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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!

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Since $-1=e^{i\pi}$.... – user63181 Jan 12 '14 at 17:59
@SamiBenRomdhane, yes I understand this, I don't understand why is the $\frac{\pi}{6}$.... – Yoar Jan 12 '14 at 18:16
Try taking the expression to the sixth power and seeing what happens if you take out the $\frac{\pi}{6}$ – Justin L. Jan 12 '14 at 22:26

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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Because $$e^{i\pi}=-1$$ Take the sixth root of both sides and you get a principal value.

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Why taking the Take the sixth root of both sides? Thank you! – Yoar Jan 12 '14 at 18:04
Because you want to find an $x$ such that $x^{6}=-1 \implies x^{6}=e^{i\pi}$ – Daniel Littlewood Jan 12 '14 at 18:18
OK, so we have $x=e^{\frac{\pi i}{6}}$ now, why we add $\frac{2 k \pi}{6}$?? – Yoar Jan 12 '14 at 18:23
Because $e^{2k\pi}=1^{k}=1$, so we can multiply each side by it and not change the equality. This gives five more roots. – Daniel Littlewood Jan 12 '14 at 18:40

Hint: If $W^n=r, ~~W=r(\cos\theta+i\sin\theta)$ then $$W=\sqrt[n]{r}\left(\cos\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2k\pi}{n}\right)\right)$$ wherein $k=0,1,2,\cdots,n-1$

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Since we know this equation has complex roots, we'll change notation a little and write:

$$ z^{6} = -1 $$

Furthermore, since $z$ is raised to the power of 6, we know that there are 6 solutions for $z$. Let $w=z^{3}$. We now have:

$$ w^{2} = -1 $$

with roots $w = {\pm}i= e^{i\pi/2 + n\pi}$ where $n=0,1,2,3...$ depending on the sign of $i$. However, note that after $n=1$ we obtain no new solutions. (When n is a multiple of 2 we simply add on a full 360 degree rotation in the complex plane; when n is odd we move through a number of full rotations plus half a rotation more, so arrive back at a root that we've already found.)

Returning to the original problem: we now take the cube root to obtain solutions for $z$:

$$ z=w^{1/3} = e^{i\pi/6 + n\pi/3} = e^{i\pi/6 + 2n\pi/6} $$

where n takes all integer values up to but not including $n=6$. At this point we have:

$$ z=e^{i\pi/6 + 2\cdot6\pi/6} = z=e^{i\pi/6 + 2i\pi} = e^{i\pi/6} $$

But this is the same solution as for $n=0$. That is, all roots with $n>5$ are in fact duplicates of the roots with $n=0,1,2,3,4,5$.

By the way, the final equality above is obtained by noticing that adding $2\pi$ radians is the same as:

$$ e^{i\pi/6 + 2i\pi} = e^{2i\pi}e^{i\pi/6}= e^{i\pi/6} $$

since $e^{2i\pi}=1$

I hope that helps.

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The reason is that equations of the form $z^n +k=0$ form $n$ points in the complex plane that can be joined to a regular $n$-gon. This is the image WolframAlpha generates for $x^6+1=0$, your case:

Six roots of $-1$

By changing the value of $k$ you merely rotate (and/or expand) the $n$-gon.

This is the picture for $x^6 -i=0$:

Six roots of $1$

This is the picture for $x^6 -1=0$:

enter image description here

EDIT: As Daniel pointed in the comments, the $\pi/6$ comes from the first approach we should try, that is, to imitate the way we take roots for real numbers. Therefore, if $x^6 = e^{i \pi}$ then $x = (e^{i \pi})^{1/6} = e^{i \pi/6}$ should be a root. However, the complex exponential is a periodic function. Therefore, if we have $e^{i \pi/6 + \pi/3} = e^{i \pi/2}$, when we take the sixth power we get

$$x^6 = e^{i \frac{\pi}{2} \cdot 6} = e^{i 3 \pi} = e^{i (\pi + 2 \pi)} = e^{i \pi} = -1.$$

The same happens with the other roots. Now you may ask: but then shouldn't it have countably many roots? The answer is yes, but there are only six distinct ones. These are the points represented in the image. If you take the integer $k$ greater than $5$, the root will coincide with one of the points.

EDIT (2): I'll delve deeper into these calculations. The fundamental fact here is that the complex exponential is periodic. This means that exists $p \in \mathbb{C}$ such that $\exp (z+p) = \exp (z)$ for all $z \in \mathbb{C}$. The complex number $p$ is $i 2 \pi$.

The passage $\exp (i \pi/2 \cdot 6) = \exp (i 3 \pi)$ is just a simplification like I said in the comments: we are writing $6/2$ as $3$. This gives us $3 \pi$. We can write $3 \pi = \pi + 2\pi$, giving us $$\exp (i 3 \pi) = \exp(i (\pi + 2\pi)) = \exp (i \pi + i 2 \pi).$$

Using the property $\exp (a+b) = \exp(a) \cdot \exp(b)$ valid for all complex $a,b$ we get $$\exp (i \pi + i 2 \pi) = \exp (i \pi) \cdot \exp (i 2 \pi).$$ Since $\exp (i 2 \pi) = 1$, we find that $\exp (i 3 \pi) = \exp (i \pi)$.

Another way is to notice the periodicity: as $\exp (z + i 2 \pi) = \exp (z)$ for all $z \in \mathbb{C}$, we get $\exp (i \pi + i 2 \pi) = \exp (i \pi)$ for $z = i \pi$. Since $\exp(i \pi) = -1$, we have shown that this is also a root. The process can be repeated for the other $k$.

Best wishes.

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Thank you, but I still not understand the $\frac{\pi}{6}$..... – Yoar Jan 12 '14 at 18:18
If we move -1, why we got the last image (BTW, it's $x^6+1=0$...). – Yoar Jan 12 '14 at 18:20
@YoavFridman I have added more things. Please tell me if it is more helpful now. Best wishes. – Mark Fantini Jan 12 '14 at 18:41
I don't understand why $e^{i \frac{\pi}{2} \cdot 6} = e^{i 3 \pi} = e^{i \pi + 2 \pi}$... Thank you! – Yoar Jan 12 '14 at 18:50
I was considering $x = \exp (i \pi/6 + \pi/3))$, which is to take $k=1$ in the formula $\exp (i (\pi/6 + (2 \pi k)/6))$ you gave. I was trying to demonstrate how the method works to show it is a root for each $k=0,1,2,3,4,5$, in particular I took $k=1$. Therefore, we have $\exp (i ( \pi/6 + (2 \pi 1)/6)) = \exp( i(\pi/6 + (2 \pi)/6)) = \exp ( i (\pi/6 + \pi/3)) = \exp (i ( (\pi + 2 \pi)/6))) = \exp (i(3 \pi/6)) = \exp (i \pi /2).$ Then I took the sixth power of each side, the sixth power of $x = \exp (i \pi/2)$. – Mark Fantini Jan 12 '14 at 19:47

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