I want to discover if there exist a solution of the following equation: $$\Im(\log(\Gamma(iy)) - y\beta = 0 $$ for $y\in\mathbb{R}\setminus \{0\}$ and $\beta >0$
I already proved that if there is a solution $y^*<0$ then also $-y^*$ is a solution, but this do not help me with the general proof.
Thanks a lot
If the function is $\ln(\Gamma(i y))$, a solution may not exist, because the imaginary part is negative for small positive $y$ and cannot be greater than $\pi$.
If $\operatorname{ln\Gamma}$ is defined as the analytic continuation of $\ln(\Gamma(z))$ from the positive real axis to $\mathbb C \backslash (-\infty, 0]$, then $\operatorname{Im} \operatorname{ln\Gamma}(+i 0) = -\pi/2$; $\operatorname{Im} \operatorname{ln\Gamma}(i y) \sim y \ln y$ at infinity; $\operatorname{ln\Gamma}(i y)$ is continuous on $(0, \infty)$.
Therefore $y = y(\beta)$ s.t. $\operatorname{Im} \operatorname{ln\Gamma}(i y) = \beta y, \, y > 0$ exists for any $\beta > 0$.