Bounds on the real and imaginary parts of the digamma function $\psi $ Let $\psi $ be the digamma function given by
$$\psi (z)=\left.\frac {d}{dt}\log\Gamma (t)\right|_{t=z}. $$
I wonder does anyone know of any lower and/or upper bounds on the real and imaginary parts of $\psi $? A Google search brings up some bounds for $ z=x $ real only, but I'm interested in the digamma function as a function of a complex variable. The region of interest is $0<\text{Re}(z)<1/2$.
EDIT
Empirical evidence suggests $$\text{Re}(\psi(z))\approx\log(\text{Im}(z)),$$ and $$\text{Im}(\psi(z))\approx\frac{\pi}{2},$$ for $\text{Im}(z)>c$, where $c$ is "not very large". Using knowledge of $\log(z)$ for complex $z$ this leads me to believe that $$\psi(z)\approx \log\left|z\right|+i\text{Arg}(z)=\log z,$$ where we take the principal argument.
Indeed, after doing a bit of digging I have found that (Karatsuba, A. A., Voronin, S. M., 1991, p.344)
$$\psi(z) - \log z \ll \frac{1}{\left|z\right|},$$
for $\left|\text{arg}(z)\right|<\pi$.
 A: We have the sum form
$$ \Psi(z) = -\gamma + \sum_{k=1}^\infty \left(\dfrac{1}{k} - \dfrac{1}{k+z-1}\right) = -\gamma + \sum_{k=1}^\infty \dfrac{z-1}{k(k+z-1)}$$
If $z = x + i y$, then e.g.  we have 
$$\text{Im} \left(\dfrac{1}{k} - \dfrac{1}{k+z-1}\right) = \dfrac{y}{(k+x-1)^2 + y^2}$$
Thus
$ \text{Im}(\Psi(z))$ and $y$ have the same sign, and
$$ \left|\text{Im}(\Psi(z))\right| < \sum_{j=-\infty}^\infty \dfrac{y}{(j+x)^2 + y^2} =  {\frac {\pi \,\cosh \left( \pi \,y \right) \sinh \left( \pi \,y
 \right) }{ \left( \cosh \left( \pi \,y \right)  \right) ^{2}- \left( 
\cos \left( \pi \,x \right)  \right) ^{2}}} \le \frac {\pi \,\cosh \left( \pi \,y \right) \sinh \left( \pi \,y
 \right) }{ \left( \cosh \left( \pi \,y \right)  \right) ^{2}- 1}
$$
Lots of other bounds can be obtained. Is there any particular region of the complex plane, or asymptotic direction, you're interested in? 
EDIT: For $0 \le x \le 1/2$ and $y > 0$, $\dfrac{y}{(k+x-1)^2 + y^2}$ is decreasing
in $k$ for $k \ge 1$, so
$$ \text{Im} \Psi(z) \ge \int_1^\infty \dfrac{y\; dt}{(t+x-1)^2 + y^2} = 
\dfrac{\pi}{2} - \arctan(x/y) $$
while
$$ \text{Im} \Psi(z) \le \dfrac{y}{x^2+y^2} + \int_1^\infty \dfrac{y\; dt}{(t+x-1)^2 + y^2} = \dfrac{\pi}{2} - \arctan(x/y) + \dfrac{y}{x^2+y^2}$$
Similarly, although somewhat less cleanly because monotonicity is not as clear,
$$ \text{Re} \Psi(z) = \sum_{k=1}^\infty \left( \dfrac{1}{k} - \dfrac{k+x-1}{(k+x-1)^2 + y^2} \right) \approx \int_1^\infty \left( \dfrac{1}{t} - \dfrac{t+x-1}{(t+x-1)^2 + y^2} \right)\; dt = \dfrac{\ln(x^2 + y^2)}{2} $$
