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I have the error function:

$$\text{erf}(x + iy) = \frac{2}{\sqrt{\pi}} \int_0^{x + iy} e^{-t^2} dt$$

I'm looking for an upper bound on $|\text{erf}(x + iy)|$ in terms of $y$. I know that for $y=0$ a trivial bound is $1$, but certainly I cannot have a constant bound since $\text{erf}$ is entire and analytic on $\mathbb{C}$.

Wolfram says that:

$$|\text{erf}(x + iy)|^2 = \text{erf}\left( x - x \sqrt{- \frac{y^2}{x^2}} \right) \text{erf}\left( x + x \sqrt{- \frac{y^2}{x^2}} \right)$$

This is the only "simplification" I could find but it's not really helpful... Does anyone know a useful bound?

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Take $z=x+iy$ and $t=vz$. Then $$\text{erf}\left(z\right)=\frac{2z}{\sqrt{\pi}}\int_{0}^{1}e^{-z^{2}v^{2}}dv$$ so $$\left|\text{erf}\left(z\right)\right|\leq\frac{2\left|z\right|}{\sqrt{\pi}}\int_{0}^{1}\left|e^{-z^{2}v^{2}}\right|dv=\frac{2\left|z\right|}{\sqrt{\pi}}\int_{0}^{1}e^{-v^{2}\text{Re}(z^{2})}dv$$ $$=\frac{2\sqrt{x^{2}+y^{2}}}{\sqrt{\pi}}\int_{0}^{1}e^{-v^{2}\left(x^{2}-y^{2}\right)}dv.$$ At this point we can estimate the integral using the erf function with real argument, getting $$\left|\text{erf}\left(z\right)\right|\leq\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}-y^{2}}}\text{erf}\left(\sqrt{x^{2}-y^{2}}\right)$$ if $x^{2}\neq y^{2}$ and with $1$ if $x^{2}=y^{2}$ or we can use the trivial bounds $$\int_{0}^{1}e^{-v\left(x^{2}-y^{2}\right)}dv \leq \begin{cases} 1, & x^{2}\geq y^{2}\\ e^{y^{2}-x^{2}}, & x^{2}<y^{2}. \end{cases}$$

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