Quadrant asymptotics of the error function (${\int }_0^z{e}^{-t^2}dt$)

For $z=r{e}^{i\theta }$ with fixed $0\le \theta \le \pi /2$, a horizontal tail gives $f(z)=\sqrt{\pi }/2-e^{-z^2}(2z{)}^{-1}[1+O(z^{-2})]$. The constant dominates through $\theta =\pi /4$; above that angle the exponential endpoint dominates. At the boundary angle its modulus is only $O(r^{-1})$, so the constant remains leading. On the imaginary axis the integral is purely imaginary and grows as $i{e}^{r^2}/(2r)$. The approximation is not a determination of exponentially subdominant constants in that growing sector.