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I have a such equation: $$ {x}^{n}-1=0 $$ I have n complex roots. For example:$${x}^{7}-1=0$$ $${x}_{1}=1; {x}_{2} = {(-1)}^{\frac{1}{7}}; {x}_{3}=-{(-1)}^{\frac{2}{7}}; ...;{x}_{7}={(-1)}^{\frac{6}{7}}$$ Then I could say, that one of this roots is primitive. Obviously, it's $${x}_{2} = {(-1)}^{\frac{1}{7}},$$ because all powers of this root enumarate rest of roots. So, I care about theorem, that roots $${x_2}^{0};{x_2}^{1};{x_2}^{2};{x_2}^{3};{x_2}^{4};{x_2}^{5};{x_2}^{6}$$ do not equal between themself. Below I am going to show a proof of the statement:

Let a is primitive root. Let $${a}^{k}={a}^{m}, k \neq m$$ Then $${a}^{k-m}=1,$$but $$(k-m)<n$$and, therefore cannot be devided on n . Consequently$${a}^{k}\neq{a}^{m}.$$ Do you find it plain, that if you raise a number to a power, you cannot get two numbers with similar powers? Why that proof is established on the such suppose? It's obvious that I cannot raise in such power the number $${x}_{2}$$ that I obtain, for example, that:$${{x}_{2}}^{2}={(-1)}^{\frac{2}{7}}$$ or that: $${{x}_{2}}^{4}={(-1)}^{\frac{4}{7}}$$ and that they will be equal. Despite my desire, it's impossible to get two equal numbers that have different powers, if I use powers less than n. How did it build the proof?

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closed as unclear what you're asking by mrtaurho, hardmath, Leucippus, Lord Shark the Unknown, ancientmathematician Feb 8 at 7:39

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ It might help to do a small example, say $n=4$ or $n=6$ to understand that while you can get some powers of the roots to be equal below $n$, not all of the roots will have the same power. A primitive root $r$ will be one that has all its powers distinct up through $r^n = 1$. $\endgroup$ – hardmath Feb 7 at 14:58
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I feel like Adwin's suggestion of using Polar coordinates may be the ``best'' approach here, but there is an alternative method: In general, a polynomial has a repeated root if and only if it has a common factor with its derivative (this is an easy exercise). Then observe that $x^n-1$ is relatively prime to its derivative, $nx^n$, so $x^n-1$ has no repeated roots.

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A quick look suggests that $(-1)^{\frac{1}{7}}$ is not a root of the equation. Neither is $x_3$.

Have you learned polar form of complex numbers? I guess that knowledge will help. https://en.m.wikipedia.org/wiki/Complex_number#Polar_complex_plane

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