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So I was thinking about roots of unity, loosely inspired by this video (17:28 ff.), in which the author, as best I understand it, says:

Suppose $z$ is a 5th root of unity. So, by definition, $z^5=1$. But positive integer powers of $z$ are also 5th roots of unity, due to the way that powers of $z$ cycle around the unit circle.

I tried to come up with an argument that justifies the above statement. I.e. the statement that $z^2, z^3, z^4$ are also 5th roots of unity if $z$ is:

If e.g. $z^2$ is a 5th root of unity, then $\left(z^2\right)^5=1$. But, due to exponent properties, this is the same as $\left(z^5\right)^2=\left(1\right)^2=1$. So $z^2$ is indeed a 5th root of unity. And similarly for $z^3$ and $z^4$.

But there seems to be a problem with this argument, right? Because what about the pseudo-argument:

If $z^{1.06}$ is a 5th root of unity, then $\left(z^{1.06}\right)^5=1$. But, due to exponent properties, this is the same as $\left(z^5\right)^{1.06}=\left(1\right)^{1.06}=1$. So $z^{1.06}$ is indeed a 5th root of unity. And similarly for $z^{1.07}$, $z^{\pi}$, and so on.

But that seems to imply that there are continuum many 5th roots of unity, whereas there are only supposed to be 5 of them. Where am I going wrong?

Thanks!

RobPratt
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    'positive integer powers of $z$...' – peek-a-boo Jun 11 '22 at 19:53
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    @peek-a-boo You appear to have missed the point - kindly read the question – Stephen Donovan Jun 11 '22 at 19:54
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    @StephenDonovan well, that comment suggests there's something specific about integers which is being exploited, and hence OP should think about which specific property is being messed up by a fraction like $1.06$. (ok but perhaps my first comment was too cryptic to indicate this) – peek-a-boo Jun 11 '22 at 19:56
  • @peek-a-boo Fair enough - I think you can understand though how it can read like you were misunderstanding OP's question as trying to show that non-integer powers of roots of unity should also be roots of unity. In any case I agree that the issue here has to do with what happens when we take non-integer powers – Stephen Donovan Jun 11 '22 at 20:01
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    Regarding the "formula" $(z^s)^t = (z^t)^s$, let me suggest that you investigate the domain of values of $s$ and $t$ on which the terms of that formula are well-defined. – Lee Mosher Jun 11 '22 at 20:02
  • This is a nice question: easy to understand, hard to think about. +1 – insipidintegrator Jun 11 '22 at 20:04
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    Not at all hard to think about @insipidintegrator : fractional powers are completely undefined except when the base is a positive real number. – Lubin Jun 11 '22 at 20:09
  • Uhh…I never knew it. Thanks! – insipidintegrator Jun 11 '22 at 20:11
  • Wait. $\sqrt[3]{-1}$ is undefined? – insipidintegrator Jun 11 '22 at 20:11
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    It sounds like you have got it swapped around a bit. Based on your characterization, the argument appears to be: Suppose $z$ is a $k$th root of unity. Then $z^k = 1$. Now consider $z^m$. Is it also a root of unity? We have $(z^m)^k = z^{mk} = (z^k)^m$ by associativity, and that must equal $1$. – Brian Tung Jun 11 '22 at 20:13
  • @insipidintegrator: It's not uniquely defined. It can be multivalued, with values equal to $-1$, $-\omega$, and $-\omega^2$, where $\omega$ is a non-real cube root of $1$. – Brian Tung Jun 11 '22 at 20:14
  • Then $\sqrt[3]{1}$ must also not be well defined. @BrianTung how do you differentiate between the two cases? Is it just a matter of convention ? – insipidintegrator Jun 11 '22 at 20:15
  • @insipidintegrator: It's more or less the situation with $i$ and $-i$. They are collectively defined as numbers who square to $-1$, so that they are, weirdly, distinct (in the sense that you can't change one for the other freely on an appearance-by-appearance basis) yet indistinguishable (we may have been calling the "wrong one" $i$ for centuries and it wouldn't make any difference). More formally, we can say that complex conjugation is a field automorphism of $\mathbb{C}$, so that (for example) the truth of an equation is unchanged if all instances of $i$ are changed to $-i$ and vice versa. – Brian Tung Jun 11 '22 at 20:24
  • My comment to OP should also include that (power) associativity only works provided we're dealing with integer powers of $z$! – Brian Tung Jun 11 '22 at 20:26
  • Regarding the cube root function, it is well-defined as a function with domain $\mathbb R$ and codomain $\mathbb R$, and indeed with that domain and codomain we have $\sqrt[3]{1}=1$ and $\sqrt[3]{-1}=-1$. But the cube root function is not well-defined as a function from $\mathbb C$ to $\mathbb C$. Indeed, with that domain and codomain even $\sqrt[3]{1}$ is not well-defined: as $z$ ranges over the complex numbers the equation $z^3=1$ has three roots $1$, $\omega=-\frac{1}{2} + \frac{\sqrt{3}}{2} i$, and $\omega^2=-\frac{1}{2} - \frac{\sqrt{3}}{2}$. – Lee Mosher Jun 12 '22 at 02:36
  • More generally for any nonzero complex number $w$ the equation $z^3=w$ has three roots, and it is not possible to pick out one of those roots for each $w$ in such a fashion as to define a continuous function $z = \sqrt[3]{w}$. Similar crap happens for other real valued exponents, and this lack of well-definedness creates all sorts of havoc for ordinary formulas like $(z^s)^t = (z^t)^s$. – Lee Mosher Jun 12 '22 at 02:43
  • @s7eqx9f74nc4 have you considered that $1^{1.06}$ may admit complex values? – insipidintegrator Jun 18 '22 at 03:00

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You write:

If $z^{1.06}$ is a 5th root of unity, then $\left(z^{1.06}\right)^5=1$. But, due to exponent properties, this is the same as $\left(z^5\right)^{1.06}=\left(1\right)^{1.06}=1$.

The problem here is not to do with issues about the well-definedness of fractional powers. The problem is your claim that $(z^5)^{1.06} = (1)^{1.06}$? Your assumption that $z^{1.06}$ is a $5$th root of unity doesn't imply that $z$ is a $5$th root of unity.

Rob Arthan
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