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I'm studying for my linear algebra exam and I came across this exercise that I can't solve.

Find all roots of polynomial $x^{6} + 1$. Hint: use De Moivre's formula.

I guessed that two roots are $i$ and $-i$, since:

$i^{6} = (i^{2})^{3} = (-1)^{3} = -1 $

therefore, $i$ is root and his complex conjugate $-i$ has to be root too. However that was just guessing. I have no idea how can I use De Moivre's formula here.

Can you help me solve this?

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up vote 6 down vote accepted

Hint: if $x^6=-1$, then $|x|^6=1$ and you can write $x=\cos\theta + i\sin\theta$.


Then the equation is, thanks to De Moivre theorem and $\cos^2 + \sin^2 =1$, equivalent to $$ \cos 6\theta =-1\\ 6\theta = \pi\mod 2\pi\\ \theta\in \frac \pi 6+\left\{0, \frac\pi 3, \frac{2\pi}3,\pi,\frac{4\pi} 3, \frac{5\pi}3 \right\}. $$

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Watch it: the equation is $\;x^6+1=0\iff x^6=\color{red}-1\;$ ... – DonAntonio Mar 29 '14 at 15:57
Check also the values of $\;\theta\;$, @mookid ... – DonAntonio Mar 29 '14 at 16:00
@mookid this doesn't yield either $i$ or $-i$, you might want to check your calculation. – Sabyasachi Mar 29 '14 at 16:02
@mookid yes it is. – Sabyasachi Mar 29 '14 at 16:05
Of course, simple and elegant. Thank you – isklenar Mar 29 '14 at 16:09

I'd like to add that this is possible to solve using well known formulas for $x^n\pm a^n$: $$\begin{align}x^6+1&=(x^3)^2-i^2=(x^3+i)(x^3-i)=(x^3-i^3)(x^3+i^3)\\&=(x-i)(x^2+ix+i^2)\cdot(x+i)(x^2-ix+i^2)\end{align}$$

And now you can use the formula for quadratic roots to get them all in an explicit form.

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Also,$x^6+1=(x^2)^3+1^3=(x^2+1)(x^4+x^2+1)=0$. Now we have to solve two quadratic equations to obtain the required roots. – rah4927 Mar 29 '14 at 16:59

If $z^6 + 1= 0$ then $z^6 = -1$. We can write $-1 = \mathrm{e}^{\mathrm{i}(\pi+2\pi n)}$ where $n \in \mathbb{Z}$. It follows that $$(-1)^{1/6} = \{\mathrm{e}^{\mathrm{i}(\pi/6+\pi n/3)} : n \in \mathbb{Z}\}$$ Putting $n=0,1,2,3,4,5$ will give you all of the solutions you need. For example, when $n=1$: $$z = \mathrm{e}^{\mathrm{i}(\pi/6+\pi /3)} = \mathrm{e}^{\mathrm{i}(\pi/2)} = \cos\tfrac{\pi}{2}+\mathrm{i}\sin\tfrac{\pi}{2} = \mathrm{i}$$

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The easiest way to do these problems is using the polar form of a complex number.

$$ x^6 + 1 = 0 \rightarrow x^6 = -1 $$

Write $-1$ as a complex number in polar form with the $+2\pi n$ (the most general form), then write $x$ as a general complex number:

$$ \left(re^{i\theta}\right)^6 = 1e^{\pi i + 2\pi n i} \\ r^6 e^{6\theta i} = 1 e^{\left(\pi + 2\pi n\right)i} $$

From here, $r = 1$ (because $x^6 = 1$ has only one real root: $x = \sqrt[6]{1} = 1$). Then you just set the angles equal:

$$ 6\theta = \pi + 2\pi n \\ \theta = \frac{\pi + 2\pi n}{6} = \frac{\pi}{6} + \frac{\pi}{3}n $$

Now just keep incrementing $n$ until you start looping (which will happen after $6$ consecutive values of $n$):

$$ \theta = \frac{\pi}{6}, \frac{\pi}{6} + \frac{\pi}{3}, \frac{\pi}{6} + \frac{2\pi}{3}, ... $$

If you want to write them as $a +bi$ then you have to convert each of those to $r\cos(\theta) + ri\sin(\theta)$.

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