Can anyone prove that gamma function is complex for complex inputs?

Question

I am learning about gamma function and just got a doubt that how can I prove that gamma function gives out a complex number when taken complex numbers as inputs and that no real number would be obtained when complex numbers are taken as inputs. I am trying to prove this,but I don't think so that I have enough information of gamma function to prove that. Thanks in advance

Answer 1

It's not true.

For the sake of contradiction, suppose $\operatorname{Im}(\Gamma(z))$ were positive on the entire upper half plane. Because of the simple poles of $\Gamma$, there exist negative real numbers $x$ with $\Gamma(x) < 0$, and by continuity there exist $z$ in the upper half plane with $\operatorname{Re}(z) < 0$ and $\operatorname{Re}(\Gamma(z)) < 0$. Using $\Gamma(z+1) = z \Gamma(z)$, we find that $$\operatorname{Im}(\Gamma(z+1)) = \operatorname{Re}(z) \operatorname{Im}(\Gamma(z)) + \operatorname{Im}(z) \operatorname{Re}(\Gamma(z)) \,.$$ By assumption, the LHS is $> 0$, and by the choice of $z$, the LHS is $ < 0$. This is a contradiction.

When $\operatorname{Im}(\Gamma(z))$ is negative on the entire upper half plane, the argument is similar: choose $z$ in the upper half plane with $\operatorname{Re}(z) < 0$ and $\operatorname{Re}(\Gamma(z)) > 0$. Then the LHS above is negative, and the RHS is positive.