname{cis}(\frac \pi 3), 2\operatorname{cis}(\pi)= 2\cdot (-1)=-2, 2\operatorname{cis}(\frac {5\pi} 3)$
You assumed that we needed $3*\theta = 2k \pi$ and you didn't realize we actually needed $3*\theta = (2k+1)\pi$.
.....
And I don't quite understand we you did $r= \sqrt{(-8)^2} 8$.  Is that the $r$ that "belongs to" $-8$ or to $(-8)^{\frac 13}$?  Either way, I'm not sure what you were doing.
To solve $z^{\frac 1m}$ you need to:

*

*Express $z = r\operatorname{cis}({\theta})$.  where $r = |z|$ and $\theta =\operatorname{Arg}(z)= \operatorname{Arg}(\frac z{|z|})$.

*$z^{\frac 1m} = \sqrt[m]{r}\operatorname{cis}(\frac \theta m + \frac {2k\pi}m)$.

so to do $1$ we have $r = |-8| = 8$ and so $\theta = \operatorname{Arg}(-8)= \operatorname{Arg}\frac{-8}{8} =  \operatorname{Arg}(-1) = \pi$.
I think somehow you assumed our Argument was $0$ and not $\pi$.
