<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] If we consider the following symmetric q-continued fraction,where $q=ab$,$|q|\lt1$ see here and here
$$\cfrac{1}{1-ab+\cfrac{(ab)(1-b^2)(1-a^2)}{1-(ab)^3+\cfrac{(ab)^2(1-ab^3)(1-a^3b)}{1-(ab)^5+\cfrac{(ab)^3(1-a^2b^4)(1-a^4b^2)}{1-(ab)^7+\cfrac{(ab)^4(1-a^3b^5)(1-a^5b^3)}{1-(ab)^9+\ddots }}}}}=\frac{(a^5b^3;(ab)^4)_\infty\,(a^3b^5;(ab)^4)_\infty}
{(ab^3;(ab)^4)_\infty\,(a^3b;(ab)^4)_\infty}$$
and let $a\rightarrow q^a$ ,$b\rightarrow q^b$ then we have
$$\cfrac{1}{1-q^{(a+b)}+\cfrac{q^{(a+b)}(1-q^{2b})(1-q^{2a})}{1-q^{3(a+b)}+\cfrac{q^{2(a+b)}(1-q^{(a+3b)})(1-q^{(b+3a)})}{1-q^{5(a+b)}+\cfrac{q^{3(a+b)}(1-q^{(2a+4b)})(1-q^{(2b+4a)})}{1-q^{7(a+b)}+\cfrac{q^{4(a+b)}(1-q^{(3a+5b)})(1-q^{(5a+3b)})}{1-q^{9(a+b)}+\ddots }}}}}=\frac{(q^{(5a+3b)};q^{4(a+b)})_\infty\,(q^{3a+5b};q^{4(a+b)})_\infty}
{(q^{(a+3b)};q^{4(a+b)})_\infty\,(q^{(3a+b)};q^{4(a+b)})_\infty}$$
now after multiplying both sides by $1-q$ ,and performing equivalence transformation
$$\cfrac{1}{\frac{1-q^{(a+b)}}{1-q}+\cfrac{q^{(a+b)}\Big(\frac{1-q^{2b}}{1-q}\Big)\Big(\frac{1-q^{2a}}{1-q}\Big)}{\frac{1-q^{3(a+b)}}{1-q}+\cfrac{q^{2(a+b)}\Big(\frac{1-q^{(a+3b)}}{1-q}\Big)\Big(\frac{1-q^{(b+3a)}}{1-q}\Big)}{\frac{1-q^{5(a+b)}}{1-q}+\cfrac{q^{3(a+b)}\Big(\frac{1-q^{(2a+4b)}}{1-q}\Big)\Big(\frac{1-q^{(2b+4a)}}{1-q}\Big)}{\frac{1-q^{7(a+b)}}{1-q}+\cfrac{q^{4(a+b)}\Big(\frac{1-q^{(3a+5b)}}{1-q}\Big)\Big(\frac{1-q^{(5a+3b)}}{1-q}\Big)}{\frac{1-q^{9(a+b)}}{1-q}+\ddots }}}}}=(1-q)\frac{(q^{(5a+3b)};q^{4(a+b)})_\infty\,(q^{3a+5b};q^{4(a+b)})_\infty}
{(q^{(a+3b)};q^{4(a+b)})_\infty\,(q^{(3a+b)};q^{4(a+b)})_\infty}$$
then apply the equality $\lim_{q\rightarrow 1}\frac{1-q^x}{1-q}=x$,see q-analog and the q-gamma function $\Gamma_q(x)=(1-q)^{1-x}\frac{(q;q)_{\infty}}{(q^x;q)_{\infty}}$
which concludes the proof of continued fraction (1)
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