How do I solve and plot the complex equation I have the following complex equation:
\begin{equation}
z^6 + 1 = 0
\end{equation}
I would like to be able to gain some intuition and understanding. I know from the fundamental theorem of algebra that any polynomial with degree n has n roots. So it is plausible to say that the above equation has six roots. However, what about repeated roots? 
Since I am not experienced with complex equations, to me all I see is that $\pm i$ are two solutions, but are there others?
The CAS Sage gave me the following output (though I never know when to trust it):
[z == 1/2*I*(-1)^(1/6)*sqrt(3) + 1/2*(-1)^(1/6), z == 1/2*I*(-1)^(1/6)*sqrt(3) - 1/2*(-1)^(1/6), z == -(-1)^(1/6), z == -1/2*I*(-1)^(1/6)*sqrt(3) - 1/2*(-1)^(1/6), z == -1/2*I*(-1)^(1/6)*sqrt(3) + 1/2*(-1)^(1/6), z == (-1)^(1/6)]

Thanks for all the help in advance
 A: They're all roots of $-1$, which has absolute value $1$, so they all have absolute value $1$ as well. That means they're found on the unit circle in $\mathbb{C}$.
As for how they're distributed around the circle, if you measure angles from the positive real axis, you want angles which, when multiplied by $6$ give you $\pi$, so you're looking at odd multiples of $\frac{\pi}{6}$. That includes $\frac{3\pi}{6}$ and $\frac{9\pi}{6}$, which correspond to the answers $\pm i$ that you noted. The other $4$ roots are $\pm \frac{\sqrt{3}}{2}\pm \frac{1}{2}i$, where the two plus/minus symbols can vary independently, giving $4$ values.
Does this help?
A: Are you familiar with the polar form for complex numbers.  You have $z^6=-1$  If you write $z=re^{i\theta},z^6=r^6e^{6i\theta}$  What does this tell you about $r, \theta?$
A: Since $e^{i2\pi k}=1,\, k\in\mathbb{Z}$, we have

$$ z^6 + 1 = 0 \implies z^6=-1=e^{i\pi} e^{i2\pi k}=e^{ i(2k+1)\pi } \implies z= e^\frac{{ i(2k+1)\pi }}{6}, k=0,1,2,3,4,5.$$

A: It's true. A good group of mathematicians don't give any credit. We solve equations like $z^6$+1=0 by use of De'Moivre's formula. The history behind such problems like this provide a better understanding of such an unnatural solution. How can a number raised to an even power plus another even number ever be zero? We also need to pay attention to the fact that a n-degree polynomial will give us n solutions and will form an n-gon in the complex plane. Also, remember cos, sin are 2$\pi$ periodic, hence adding 2$\pi$k. 
All other algebra done in previous posts are correct, i just wanted to give some insight.
http://en.wikipedia.org/wiki/De_Moivre%27s_formula
A: $$z^6=-1=\cos\pi=\text{cis}{\pi}$$
Then,
$$z_k=\text{cis}{\frac{\pi+2k\pi}{6}}$$
the six roots obtained with setting $k=0,1,2,3,4,5.$
Further reading: Paul's Online Math Notes - Roots of complex numbers
A: z^6 = -1
In polar form:
(re^(i theta))^6 = -1
r^6 e^(6 i theta) = e^(i (pi + 2*pi*k))
r = 1.
e^(6 i theta) = e^(i (pi + 2*pi*k))
6 i theta = i (pi + 2*pi*k)
6 theta = pi + 2*pi*k
theta = (2k+1) pi / 6.
Then
z = cos ((2k+1) pi / 6) + i sin((2k+1) pi / 6).
Different ones are produced when k = 0,1,2,3,4,5.

