How to find this $\prod_{k=-\infty}^{+\infty}\frac{x^2+(4k+1-y)^2}{x^2+(4k+3-y)^2}$ let $x,y$ be real numbers,show that
$$\prod_{k=-\infty}^{+\infty}\dfrac{x^2+(4k+1-y)^2}{x^2+(4k+3-y)^2}=\dfrac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}$$
This is a question that a physicist asked me. He said that he could prove the equation in a physical way, but not in a mathematical way. I thought about it for a while and might need to use the following conclusion to solve it, but it didn't work out
$$\prod_{k=1}^{+\infty}\left(1+\dfrac{1}{k^2}\right)=\dfrac{\sinh{\pi/2}}{\pi/2}$$？
 A: Firstly, define the complex number $z=x+yi$. Then, we are asked to prove that
$$
\prod_{k\in\mathbb{Z}}\left|\frac{z-(4k+1)i}{z-(4k+3)i}\right|^2=\frac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}.
$$
Now, let's rewrite the right-hand side of this equality:
$$
\frac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}
=
\frac{\left(e^{\frac{\pi}{2}x}-\sin\dfrac{\pi}{2}y\right)^2+\cos^2\dfrac{\pi}{2}y}{\left(e^{\frac{\pi}{2}x}+\sin\dfrac{\pi}{2}y\right)^2+\cos^2\dfrac{\pi}{2}y}
=
\left|\frac{e^{\frac{\pi}{2}x}+i\left(\cos\dfrac{\pi}{2}y+i\sin\dfrac{\pi}{2}y\right)}{e^{\frac{\pi}{2}x}-i\left(\cos\dfrac{\pi}{2}y+i\sin\dfrac{\pi}{2}y\right)}\right|^2=
\\
=\left|\frac{e^{\frac{\pi}{2}x}+ie^{i\frac{\pi}{2}y}}{e^{\frac{\pi}{2}x}-ie^{i\frac{\pi}{2}y}}\right|^2=\left|\frac{e^{\frac{\pi}{2}(x-iy)}+i}{e^{\frac{\pi}{2}(x-iy)}-i}\right|^2
=\left|\frac{e^{\frac{\pi}{2}\overline{z}}+i}{e^{\frac{\pi}{2}\overline{z}}-i}\right|^2=\left|\frac{e^{\frac{\pi}{2}z}-i}{e^{\frac{\pi}{2}z}+i}\right|^2.
$$
Thus, we need to prove that
$$
\prod_{k\in\mathbb{Z}}\left|\frac{z-(4k+1)i}{z-(4k+3)i}\right|=\left|\frac{e^{\frac{\pi}{2}z}-i}{e^{\frac{\pi}{2}z}+i}\right|.
$$
Let $w=z-i$, then the last equality can be rewritten as
$$
\prod_{k\in\mathbb{Z}}\left|\frac{w-4ki}{w-(4k+2)i}\right|=\left|\frac{e^{\frac{\pi}{2}w}-1}{e^{\frac{\pi}{2}w}+1}\right|.
$$
Update. As David mentioned in the comments, in order to compute the product $\prod_{k\in\mathbb{Z}}\frac{t-4k}{t-(4k+2)}$ we can use the following formula:
$$
\frac{\sin\pi z}{\pi z}=\prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2}\right),
$$
or
$$
\sin\pi z=\pi z\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{k-z}{k}.
$$
For $z=t/4$ and $z=(t-2)/4$ we have
$$
\sin\frac{\pi t}{4}=\frac{\pi t}{4}\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{4k-t}{4k}
$$
and
$$
\sin\frac{\pi (t-2)}{4}=\frac{\pi (t-2)}{4}\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{4k+2-t}{4k},
$$
respectively. Hence,
$$
\frac{\sin\pi t/4}{\sin\pi (t-2)/4}=\prod_{k\in\mathbb{Z}}\frac{t-4k}{t-(4k+2)}.
$$
Therefore,
$$
\prod_{k\in\mathbb{Z}}\left|\frac{w-4ki}{w-(4k+2)i}\right|=[t=w/i]=\prod_{k\in\mathbb{Z}}\left|\frac{t-4k}{t-(4k+2)}\right|=\left|\frac{\sin\pi t/4}{\sin\pi (t-2)/4}\right|=
\\
=\left|\frac{e^{i\pi t/4}-e^{-i\pi t/4}}{e^{i\pi (t-2)/4}-e^{-i\pi (t-2)/4}}\right|=\left|\frac{e^{i\pi t/4}-e^{-i\pi t/4}}{-ie^{i\pi t/4}-ie^{-i\pi t/4}}\right|=\left|\frac{e^{\pi w/2}-1}{e^{\pi w/2}+1}\right|,
$$
so we're done.
