This problem appeared in Smith's prize exam 1875.
Evaluate the modulus of
\begin{equation} \Gamma\left(\frac{1}{2}+\sqrt{-1}a\right) \end{equation}
If we use the corollary
\begin{equation} \Gamma(x)\Gamma\left(x+\frac{1}{2}\right) = \frac{\pi^{1/2}}{2^{2x-1}}\Gamma(2x) \end{equation}
we obtain
\begin{equation} \Gamma\left(\frac{1}{2}+\sqrt{-1}a\right)\Gamma(\sqrt{-1}a) = \frac{\pi^{1/2}}{2^{2\sqrt{-1}a-1}}\Gamma(2\sqrt{-1}a) \end{equation}
Then \begin{equation} \Gamma\left(\frac{1}{2}+\sqrt{-1}a\right) = \frac{\pi^{1/2}}{2^{2\sqrt{-1}a-1}}\frac{(2\sqrt{-1}a-1)!}{(\sqrt{-1}a-1)!} \end{equation}
This does not lead anywhere. Can anybody help in solving the problem.