# Limit of error function in complex argument

How to calculate the limit \begin{align*} \lim_{b\to \infty}\frac{\vert \text{erf}(\sqrt{a+\mathrm{i} b})\vert^{2}}{\sqrt{a^2+b^2}} \end{align*} where \begin{align*} \text{erf}(a+\mathrm{i} b)=\frac{2}{\sqrt{\pi}}\int_{0}^{a+\mathrm{i} b}e^{-t^2}~dt \end{align*} is the error function. The Mathematica shows that as $$b$$ grows, $$\vert \text{erf}(\sqrt{a+\mathrm{i} b})\vert^{2}$$ approaches to $$1$$ and consequently \begin{align*} \frac{\vert \text{erf}(\sqrt{a+\mathrm{i} b})\vert^{2}}{\sqrt{a^2+b^2}} \end{align*} becomes very small. If I try to simplify the error function, then I find an expression in terms of $$\cos$$ and $$\sin$$ and that diverge. How to proves that the above limit converges to $$0$$ (as Mathematica shows).

• Hint: try to, by substitution, take the $a+bi$ from the limit down to the integrand. Jun 7 '18 at 9:12

$$\text{erf}\,(z) \sim 1 - \frac{e^{-z^2}}{\sqrt{\pi\,z}} \text{ with } z=\sqrt{a+ib}=(a^2+b^2)^{1/4}\exp{\big(\frac{i}{2}\,\arctan{(b/a)} \big)}.$$
With $b \to \infty$ the argument of the exponential goes to $\pm i\pi/4,$ with the positive sign for positive $a$. There are more terms multiplying the exponential in the asymptotic expansion, but an asymptotic text will tell you that they are all bounded for $|ph(z)| \le \pi/4,$ and your case is right on the boundary, so the expansion works for you. Since $\exp(-r(1+i)) \to 0$ for positive $r=\sqrt{2(a^2+b^2)},$ all that remains is 1. The limit is zero entirely by the behavior of $\sqrt{a^2+b^2}$ in the denominator.
• Thanks. I did not understand the last part $exp(-r(1+i)\to 0$. When $r$ approaches $\infty$ and $\infty(1+i)$ has no value? Jun 8 '18 at 18:51
• @skorpion $|\exp(-r -ir)| = |\exp(-r)||\exp(-ir)| = |\exp(-r)|$, since $r$ is real. So indeed it goes to 0 as $r$ gets bigger. Jun 19 '18 at 14:33