<s>[INST] conjectured general continued fraction for the quotient of gamma functions Given complex numbers $a=x+iy$, $b=m+in$ and a gamma function $\Gamma(z)$ with $x\gt0$ and $m\gt0$, it is conjectured that the following general continued fraction which is symmetric on $a$ and $b$ is true
$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}\tag{1}$$
And can be further generalised to
$$\frac{1}{\sqrt{c}}\tan\left(\frac{b-a}{b+a}\tan^{-1}\left(\frac{\sqrt{c}}{d}\right)\right)=\cfrac{(b-a)}{d(a+b)+\cfrac{c(2a)(2b)} {3d(a+b)+\cfrac{c(3a+b)(a+3b)}{5d(a+b)+\cfrac{c(4a+2b)(2a+4b)}{7d(a+b)+\ddots}}}}\tag{2}$$   
Corollaries:
(i) Specializing to $a=1/2$ and $b=2z/n+1/2$,we obtain
the continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$ found in this post after applying the functional equation of the gamma function
(ii) and specializing further to $a=-1/2$ and $b=2z/n+3/2$,we obtain immediately the continued fraction for $\cot\left(\frac{z\pi}{4z+2n}\right)$ found here after applying the functional equation of the gamma function.This continued fraction was proved by @GEdgar.
(iii) letting $2a=m-n$ and $2b=m+n$,we find $$\displaystyle\tan\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2-n^2} {3m+\cfrac{4m^2-n^2}{5m+\cfrac{9m^2-n^2}{7m+\cfrac{16m^2-n^2}{9m+\ddots}}}}}$$
From which we obtain its hyperbolic companion $$\displaystyle\tanh\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2+n^2} {3m+\cfrac{4m^2+n^2}{5m+\cfrac{9m^2+n^2}{7m+\cfrac{16m^2+n^2}{9m+\ddots}}}}}$$
(iv)and if $2a=-n$ and $2b=2m+n$,then it follows that
$$\displaystyle\tan\left(\frac{\pi(m+n)}{4m}\right)=\frac{1+\tan\Big(\frac{\pi n}{4m}\Big)}{1-\tan\Big(\frac{\pi n}{4m}\Big)}=\cfrac{m+n}{m+\cfrac{(-n)(2m+n)} {3m+\cfrac{(m-n)(3m+n)}{5m+\cfrac{(2m-n)(4m+n)}{7m+\cfrac{(3m-n)(5m+n)}{9m+\ddots}}}}}$$   
Q: Is the conjectured general continued fraction true (for all complex numbers $a$,$b$ with $x\gt0$ and $m\gt0$)? 
 [/INST] Continued fraction (2) can be simplified as
$$
\tan\left(\alpha\tan^{-1}z\right)=\cfrac{\alpha z}{1+\cfrac{\frac{(1^2-\alpha^2)z^2}{1\cdot 3}} {1+\cfrac{\frac{(2^2-\alpha^2)z^2}{3\cdot 5}}{1+\cfrac{\frac{(3^2-\alpha^2)z^2}{5\cdot 7}}{1+\ddots}}}}\tag{2a}
$$
This is a special case of the following continued fraction due to N$\ddot{\text{o}}$rlund (B.Berndt, Ramanujan's notebooks, vol.2)

To obtain (2a) from (21.6) set $\beta=-\alpha,\gamma=\frac{1}{2},\frac{x(1-x)}{(1/2-x)^2}=z^2$. Now one can apply the following formulas
$$
\, _2F_1\left(a,-a;\frac{1}{2};x\right)=\cos \left(2 a \sin ^{-1}\left(\sqrt{x}\right)\right)
$$
$$
-2a^2\, _2F_1\left(1+a,1-a;\frac{3}{2};x\right)=\frac{d}{dx}\left[\, _2F_1\left(a,-a;\frac{1}{2};x\right)\right]
$$
$$
2 \sin ^{-1}\left(\sqrt{x}\right)=\tan^{-1}\frac{\sqrt{x(1-x)}}{1/2-x}
$$
to complete the proof.
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