Imaginary part of $\frac{\Gamma'(z)}{\Gamma(z)}$

Question

I want to prove that:

$$ \Im\left(\frac{\Gamma'(z)}{\Gamma(z)}\right) = 0 \qquad \textit{iff} \qquad \Im(z) = 0$$

I tried to plot the function on mathematica and it seems true, excluding the poles on the negative real axis. Thanks

Answer 1

We have $$\psi(z) = -\gamma - \sum_{k \geq 1} \left( \frac 1 {k + z - 1} - \frac 1 k \right), \\ \operatorname{Im} \frac 1 {k + z - 1} = -\frac {\operatorname{Im} z} {(\operatorname{Re} z + k - 1)^2 + \operatorname{Im}^2 z},$$ therefore $\operatorname{sgn} \operatorname{Im} \psi(z) = \operatorname{sgn} \operatorname{Im} z$.

Answer 2

By definition of derivative of a complex function $$ f'(z) = \mathop {\lim }\limits_{\Delta \,z\; \to \;0} = {{f(z + \Delta z) - f(z)} \over {\Delta z}} $$ the limit must exist and be the same for whichever $$ \Delta z = \left| {\Delta z} \right|e^{\,i\,\arg \left( {\Delta z} \right)} $$ i.e. for whichever value of its argument, i.e. whichever direction in the complex plane.

So if $f(z)$ is analytic in $z=r \in \mathbb R$ and $f(r) \in \mathbb R$ then its derivative exist and is equal to the derivative in $r$, and viceversa the derivative in $r$ is equal to the derivative in $z$ (at $z=r$).
And Gamma respects such conditions.