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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!

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    $\begingroup$ 'positive integer powers of $z$...' $\endgroup$
    – peek-a-boo
    Jun 11, 2022 at 19:53
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    $\begingroup$ @peek-a-boo You appear to have missed the point - kindly read the question $\endgroup$
    – Stephen Donovan
    Jun 11, 2022 at 19:54
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    $\begingroup$ @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) $\endgroup$
    – peek-a-boo
    Jun 11, 2022 at 19:56
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    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Jun 11, 2022 at 20:02
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    $\begingroup$ Not at all hard to think about @insipidintegrator : fractional powers are completely undefined except when the base is a positive real number. $\endgroup$
    – Lubin
    Jun 11, 2022 at 20:09

1 Answer 1

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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.

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  • $\begingroup$ Sorry, maybe I phrased it in a confusing way, what I intended to mean was the argument as phrased by Brian Tung here math.stackexchange.com/questions/4470544/…. My basic question is what (rule, axiom, blabla) prevents me from making $m$ non-integer? Because if I can make it non-integer, then it seems like I can end up with more than five 5th roots of unity. $\endgroup$
    – s7eqx9f74nc4
    Jun 12, 2022 at 21:49
  • $\begingroup$ @RobArthan the OP says that i z is a fifth rot of unity, then $z^{1.06} $is also. $\endgroup$
    – insipidintegrator
    Jun 18, 2022 at 2:57

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