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
Best Answer
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?