Complex exponential function vs Real Dirichlet (popcorn)

Solution 1:

Without loss of generality, $$\ln b=\operatorname{Log} b+2n\pi i$$ where the principal logarithm is taken and $n\in\mathbb Z$.

Then, $$e^{a\ln b}=e^{a\operatorname{Log} b}\cdot e^{2an\pi i}$$

The first part is obviously single-valued, so let’s discuss the single-valued-ness of the second part.

When $a\in\mathbb Z$:

$$an\in\mathbb Z\implies \cos(2an\pi)=1, \sin(2an\pi)=0 \implies e^{2an\pi i}=1\,\,\forall n$$ Thus the expression in single-valued.

When $a\not\in\mathbb Z$:

We will show that $e^{2an\pi i}- e^{2a(n+1)\pi i}\ne0$.

Considering real part: $$\cos(2an\pi)-\cos(2a(n+1)\pi=-2\sin(a\pi(2n+1))\sin(-2a\pi)$$

Since $(2n+1)a, -2a\not\in\mathbb Z$, the difference of two cosines is non-zero.

This shows that a change of choice of $n$ would result in the change of value of the expression, thus the expression is multi-valued.

In conclusion the iff statement is proved.