This may seem absurd but what is wrong with the next reasoning about $n$th roots of unit?. For $k,l\in\mathbb Z$ such that $0 \leq k < l \leq n-1$: $$ e^{2\pi i k/n} = (e^{\pi i})^{2 k /n} \overset{\text{Euler's identity}}{=} (-1)^{2 k /n} = ((-1)^2)^{k/n} = 1^{k/n} = 1 $$ so $e^{2\pi i k /n} = e^{2\pi i l /n}$ since it is the same reasoning for $l$. Thanks for your help, I actually don't see what is wrong.
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3For non-integral exponents, how do you define the power? The identity $a^{bc} = (a^b)^c$ doesn't unconditionally hold. – Daniel Fischer Jul 06 '13 at 21:41
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@DanielFischer The use of the letters $n$ and $k$ usually imply that both $n$ and $k$ are integers. – Fly by Night Jul 06 '13 at 21:45
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2@FlybyNight: he is actually "proving" that $z_k=1$. – Martin Argerami Jul 06 '13 at 21:49
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@MartinArgerami Indeed he is. Well-spotted. – Fly by Night Jul 06 '13 at 21:50
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@FlybyNight But $2k/n$ isn't an integer. – Daniel Fischer Jul 06 '13 at 21:53
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@SparkSm. Your question refers to a post make by Martin. Perhaps ask him? – Fly by Night Jul 06 '13 at 21:58
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@DanielFischer I didn't say it was, did I? – Fly by Night Jul 06 '13 at 21:59
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1@FlybyNight No, but then I don't see how the fact that $n$ and $k$ are usually used for integers is relevant at all. – Daniel Fischer Jul 06 '13 at 22:02
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@MartinArgerami This confuses me, you say that I am "proving" that $z_k = 1$ but in your answer you explain why this relation does not hold. What I'm proving then? Obviously this reasoning is wrong. – Spark Sm. Jul 06 '13 at 22:03
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Yes, that's why I wrote "proving" between quotes. You wrote a proof of equality $z_k=1$. The proof is wrong, and moreover it is not true that $z_k=1$ in general. – Martin Argerami Jul 06 '13 at 22:04
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Here is a simpler example:
$$
-1=(-1)^{\frac22}=\left((-1)^{2}\right)^{\frac12}=1^{\frac12}=1.
$$
Do you see the problem here? The problem is that $1^{\frac12}$ is the solution of the equation $x^2=1$ which is not unique and you are choosing the wrong solution.
Is like we are having a function $f$ which is not 1-1 (in my example $f=x^2$), lets say that $f(a_1)=f(a_2)=r$ and $a_1\neq a_2$, and we are arguing that $$a_1=f^{-1}\left(f(a_1)\right)=f^{-1}(r)=a_2$$ which is wrong.
In your case you are using the function $f=x^n$ to deduce that $f^{-1}(1)=1^{\frac1n}=1$ which is wrong. One of the solutions of $f^{-1}(1)$ is $e^{2\pi i k/n}$ and you are choosing $f^{-1}(1)=1$.
P..
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There is no canonical way of defining non-integer powers of complex numbers other than the non-negative reals. So the problem lies when you want to write $z^{st}=(z^s)^t$. You are actually providing an example that such equality does not hold in general. To make the example even simpler, you could have written $$ e^{2i\pi r}=(e^{2\pi i})^r=1^r=1. $$ This of course not true, and it just shows that the relation does not hold.
Martin Argerami
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Could you go into more detail please? What is a non-positive complex number? Do you mean complex numbers with a non-positive real part? – Fly by Night Jul 06 '13 at 21:55
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In the case $(e^{2\pi i k/n})^n$ with $k\in\mathbb Z^+\cup{0}$ it holds, right? Should clarify that in your example $r\in\mathbb R\setminus(\mathbb Z^+ \cup {0})$? – Spark Sm. Jul 06 '13 at 22:11
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@SparkSm. When $r$ is a positive integer, the equality I wrote is true. When it is not, it isn't. The point is that if you assume that you can do $z^{st}=(z^s)^t$ you get the equality I wrote, which is false in most cases. – Martin Argerami Jul 06 '13 at 22:13
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@FlybyNight: ok. It is very natural to me. The complex numbers are an ordered field, where the order is given by the cone of the non-negative real numbers. When you have an order, "$z\geq0$" is generally referred to as "positive" (or, more properly, "non-negative"). In any case, I have edited that. – Martin Argerami Jul 06 '13 at 22:15
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@MartinArgerami This is getting really confusing. You wrote a reply saying "No" and mentioning cones. So I replied. By the time I replied you'd deleted that and simply typed "Yes". So I deleted my reply to your deleted reply. Now you have replied to my deleted reply to your deleted reply. Do you feel dizzy? I know that I do :-\ – Fly by Night Jul 06 '13 at 22:17
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I'm not deleting anything else :D I first answered "no" because I misread your statement, as I'm not used to explicitly think on what "non-positive" complex numbers look like. But you had it right. – Martin Argerami Jul 06 '13 at 22:38
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1@Martin, "the complex numbers are an ordered field"? Not according to the definitions with which I am familiar. I hate to cite Wikipedia as an authority, but it's all I can access right now: http://en.wikipedia.org/wiki/Ordered_field – Gerry Myerson Jul 07 '13 at 09:36
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The complex numbers have a very natural order, given by the cone of non-negative real numbers. So we can see the complex numbers as a field with an order, and I would personally call that an "ordered field". Of course the order is not total, which is what the Wikipedia definition requires. – Martin Argerami Jul 07 '13 at 16:37