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Question

claim: $w_0,w_1,w_2.....w_{n-1}$ is n of many nth rot of unity when $w_0=1$, prove $w_k=(w_1)^k$

What i have so far

well we can first plug and chug some random numbers:

let's say that $k=0$

then we get the following: $w_0=(w_1)^0=1$

and plugging in 1 as k: $w_0=(w_1)^1=w_1$

But now I don't know how to go from here to prove the statement

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    How are $w_k$ defined? The statement does not hold true without additional conditions, take for example $n=4$ and $w_0=1\,$, $w_1=-1\,$, $w_2=i\ne w_1^2\,$. – 2017-02-05
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    I suspect you have already been given $\omega_k = \cos \frac {2k\pi}n + i\sin \frac {2k\pi}n$. If so, then this problem refers specifically to that expression, which avoids the ordering problem dxiv is referring to. You can prove the result with De Moivre's theorem: $\cos n\theta + i\sin n\theta = (\cos \theta + i\sin \theta)^n$. (If you were given the formula $\omega_k = e^{i\frac {2k\pi}n}$, then the problem is a trivial application of the rules of exponents.) – 2017-02-05
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    @dxiv actually i did set up statements and in know that this is solvable because it came from a textbook. – 2017-02-05
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    @JohnRawls This doesn't answer my question: "*how are $w_k$ defined?*". Without that information the statement is false, and I gave you an explicit counterexample in my first comment. – 2017-02-05
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    @dxiv are you sure because it came straight from a textbook, i just dont know how to solve it – 2017-02-05
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    @JohnRawls There is something missing, either in the question, or in the translation. If you don't know which root $w_k$ is, then the problem cannot be solved. If you have additional information (such as, for example, that $\arg w_k = 2 k \pi / n$), then see Paul Sinclair's comment. – 2017-02-05

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