Obviously, $x=1$ is one of the roots. There must be other roots who are complex number. I guess they are related to exponent. I don't know how to get them and how to prove it.
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2Do you know about complex numbers? Take a look at https://en.wikipedia.org/wiki/Root_of_unity – Jose Brox Nov 07 '17 at 15:50
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2Use polar coördinates, i.e. solve $(re^{i\theta})^n=1$ where $r>0$ and $\theta\in[0,2\pi)$. – drhab Nov 07 '17 at 15:53
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Complex numbers can be represent by 2 things, length and angle. We can write complex number $z$ like this: $z=re^{i\theta}$ where $r$ is the length and $\theta$ is the angle. So $z^n=\left(re^{i\theta}\right)^n=r^ne^{i\theta n}$
Now, because $r$ of $1$ is $1$ then $r^n=1^n=1$
Now because angle of $2\pi k,k\in \Bbb Z$ is the same for all $k$ I can have the equation $2\pi=\theta n\implies \frac{2\pi}{n}=\theta$. You can solve this equation with different k's (from k=0 to k=n-1) and this is the way to found all of the angles, now just plug them into $z=re^{i\theta}$ as $\theta$ and put $1$ as $r$ and done
ℋolo
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Hint: Write $z=x+iy=re^{i\phi}$ where $r$ is the absolute value of $z$ and $\phi$ its argument. Then think where $e^{i\phi}$ takes the value $1$.
Jakob Elias
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Observe that, since $e^{2\pi\cdot i}=1$, $e^{\frac{2\pi i}n}$ is a root of $x^n-1$. Therefore $$\left(e^{\frac{2\pi i}n}\right)^k=e^{\frac{2k\pi i}n}$$ is also a root. You can check that these are different for $0\leq k\leq n-1$, and since $x^n-1$ has $n$ roots in $\mathbb{C}$, those are indeed all the roots of $x^n-1$.
Jose Brox
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