Cfrgtkky

I'm working on something that involves the error function, except with complex inputs, i.e. erf(z) with $$zinmathbb{C}$$. In particular, at this point, I need a formula for the zeros of this function. I've found the paper "Complex Zeros of the Error Function and of the Complementary Error Function" by Fettis, Caslin, and Kramer (mathematics of computation V. 27 num 122 April 1973, pp 401-407). They provide approximate zeros, but I'm more interested in understanding the original equation they are trying to solve, because I only need a formula not actual values. In working through their paper I'm confused about a quick deduction they make, fairly off casually.

To explain further, they are working with a function denoted $$w(z)=e^{-z^2}mbox{erfc}(imath z)=e^{-z^2}(1-mbox{erf}(imath z))$$ First they point out that (sym prop 1): $$w(-x+imath y)=w^*(x+imath y)$$ and that (sym prop 2): $$w(x-imath y)=2e^{-(x-imath y)^2}-w^*(x+imath y)$$ (where * denotes complex conjugation). Then that it can be expressed as $$w(z)=frac{1}{pi}int_{-infty}^infty e^{-t^2}frac{[y+imath(x-t)]}{(x-t)^2+y^2}ds$$ which after a substitution of $$x-t=s$$ can be written in the form $$frac{2e^{-x^{2}}}{pi}Big{[}yint_0^{infty}e^{-s^{2}}frac{mbox{cosh}(2xs)}{y^2+s^2}ds+imathint_0^infty e^{-s^{2}}frac{mbox{sinh}(2xs)}{y^2+s^2}s;dsBig{]}=u(x,y)+imath(v(x,y)$$ at this point they make the following series of comments:

from which it is clear that $$u(x,y)>0$$, $$v(x,y)>0$$ for
$$x>0$$, $$y>0$$. This together with the symmetry property
(sym prop1) shows that any zeros of w(z) must lie in the lower half plane, and these according to (sym prop 2) will be
determined from the equations
$$2e^{y^2-x^2}cos(2xy) = u(x,y)$$ and $$2e^{y^2-x^2}sin(2xy)=v(x,y)$$

Then there's a footnote saying that from now on y will be positive (because in fact for the complex error functions there are zeros in the first quadrant). They continue:

Since $$u(x,y) mbox{ and } v(x,y)rightarrow 0 mbox{ for } abs{z}rightarrow 0$$ it is evident that asymptotically $$abs{y}<x$$. Further since $$u mbox{ and } v mbox{ are both positive if } x mbox{ and } y mbox{ are positive }$$ it follows that

So up to now, everything is fine. Some of its a bit above my pay grade, but ok, other references concur on these points, so ok. Here comes the hard bit. They now conclude:

$$cos(2xy)>0 mbox{ and }sin(2xy)<0$$, and hence that $$2xy=2npi-beta mbox{ for } n=1,2,3,ldots mbox{ where } 0leqbeta<frac{pi}{4}$$

This has me flumoxed. It would seem that this last equation is the equation they are going to approximate in order to find approximate roots of the complex error function. But if u(x,y) and v(x,y) are the real and imaginary parts of the solution, then the zeros should be where u and v are both simultaneously zero, which doesn't seem like it can happen, because $$cos(theta)mbox{ and }sin(theta)$$ are never simultaneously zero, and $$e^{y^2-x^2}>0;forall x,yinmathbb{R}$$. My only guess is that they are using the norm squared of $$u(x,y)+imath v(x,y)$$, but they very carefully pointed out that both functions are positive for positive x,y, although that seems to be false since sin(x) clearly can be negative for positive x. So after all that :) is that what they are doing? Taking the norm squared? And if so which of them is negative, u or v? Or am I missing something entirely...

Thanks in advance.