Powers of roots of unity are also roots of unity?

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!

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

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.
• 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. Jun 12, 2022 at 21:49
• @RobArthan the OP says that i z is a fifth rot of unity, then $z^{1.06}$is also. Jun 18, 2022 at 2:57