Asymptotic expression for complex root of $\text{erf}(z)$

Question

I tried to figure out if the complex roots of $\text{erf}$ function follows some kind of rule:
First of all I wrote it with integral representation:
$$\text{erf}(z)=\text{erf}(x+iy)=0 \quad \Leftrightarrow \quad\begin{cases}\displaystyle\int_{0}^{1}e^{-t^2\left(x^2-y^2\right)}\cos\left(2xyt^2\right)\mathrm{d}t=0\\ \displaystyle\int_{0}^{1}e^{-t^2\left(x^2-y^2\right)}\sin\left(2xyt^2\right)\mathrm{d}t=0\end{cases}\qquad \text{erf}(x(1+i))=0\text{ for }x\to \infty$$ Then I calculated the first $100$ roots:
$$\begin{array}{|c|c|c|c|} \hline \text{from }z_{01}\text{ to }z_{25}&\text{from }z_{26}\text{ to }z_{50}&\text{from }z_{51}\text{ to }z_{75}&\text{from }z_{76}\text{ to }z_{100} \\ \hline 1.4506161632+1.8809430002 i&8.9298047326+9.1027134002 i&12.574098205+12.710780755 i&15.380053048+15.498447836 i\\ 2.2446592738+2.6165751407 i&9.1034740038+9.2741670843 i&12.698211423+12.833951354 i&15.481740841+15.599573701 i\\ 2.8397410469+3.1756280996 i&9.2739111487+9.4424906194 i&12.821126978+12.955947571 i&15.582766461+15.700046903 i\\ 3.3354607354+3.6461743764 i&9.4412898415+9.6078497166 i&12.942878851+13.076802401 i&15.683142668+15.799879926 i\\ 3.7690055670+4.0606972339 i&9.6057687550+9.7703959481 i&13.063499448+13.196547322 i&15.782881821+15.899084862 i \\ \hline 4.1589983998+4.4355714442 i&9.7674933130+9.9302683819 i&13.183019698+13.315212392 i&15.881995887+15.997673426 i\\ 4.5163193996+4.7804476442 i&9.9265971882+10.087594981 i&13.301469148+13.432826332 i&15.980496466+16.095656972 i\\ 4.8479703092+5.1015880435 i&10.083203589+10.242493806 i&13.418876045+13.549416616 i&16.078394805+16.193046511 i\\ 5.1587679075+5.4033326428 i&10.237426372+10.395074059 i&13.535267418+13.665009537 i&16.175701808+16.289852724 i\\ 5.4521922011+5.6888374370 i&10.389371003+10.545436985 i&13.650669147+13.779630282 i&16.272428058+16.386085974 i\\ \hline 5.7308535991+5.9604833491 i&10.539135402+10.693676664 i&13.765106033+13.893302996 i&16.368583822+16.481756323 i\\ 5.9967692808+6.2201195193 i&10.686810672+10.839880698 i&13.878601857+14.006050840 i&16.464179072+16.576873539 i\\ 6.2515360724+6.4692163130 i&10.832481752+10.984130824 i&13.991179441+14.117896045 i&16.559223491+16.671447114 i\\ 6.4964435534+6.7089659323 i&10.976227978+11.126503445 i&14.102860698+14.228859967 i&16.653726485+16.765486267 i\\ 6.7325508434+6.9403510392 i&11.118123591+11.267070108 i&14.213666686+14.338963132 i&16.747697200+16.858999964 i\\ \hline 6.9607403702+7.1641930155 i&11.258238185+11.405897925 i&14.323617649+14.448225279 i&16.841144522+16.951996920 i\\ 7.1817565249+7.3811867958 i&11.396637098+11.543049948 i&14.432733065+14.556665407 i&16.934077097+17.044485613 i\\ 7.3962340888+7.5919265791 i&11.533381775+11.678585507 i&14.541031681+14.664301807 i&17.026503332+17.136474290 i\\ 7.6047195550+7.7969251790 i&11.668530078+11.812560508 i&14.648531557+14.771152104 i&17.118431411+17.227970980 i\\ 7.8076874012+7.9966288395 i&11.802136576+11.945027705 i&14.755250094+14.877233285 i&17.209869298+17.318983496 i\\ \hline 8.0055527008+8.1914287507 i&11.934252802+12.076036943 i&14.861204074+14.982561735 i&17.300824747+17.409519449 i\\ 8.1986810309+8.3816701210 i&12.064927481+12.205635379 i&14.966409685+15.087153266 i&17.391305311+17.499586252 i\\ 8.3873963541+8.5676594131 i&12.194206741+12.333867680 i&15.070882551+15.191023142 i&17.481318347+17.589191126 i\\ 8.5719873564+8.7496701764 i&12.322134304+12.460776202 i&15.174637762+15.294186108 i&17.570871025+17.678341112 i\\ 8.7527125975+8.9279477969 i&12.448751656+12.586401155 i&15.277689897+15.396656414 i&17.659970332+17.767043069 i\\ \hline \end{array}$$ Is this deduction $\text{erf}(x(1+i))=0\text{ for }x\to\infty$ correct or is there some mistake?

Edit

For $z=x+ix$ I have:

$\begin{cases}\displaystyle\int_{0}^{1}e^{-t^2\left(x^2-x^2\right)}\cos\left(2x xt^2\right)\mathrm{d}t=\int_{0}^{1}\cos\left(2x^2t^2\right)\mathrm{d}t=\dfrac{\sqrt{\pi}}{2x}C\left(\frac{2x}{\sqrt{\pi}}\right)\to 0\\ \displaystyle\int_{0}^{1}e^{-t^2\left(x^2-x^2\right)}\sin\left(2x xt^2\right)\mathrm{d}t=\int_{0}^{1}\sin\left(2x^2t^2\right)\mathrm{d}t=\dfrac{\sqrt{\pi}}{2x}S\left(\frac{2x}{\sqrt{\pi}}\right)\to 0\end{cases}$ for $x\to\infty$

Where $C(z)$ and $S(z)$ are the Fresnel functions.

My question is:
Is it correct to assume $y=x$ or it's more correct to assume $y=ax+b+o(1)$?
Where $a,b\in\mathbb{R}$

Answer 1

Your conjecture seems correct. Without having to reinvent the wheel all the time, the complex zeros of the error function have already been investigated (though perhaps not as much as I would have thought).

You can find a lot of information on this web page. Using the expression found there for the n-th root of $\mathrm{erf} (z)$, such that $\mathrm{erf} (x_n +i y_n)=0$, one has, expanding in $1/n$,

\begin{align} x_n & = \sqrt{\pi n} +O \left ( \frac{1}{\sqrt{n}} \right ) \\ y_n & = \sqrt{\pi n} +O \left ( \frac{1}{\sqrt{n}} \right ). \end{align}

Answer 2

Taken from the Handbook of mathematical functions by Abramowitz and Stegun (equation $7.1.29$)

$$\operatorname*{erf}(x+i y) = \operatorname*{erf}(x) + \frac{e^{-x^2}}{2 \pi x} \big[(1-\cos(2 x y))+i \sin(2 x y)\big] +$$ $$\frac{2 e^{-x^2}}{\pi} \sum_{k=1}^{\infty} \frac{e^{-\frac{k^2}4}} {k^2+4 x^2}\big[f_k(x,y)+i\, g_k(x,y)\big]$$ where $$f_k(x,y) = 2 x (1-\cos(2 x y) \cosh( k y)) + k\sin(2 x y) \sinh(k y)$$ $$g_k(x,y) = 2 x \sin(2 x y) \cosh(k y) + k\cos(2 x y) \sinh(k y)$$lead to $$\color{red}{\large 2 x\,y =2n \pi- c}$$ where $0 \leq c\leq \frac \pi 4$ (this has to be worked but asymptotically (not proved),$c\to \frac \pi 4$).

Edit

In fact, the approximate zeros are (have a look here)

$$x_n \sim \lambda -\frac{\mu }{4 \lambda }+\frac{\mu ^2-2 \mu +2}{32 \lambda^3}+\cdots$$ $$y_n \sim \lambda +\frac{\mu }{4 \lambda }+\frac{\mu ^2-2 \mu +2}{32 \lambda^3}+\cdots$$ where $$\lambda= \sqrt{\frac{8n-1}{8}} \qquad \qquad \mu=\log(\lambda \sqrt{2\pi})$$ Rewriting them using $\color{blue}{m=\sqrt{\left(n-\frac{1}{8}\right) \pi}} $ $$\color{red}{\large x_n \sim m \left(1-\frac{\log \left(2 \pi m^2\right)}{8 m^2} +O\left(\frac{1}{m^4}\right)\right) }$$ $$\color{red}{\large y_n \sim m \left(1+\frac{\log \left(2 \pi m^2\right)}{8 m^2} +O\left(\frac{1}{m^4}\right)\right) }$$ For $n=100$, this gives $$17.6599207484\, +\, 17.7669936686 \, i$$ almost identical to the numbers in your table.