It is known that if $\alpha$ is a complex number, then for example, the equation $x^2 = \alpha$ has $2$ solutions. In general, there are $n$ values for the $n$-th roots of a number. In other words, if we wanted to evaluate, say $\alpha^{\frac{12}{17}}$, we would have, again, $17$ different values.

However, what if we wanted to evaluate $\alpha ^ \beta$, where $\beta$ is an irrational number ($\beta \in \mathbb{C} \setminus \mathbb{Q}$)? Are there infinitely many numers $z$ such that $\sqrt[\leftroot{-2}\uproot{2}\beta]{z} = \alpha$

I am aware that, for complex numbers, one defines $\alpha^\beta = e^{\beta\log{\alpha}}$,and by choosing different values of the logarithm (as it is a multivalued function), we can get different values for our power.

My guess is: if we raise a number $\alpha$ to a rational exponent $\beta$, then the different values of $\beta \log \alpha$ will eventually begin to repeat themselves, since $\beta \in \mathbb{Q}$. More specifically, we can put $\log \alpha= \log |\alpha| +i\theta + 2k\pi i$, for $k \in \mathbb{Z}$, so when we evaluate $\alpha^\beta = e^{{\beta\log{\alpha}}} = e^{\beta (\log |\alpha| + i\theta + 2k\pi i)}$ as $k$ runs over the integers, it will only create a finite set of values. However, when $\beta$ is irrational (or at least one of $\Re{(\beta)}$ or $\Im{(\beta)}$ is), this set of values cannot be finite.

Is my interpretation correct? Could this type or argument work for showing that certain "polynomials" with irrational powers have an infinite number of solutions (i.e $z^\sqrt{2} + z^\sqrt{3} + 1 = 0$)?

  • $\begingroup$ The last question is ambiguous - as functions let $f(z)=z^\sqrt{2} + z^\sqrt{3} + 1$; find its roots, then you must specify the domain of definition (need to take out a piece of the plane like a ray at the minimum) and the branch of $\log z$ you use there since $z^\sqrt{2}=\exp{(\log z)(\sqrt{2})}$ by definition; but you can also think in terms of pointwise sets, in the sense that $z^\sqrt{2}$ is an infinite set in general for any $z \ne 0$, same with $z^\sqrt{3}$, so it makes sense to talk about the infinite set $z^\sqrt{2} + z^\sqrt{3} + 1$ and find for which $z$ it contains $0$ $\endgroup$
    – Conrad
    May 15, 2019 at 1:30

1 Answer 1


Note that to have $$ e^{\beta (\log |\alpha| + i\theta + 2k\pi i)}=e^{\beta (\log |\alpha| + i\theta + 2j\pi i)} $$ you need to have $$ 2k\beta\pi = 2j\beta\pi+ 2m\pi $$ for some $m\in\mathbb Z$. Then you would have $$ \beta=\frac m{k-j}\in\mathbb Q. $$ That is, when $\beta$ is irrational you get infinitely many different roots.

The question about the "irrational polynomial" is a good one, but I don't see an obvious method to find its roots.


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