Closed form for $\Gamma (a+bi)\Gamma(a-bi)$

Question

I noticed that $$\Gamma (3+2i)\Gamma (3-2i)=\frac{160\pi}{e^{2\pi}-e^{-2\pi}}$$ and $$\Gamma (2+5i)\Gamma (2-5i)=\frac{260\pi}{e^{5\pi}-e^{-5\pi}}.$$ Is there a closed form for $\Gamma (a+bi)\Gamma (a-bi)$ in general?

Answer 1

The closed form you seek exists if $2a\in\Bbb Z$.

Since $\Gamma(s)\Gamma(1-s)=\pi\csc\pi s$, $\Gamma(is)\Gamma(1-is)=-i\pi\operatorname{csch}\pi s$, so $\Gamma(is)\Gamma(-is)=\frac{\pi}{s}\operatorname{csch}\pi s$. Then$$\frac{\Gamma(1+is)\Gamma(1-is)}{\Gamma(is)\Gamma(-is)}=s^2,\,\frac{\Gamma(2+is)\Gamma(2-is)}{\Gamma(1+is)\Gamma(1-is)}=1+s^2,\,\frac{\Gamma(3+is)\Gamma(3-is)}{\Gamma(2+is)\Gamma(2-is)}=4+s^2.$$