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!
Best Answer
You write:
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.