Firstly, proof verification on the values of $e^{b \ln a}$:
(Exer 3.51) Part I: Prove $e^{b \ln(a)}$ is single valued $\iff b \in \mathbb Z$. Part II: What if $b \in \mathbb Q$?
I tried:
Part I ($b \in \mathbb Z$)
Pf:
By definition of $a^b$:
$$a^b:=e^{bLn(a)} = e^{b(\ln|a|+iArg(a))} = e^{b\ln|a|}e^{ibArg(a)}$$
By definition of $e^{b\ln(a)}$, $\exists k \in \mathbb Z$ s.t.
$$e^{b\ln(a)} = e^{b(\ln|a|+i\arg(a))} = e^{b\ln|a|}e^{ibArg(a)}e^{2kbi\pi} =: a^be^{2kbi\pi}$$
$\therefore, e^{b\ln(a)}$ is single valued $\iff e^{b\ln(a)} = a^b \iff 1 = e^{2kbi\pi}$
Now $$1 = e^{2kbi\pi} = \cos(2kbi\pi) + i\sin(2kbi\pi) \iff$$
$$\cos(2kbi\pi) = 1 \wedge \sin(2kbi\pi) = 0 \iff bk \in \mathbb Z \iff b \in \mathbb Z$$
$\therefore, e^{b\ln(a)}$ is single valued $\iff e^{b\ln(a)} = a^b \iff b \in \mathbb Z$
QED
Question 1. Where have I gone wrong for Part I, if anywhere?
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Part II ($b \in \mathbb Q$)
Let $b:=\frac pq$, where $p,q$ are coprime positive integers (For any negative, $\cos$ is even and $\sin$ is odd). Now consider $\cos$:
$$\cos(2kbi\pi) := \cos(2k \frac pq i\pi)$$
whose value is different depending where $p$ is in $\{\overline{0},\overline{1}, \dots, \overline{q-1}\}$
$\therefore, e^{b\ln(a)}$ has $q$-values because $p$ has $q$-values $\mod \ q$.
Question 2. Where have I gone wrong for Part II, if anywhere?
Now that that's over with:
Question 3. (Something I noticed on my own in re Exer 3.51 but not asked in Exer 3.51)
What's the connection to the popcorn function aka Thomae's function aka modified Dirichlet $\frac1q 1_{x=\frac pq}$?
In Part II of Exer 3.51, $q$ is extracted from $\frac pq$.
In the popcorn function from elementary real analysis, $\frac1q$ is extracted from $\frac pq$. It seems the popcorn function is computing number of solutions $q$ to a complex equation and then inverting the computation $\frac 1 q$.
Question 4. (Also something I noticed on my own in re Exer 3.51 but not asked in Exer 3.51)
What exactly is the connection to solving the roots of unity?
The roots of unity are the solutions of $z^n=1$, which turn out to be $z=e^{2k\pi\frac{m}{n}}$ whose value is different depending where $m$ is in $\{\overline{0},\overline{1}, \dots, \overline{n-1}\}$. This looks similar to Part II.
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.