Pole expansion and Fourier series of lemniscate sine function

Question

Given that $$\frac{\varpi}{\text{sl}(\varpi z)}=\sum_{n,k\in\mathbb{Z}}\frac{(-1)^{n+k}}{z+n+ik}$$

It can be deduced that, for $-1<\text{Im}(z)<1$:

$$\frac{1}{\text{sl}(\varpi z)}=\frac{\pi}{\varpi}\left[\frac{1}{\sin \pi z}-4\sum_{n=0}^{\infty} \frac{\sin((2n+1)\pi z)}{e^{\pi(2n+1)}+1}\right]$$

The proof is here

$$\sum_{n,k\in\mathbb{Z}}\frac{(-1)^{n+k}}{z+n+ik}=\sum_{k\in\mathbb{Z}}(-1)^k \sum_{n\in\mathbb{Z}}\frac{(-1)^{n}}{z+n+ik}$$

Note $\frac{\pi}{sin(\pi z)}=\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{z+n}$

$$\implies \sum_{k\in\mathbb{Z}}(-1)^k \frac{\pi}{\sin(\pi (z+ik))}=\frac{\pi}{\sin(\pi z)}+\sum_{k\leq -1}\frac{2 \pi i (-1)^k}{e^{i\pi (z+ik)}-e^{-i\pi (z+ik)}} +\sum_{k\geq 1} \frac{2\pi i (-1)^k}{e^{i\pi (z+ik)}-e^{-i\pi (z+ik)}} $$

$$=\frac{\pi}{\sin(\pi z)}+\sum_{k\geq 1}\frac{2 \pi i (-1)^k}{e^{i\pi (z-ik)}-e^{-i\pi (z-ik)}} +\sum_{k\geq 1} \frac{2\pi i (-1)^k}{e^{i\pi (z+ik)}-e^{-i\pi (z+ik)}} $$ $$=\frac{\pi}{\sin(\pi z)}+\sum_{k\geq 1}\frac{2 \pi i (-1)^ke^{-\pi(k+iz)}}{1-e^{-2\pi (k+iz )}} +\sum_{k\geq 1} \frac{2\pi i (-1)^{k+1} e^{-\pi (k-iz)}}{1-e^{-2\pi (k-iz)}}$$

For $-1<\text{Im}(z)<1$, the fractions can be written as geometric series as follows: $$=\frac{\pi}{\sin(\pi z)}+\sum_{k\geq 1}2\pi i(-1)^k e^{-\pi (k+iz)}\sum_{j\geq 0}e^{-2\pi j(k+iz)}-\sum_{k\geq 1}2\pi i(-1)^k e^{-\pi (k-iz)}\sum_{j\geq 0}e^{-2\pi j(k-iz)}$$ $$=\frac{\pi}{\sin(\pi z)}+\sum_{j\geq 0}2\pi i e^{-2\pi i j z}e^{-i\pi z}\sum_{k\geq 1}(-1)^k e^{-\pi k (2j+1)}-\sum_{j\geq 0}2\pi i e^{2\pi i j z}e^{i\pi z}\sum_{k\geq 1}(-1)^k e^{-\pi k (2j+1)}$$ $$=\frac{\pi}{\sin(\pi z)}+\sum_{j\geq 0}2\pi i \left[e^{-\pi i z(2j+1)}\Big(\frac{-e^{-\pi(2j+1)}}{1+e^{-\pi(2j+1)}}\Big)-e^{\pi i z(2j+1)}\Big(\frac{-e^{-\pi(2j+1)}}{1+e^{-\pi(2j+1)}}\Big)\right]$$

$$=\frac{\pi}{\sin(\pi z)}+\sum_{j\geq 0}(2i)^2 \pi \left[\frac{e^{\pi i z(2j+1)}-e^{-\pi i z(2j+1)}}{2i} \frac{1}{1+e^{\pi (2j+1)}}\right]$$ $$=\frac{\pi}{\sin(\pi z)}-4\pi \sum_{j\geq 0} \left[\frac{\sin((2j+1)\pi z)}{1+e^{\pi (2j+1)}}\right]$$

$$\implies \sum_{n,k\in\mathbb{Z}}\frac{(-1)^{n+k}}{z+n+ik}=\frac{\pi}{\sin(\pi z)}-4\pi \sum_{j\geq 0} \left[\frac{\sin((2j+1)\pi z)}{1+e^{\pi (2j+1)}}\right]$$

This finishes the proof

However, I wasn't able to show that $$\frac{\varpi}{\text{sl}(\varpi z)}=\sum_{n,k\in\mathbb{Z}}\frac{(-1)^{n+k}}{z+n+ik}$$

sl is the lemniscate analog of sine, and the pole expansion looks like the 2D version of $1/\sin$

I am trying to use Mittag Leffler expansion however I ran into some difficulties:

(1). I do not know where the zeros of sl are on $\mathbb{C}$. Hence, I do not know where the residues are at, nor the value of residues

(2). Apparently, sl is doubly periodic with the periods $\{(1+i)\varpi,(1-i)\varpi\}$, but I also am unable to prove this fact yet. For this, I already have an idea of using the argument sum formulas (this post).But again, I don't know the values of sl$(\varpi(1\pm i))$ and cl$(\varpi(1\pm i))$, so this idea os pended for now

Any help is appreciated

Answer 1

The fundamental domain of sl (now proved) is $\{(1+i)\varpi,(1-i)\varpi\}$ and the since sl$(0)=0$. It is true that $\text{sl}((n+m)\varpi +(n-m)i\varpi)=0$ for all $n,m \in \mathbb Z$. Furthermore, $\text{sl}(\varpi)$ is also $0$ which we didn't include from the above and can extend this to $\text{sl}((n+m+1)\varpi +(n-m)i\varpi)=0$, giving:

Zeros of sl: sl$(z\varpi)$ has zeros on $z\in\mathbb{Z}[i]=\{(a+bi):a,b\in\mathbb{Z}\}$, aka the Gaussian integers.

Now we consider the Residues of $\frac{\varpi}{\text{sl}(\varpi z)}$: Res$(\frac{\varpi}{\text{sl}(\varpi z)})$ for $z\in\mathbb{Z}[i]$

Case 1: $a+b$ is even, then sl$(z\varpi +\varpi (a+bi))=\text{sl}(\varpi z)$

$$\textbf{Res}\Big(\frac{\varpi}{\text{sl}(\varpi z)},z=a+bi \Big)=\lim_{z\to a+bi}\frac{\varpi}{\text{sl}(\varpi z)}(z-a-bi)=\lim_{z\to 0} \frac{\varpi z}{\text{sl}(\varpi z)}=1$$

Case 2 $a+b$ is odd, then sl$(z\varpi +\varpi (a+bi))=-\text{sl}(\varpi z)$

$$\textbf{Res}\Big(\frac{\varpi}{\text{sl}(\varpi z)},z=a+bi \Big)=\lim_{z\to a+bi}\frac{\varpi}{\text{sl}(\varpi z)}(z-a-bi)=\lim_{z\to 0} \frac{\varpi z}{-\text{sl}(\varpi z)}=-1$$

In other words $\textbf{Res}\Big(\frac{\varpi}{\text{sl}(\varpi z)},z=a+bi\Big)=(-1)^{a+b}$

Collecting these results, the Mittag pole expansion of $\frac{\varpi}{\text{sl}(\varpi z)}$ would probably look something like

$$\sum_{(a,b)\in \mathbb{Z}^2}\frac{(-1)^{a+b}}{z+a+bi}$$

The double period matches with $\frac{\varpi}{\text{sl}(\varpi z)}$.

We are almost there, consider $$\frac{\varpi}{\text{sl}(\varpi z)}-\sum_{(a,b)\in \mathbb{Z}^2}\frac{(-1)^{a+b}}{z+a+bi}$$

Consider ONLY the fundamental domain, since the poles are canceled out of each other, we are left with a bounded holomorphic function. However, we also know that $\frac{\varpi}{\text{sl}(\varpi z)}-\sum_{(a,b)\in \mathbb{Z}^2}\frac{(-1)^{a+b}}{z+a+bi}$ is doubly periodic, hence we may extend the boundness onto $\mathbb C$.

By Liouville's theorem, a bounded holomorphic function must be a constant. To find what constant is this, we use the formula from the question:

$$\sum_{n,k\in\mathbb{Z}}\frac{(-1)^{n+k}}{z+n+ik}=\frac{\pi}{\sin(\pi z)}-4\pi \sum_{n=0}^{\infty}\frac{\sin((2n+1)\pi z)}{e^{\pi (2n+1)}+1}$$ for $-1<\text{Im}(z)<1$

$$\implies c= \frac{\varpi}{\text{sl}(\varpi z)}-\Big(\frac{\pi}{\sin(\pi z)}-4\pi \sum_{n=0}^{\infty}\frac{\sin((2n+1)\pi z)}{e^{\pi (2n+1)}+1}\Big)$$

I would like to take $z\longrightarrow 0$.

Note their Taylor series:

sl$(z)=z+O(z^5)$ and sin$(z)=z+\frac{z^3}{6}+O(z^5)$

$$\implies c= \lim_{z\to 0} \frac{\varpi }{\text{sl}(\varpi z)}-\frac{\pi}{\sin(\pi z)}+0= \lim_{z\to 0}\frac{1}{z+O(z^5)}-\frac{1}{z-O(z^3)}=\lim_{z\to 0}\frac{1}{z}O(z^3)=0$$

This completes the Mittag pole expansion of $\frac{\varpi}{\text{sl}(\varpi z)}$:

$$\frac{\varpi}{\text{sl}(\varpi z)}=\sum_{n,k\in\mathbb{Z}}\frac{(-1)^{n+k}}{z+n+ik}$$