How do we prove this 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 continued fraction holds
$$\frac{\displaystyle4\Gamma\left(\frac{2a+3}{4}\right)\Gamma\left(\frac{2b+3}{4}\right)}{\displaystyle\Gamma\left(\frac{2a+1}{4}\right)\Gamma\left(\frac{2b+1}{4}\right)}=\cfrac{(2a+1)(2b+1)}{a+b+2+\cfrac{(a-b+1)(b-a+1)} {a+b+4+\cfrac{(2a+3)(2b+3)}{a+b+6+\cfrac{(a-b+3)(b-a+3)}{a+b+8+\ddots}}}}\tag{1a}$$
Corollary
$$\frac{4}{\pi}=\cfrac{(2)^2}{3+\cfrac{(1)^2}{5+\cfrac{(4)^2}{7+\cfrac{(3)^2}{9+\ddots}}}}\tag{1b}$$
Q: How do we prove the continued fraction $(1a)$ rigorously?

This can be deduced from Gauss' continued fraction
$$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}$$
which evaluates the continued fraction in $(1a)$ as
$$\frac{(2a+1)(2b+1)}{a+b+2}\frac{_2 F_1\left(b+\frac32,\frac{b-a+1}{2};\frac{a+b}{2}+2;-1\right)}{_2 F_1\left(b+\frac12,\frac{b-a+1}{2};\frac{a+b}{2}+1;-1\right)},$$
and Kummer's formula $_2 F_1(a,b;a-b+1;-1)=\dfrac{\Gamma(a/2+1)\Gamma(a-b+1)}{\Gamma(a+1)\Gamma(a/2-b+1)}$.
After all the necessary substitutions and cancellations, this yields exactly the expected result. The proofs are sketched in the linked articles, but you may want to follow further references for a deeper treatment.
