Zeroes of trigonometric-like function

Question

Consider a function $f(z)=\cos(z)\cosh(az)+\sin(z)\sinh(bz)$ for $z\in \mathbb{C}, a,b \in \mathbb{R}$. Denote $D\subseteq \mathbb{R}^2$ being the set of such pairs $(a,b)$ of parameters so that NOT ALL zeroes of the corresponding $f(z)$ belong to coordinate axes in $z$. Formally

$$ D = \lbrace (a,b) \, \small| \exists z\in \mathbb{C} \, f(z) =0 \, \wedge \, {\rm arg} (z) \neq 0,\pi/2,\pi,3\pi/2 \rbrace. $$

Obviously the line $b=0$ is not in $D$, as zeroes of $\cos(z)\cosh(az)=0$ belong to axes $Re z=0, Im z=0$.

Problem: define geometrically the set $D$.

Calculations show that a set $D$ is finite near the origin. How to describe it more exactly? Is it possible to find a ball centred at origin for which $D$ is inside this ball? Is it possible to find estimates for a radius of this ball? More suggestions?

I knew of this problem some years ago from Prof. A.P.Soldatov. This problem occurs as solvability condition of some problems for Lavrent'ev-Bitsadze mixed type differential equation.

Answer 1

It looks to me like $D$ is not bounded. I have produced an animation using Maple. For $a$ running from $1$ to $40$ we take $b = 5a$, and plot the curves $\text{Re}(f(x+iy))=0$ (blue) and $\text{Im}(f(x+iy))=0$ (red) for $0 < x,y < 1/2$. The intersections of red and blue curves are off-axis zeros of $f$. It certainly appears that there are such zeros, and thus that the ray $b=5a$, $a > 1$, is contained in $D$. I suspect that it is possible to prove this using the Argument Principle.

EDIT: For various points in the first quadrant of the $(a,b)$ plane, I computed the number of zeros in the rectangle $0.01 \le \text{Re}(z) \le 2, 0.01 \le \text{Im}(z) \le 2$, by numerically evaluating the integral of $(2 \pi i)^{-1} f'(z)/f(z)$ around the boundary of this region. Here is the result, showing the points for which there is at least one such zero. There may be some tantalizing hints of structure here. What seems to happen is that off-axis solutions bifurcate from zeros of multiplicity $2$ on the imaginary axis at certain $(a,b)$ values.

Answer 2

Robert Israel's calculations suggest that if $0<a<b$ then $f(z)$ has infinitely many zeros near but not on the imaginary axis, at least on certain rational rays $b=ra$ such as Robert's example with $r=5$. Here's an explanation.

If $z=iy$ then $f(z) = \cos ay \cosh y - \sin by \sinh y$. For large (positive) $y$ this is very nearly $$ g(y) := \frac12 (\cos ay - \sin by) e^y = \sin \left(\frac{b+a}{2}y - \frac\pi4\right) \, \sin \left(\frac{b-a}{2}y - \frac\pi4\right) \, e^y, $$ which has zeros on two arithmetic progressions (AP's). For suitable values of $r$ (seems to be $r=m/n$ with $m,n$ odd and congruent mod~$4$), these AP's intersect in an AP of double zeros of $g$, where $\cos ay = \sin by = \pm 1$. But the approximation is not exact: at this double near-zero, $\cosh y$ slightly exceeds $\sinh y$, while $\sin by$ moves faster than $\cos ay$ (because $b>a$), so the actual $f(iy)$ does not vanish at or near the double zero of $g(y)$. If we now fix $a$ but slightly change say $b$, the resonance disappears, and $f(iy)$ has two simple zeros near those of $g(y)$. As $b$ moves through $ra$, the zeros merge and disappear, then reappear when $b$ has moved far enough to the other side of $ra$. At and near $b=ra$, then, those zeros must have taken a detour through $\bf C$ as a pair of distinct complex conjugate roots of $f(iy)$.