# Question on roots of unity

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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For non-integral exponents, how do you define the power? The identity $a^{bc} = (a^b)^c$ doesn't unconditionally hold. – Daniel Fischer Jul 6 '13 at 21:41
@DanielFischer The use of the letters $n$ and $k$ usually imply that both $n$ and $k$ are integers. – Fly by Night Jul 6 '13 at 21:45
@FlybyNight: he is actually "proving" that $z_k=1$. – Martin Argerami Jul 6 '13 at 21:49
@MartinArgerami Indeed he is. Well-spotted. – Fly by Night Jul 6 '13 at 21:50
@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 6 '13 at 22:02

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 $f^{-1}(1)=1$.

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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.

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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 6 '13 at 21:55
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 6 '13 at 22:11
@FlybyNight: yes. – Martin Argerami Jul 6 '13 at 22:11
@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 6 '13 at 22:13
@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: en.wikipedia.org/wiki/Ordered_field – Gerry Myerson Jul 7 '13 at 9:36