# Complex exponential function vs Real Dirichlet (popcorn)

## 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?

-

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.

• Please mark places from the book Aug 13, 2018 at 20:27
• @YuriNegometyanov What do you mean by 'mark places' please?
– BCLC
Aug 14, 2018 at 5:30
• To understand the question, I need firstly to understand the logic of the author of the book. For this, I need to see which part of the text is a quote from the book and which one is not. Aug 14, 2018 at 6:04
• @YuriNegometyanov Ok I added back the link to the text for Exer 3.51. That's about it. Thanks
– BCLC
Aug 14, 2018 at 6:08

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.

• Thanks Szeto, but (1) where exactly did I go wrong, if anywhere? (2) What's the connection to the popcorn function? (3) What is the relationship of proving this result and solving for the roots of unity?
– BCLC
Aug 5, 2018 at 11:50
• @BCLC Sorry, but what is popcorn function? Aug 5, 2018 at 12:10
• Szeto, no need to be sorry: Thomae's function, named after Carl Johannes Thomae, has many names: the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function, the Riemann function, or the Stars over Babylon (John Horton Conway's name).
– BCLC
Aug 5, 2018 at 12:24
• @BCLC Thanks! Actually, why/where do you think you go wrong? Aug 5, 2018 at 12:27
• Szeto, just checking. Nowhere is an answer :p Do you have any answers for 2,3 or 4? I edited question to hopefully further clarify.
– BCLC
Aug 14, 2018 at 6:10