<s>[INST] I'd like to get explain about complex roots If $x^6+1=0$ so $x^6=-1$, then we have to find the roots at $\mathbb{C}$.
I saw that the roots are $$\Large{e^{(\frac{\pi}{6}+\frac{2k\pi}{6})i}}\;\small{k=0,1,2,3,4,5}$$
this what I understand. maybe I wrong...
My question is why we are putting the $\frac{\pi}{6}$?
Thank you!
 [/INST] As you said, $x^6 = -1$. We want to find all complex solutions. You may have been exposed to complex numbers as things of the form $a + bi$, where $a,b$ are real. However, there is another form you can write them in: $re^{i\theta}$, where $r,\theta$ are real, and $r \ge 0$.
The geometric interpretation of $a + bi$ is rectangular; you go $a$ units right and $b$ units up, and there you are. But just like coordinates, you can also express it in a polar fashion: $\theta$ is what direction you point, and $r$ is how far out you go (the diagram uses $\varphi$ instead of $\theta$, but the latter is more common, as far as I know).
Why does this work? One can show, by a variety of methods, that $e^{i\theta} = \cos \theta + i \sin \theta$ (Euler's formula). This corresponds to a point on the unit circle: $\cos \theta$ to the right and $\sin \theta$ up. Scaling by $r$ gives you a vector further away, but in the same direction.
So, back to the problem. Since $x$ is complex, we can represent it as $re^{i\theta}$. The problem is now $-1 = \left( r e^{i \theta} \right)^6 = r^6 e^{i \cdot 6\theta}$. Convince yourself, geometrically or algebraically*, that $r$ must be $1$. Thus, we are looking for all $\theta$ such that $e^{i\cdot 6\theta} = -1$.
We have to use Euler's formula here: $-1 = e^{i\cdot 6\theta} = \cos 6 \theta + i \sin 6 \theta$. Clearly, $\cos 6 \theta$ is $-1$ and $\sin 6 \theta$ is $0$. Well, it seems plausible that $6 \theta = \pi$, right? So $\theta = \frac{\pi}{6}$.
But wait! We know that going around the unit circle doesn't change sine or cosine. So what if $6 \theta = \pi + 2\pi$ (i.e., $3 \pi$)? Then $\theta = \frac{\pi}{2}$. But we can keep going! We can say $6 \theta = \pi + 2 k \pi$ for any $k \in \mathbb{Z}$, so $\theta = \frac{\pi}{6} + \frac{2 k \pi}{6}$.
There are only finitely many, thankfully. As we just noted, going around the unit circle doesn't change sine or cosine. Therefore, it doesn't change $e^{i\theta}$, or our $x$. So if any of our $\theta$ differ by $2\pi$, we can toss one out, because we already counted that solution.


*

*$k = 0, \ \theta = \frac{\pi}{6}$

*$k = 1, \ \theta = \frac{3\pi}{6}$

*$k = 2, \ \theta = \frac{5\pi}{6}$

*$k = 3, \ \theta = \frac{7\pi}{6}$

*$k = 4, \ \theta = \frac{9\pi}{6}$

*$k = 5, \ \theta = \frac{11\pi}{6}$

*$k = 6, \ \theta = \frac{13\pi}{6} = \frac{\pi}{6} + 2\pi$

*$k = 7, \ \theta = \frac{15\pi}{6} = \frac{3\pi}{6} + 2\pi$

*$k = 8, \ \theta = \frac{17\pi}{6} = \frac{5\pi}{6} + 2\pi$

*$\vdots$


So, at $k = 6$, they start repeating. You can check that $k = -1$ is the same as $k = 5$, $k = -2$ the same as $k = 4$, and so on. The only ones we care about are $k = 0,1,2,3,4,5$ (can you generalize this to $n$th roots?), and so the $6$ solutions are:
$$ x = {\Large e^{i \left(\frac{\pi}{6} + \frac{2 k \pi}{6} \right)}}, k = 0,1,2,3,4,5 $$

*The geometric approach is fast: $-1$ has distance $1$ from the origin, so it must have $r = 1$. The algebraic one is slower, but more rigorous, and shows why polar form works the way it does. Note that $|e^{i\alpha}| = |\cos \alpha + i \sin \alpha| = \sqrt{\cos^2 \alpha + \sin^2 \alpha} = 1$ for any $\alpha$. So, because $-1 = r^6 e^{i \cdot 6 \theta}$, we take norms on both sides: $$|-1| = |r^6 e^{i \cdot 6 \theta}| = |r^6| |e^{i \cdot 6 \theta}| = |r^6| \cdot 1$$
And since $|-1| = 1$, $|r^6| = 1$, and from there, it is quick to conclude that $r = 1$.
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