Limits of complex error and gamma functions in the complex plane?

Question

What are the following one-sided limits in the complex plane (in the form $x+iy$):

For the complex error function:

• $\lim_{x \to 0^+, y \to 0^+}\text{erf}\left(x+iy\right) = $
• $\lim_{x \to +\infty, y \to 0^+}\text{erf}\left(x+iy\right) = $
• $\lim_{x \to 0^+, y \to +\infty}\text{erf}\left(x+iy\right) = $
• $\lim_{x \to +\infty, y \to +\infty}\text{erf}\left(x+iy\right) = $

For the complex gamma function:

• $\lim_{x \to -\infty, y \to +\infty}\Gamma\left(x+iy\right) = $
• $\lim_{x \to 0^-, y \to +\infty}\Gamma\left(x+iy\right) = $
• $\lim_{x \to -\infty, y \to 0^+}\Gamma\left(x+iy\right) = $
• $\lim_{x \to 0^-, y \to 0^+}\Gamma\left(x+iy\right) = $
• $\lim_{x \to 0^+, y \to 0^+}\Gamma\left(x+iy\right) = $
• $\lim_{x \to +\infty, y \to 0^+}\Gamma\left(x+iy\right) = $
• $\lim_{x \to 0^+, y \to +\infty}\Gamma\left(x+iy\right) = $
• $\lim_{x \to +\infty, y \to +\infty}\Gamma\left(x+iy\right) = $

I am searching only for limits that are well defined from an analytical point of view.

Answer 1

$\displaystyle \text{erf}(z)=\frac{2\sqrt2}{\tau}\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{n!(2n+1)}$ for all complex $z$. (taylor series of erf)
Erf is entire, therefore the limit as, for example, $x\to 0$ is the same as the limit at $=0$.

1. Obviously $0$.
2. For real $z>0$, $\displaystyle \text{erf}(z)=\frac{2\sqrt{2}}{\sqrt{\tau}}\int_{0}^{z} e^{-t^2} \mathrm{d}t \ \ \ $ which is the probability that a variable from a normal distribution will be less than $z\sqrt 2$ standard deviations from the mean, which tends to $1$ as $z\to\infty$.
3. When $x=0$, $y\to =\infty$, $z^{2n+1}=(-1)^ny^{2n+1}$, the $(-1)^n$s cancel to make all the terms positive and erf diverges.
4. If $y$ grows way faster than $x$ but both tend to infinity, erf will still diverge like before.
   
   
   $\displaystyle \Gamma(z)=\int_0^{\infty}t^{z-1}e^{-t}\mathrm{d}t$ in the upper half-plane with an analytic continuation to $\mathbb{C}\setminus \{0\}$ following from $\Gamma(z+1)=z\Gamma(z)$. Which side limits are taken on is irrelevant since $\Gamma$ is meromorphic. If $x\to \infty$ $\Gamma$ blows up spectacularly. If $\Gamma(z)$ cannot have a finite limit $\Gamma(0)$ as $x,y \to 0,0$ because then $1=\Gamma(1)=0\Gamma(0)$.