Let $z\in\overline{\mathbb C}$. I want to compute $$L \equiv \lim_{\mathbb{R} \ni a \to \infty}\operatorname{erf}(a+z).$$
Can I use the fact that $z + \infty = \tilde{\infty}$ where $\tilde{\infty}$ is complex infinity to conclude that$$L=\displaystyle\lim_{\mathbb{R} \ni a\to\infty}\operatorname{erf}a=1?$$
You can massage equations (5) to (7) of [1] to show that for any real numbers $x$ and $y$ $$ \operatorname{erf}(x+iy) = \operatorname{erf}(x) + f(x,y) $$ where $\lim_{x \rightarrow \infty} f(x,y) = 0$ for each $y$. It follows that $\lim_{x \rightarrow \infty} \operatorname{erf}(x+iy) = 1$, as desired.
[1] Salzer, H. E. "Formulas for calculating the error function of a complex variable." Mathematical Tables and Other Aids to Computation 5.34 (1951): 67-70.
For posterity, I include the relevant part of [1] below. Note that the author defines $\Phi(Z) \equiv \int_0^Z e^{-u^2} du$ (i.e., the error function integral without the usual scalar multiplier).