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

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

** 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 $n=2$,then we have the square root of $2$ $$\begin{aligned}{1+\cfrac{1}{2+\cfrac{1} {2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}}}=\sqrt{2}\end{aligned}$$

** Q**: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $x\gt0$?

** Update**:I initially defined the continued fraction $\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.