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# a conjectured continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$

Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)$ is true

$$\begin{aligned}\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\frac{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}=\cfrac{2z}{2z+n+\cfrac{(n)(4z+n)} {3(2z+n)+\cfrac{(2z+2n)(6z+2n)}{5(2z+n)+\cfrac{(4z+3n)(8z+3n)}{7(2z+n)+\ddots}}}}\end{aligned}$$

Corollaries:

By taking the limit(which follows after abel's theorem) $\begin{aligned}\lim_{z\to0}\frac{\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)}{2z}=\frac{\pi}{4}\end{aligned}$, we recover the well known continued fraction for $\pi$

$$\begin{aligned}\cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}}=\pi\end{aligned}$$

If we let $z=1$ and $n=2$,