Suppose $a$ and $b$ are positive real constants. Consider the function $f: \mathbb C \to \mathbb C$ defined as $$ f(z)=1+z+ae^{-bz}.$$
My question is:
Can we find a "useful" sufficient condition on $a$ and $b$ that guarantees that $f$ has no zeroes on the closed positive half-plane $R=\{z \in \mathbb C : {\rm Re}(z) \geq 0\}$?
Some initial observations:
$a < 1$ is a sufficient condition that ensures that $f$ has no zeroes on $R$. Indeed if $a < 1$, then $|ae^{-bz}| \leq a < 1 \leq |1+z|$ for all $z\in R$.
$b = 0$ is a sufficient condition that ensures that $f$ has no zeroes on $R$. Indeed, if $b=0$, then $f(z)= 1+z$, whose only zero lies at $z = -1$.
But ideally, I would prefer to have a statement that says, "If $a$ and $b$ together obey such-and-such inequality, then $f$ is guaranteed to have no zeroes on $R$". This inequality should have $a < 1$ and $b = 0$ as special cases. Please let me know if you have any ideas.
[This question arose when studying a Laplace transform representation of solutions to a class of integral equations. Any progress on this question will improve my understanding of how the asymptotic behaviour of my solutions depends on parameters.]
Update: I've made some progress with this, but my argument isn't really a complex analysis argument, and I feel like there is an even stronger sufficient condition out there...
So here is what I've done. We already know what happens when $a < 1$, so we'll focus on the $a \geq 1$ case. I claim that if $$ b < \frac{\cos^{-1}(-\tfrac 1 a)}{\sqrt{a^2 - 1}},$$ then $f$ has no zeros on the positive half-plane $R$.
Indeed, suppose $z=x+iy$ is a zero of $f$, with $x \geq 0$. Then we have $$ \begin{align} 1+x+ae^{-bx}\cos by = 0 \\ y - ae^{-bx}\sin by = 0\end{align}$$ To satisfy the first equation, we must have $$ \cos b y \leq -\tfrac 1 a,$$ which implies that $$ |y| \geq \frac{\cos^{-1}(-\tfrac 1 a)}{b}$$ and also that $$ |ae^{-bx} \sin by| \leq \sqrt{a^2 - 1}$$ Hence the second equation is inconsistent with the inequality $b < \frac{\cos^{-1}(-\tfrac 1 a)}{\sqrt{a^2 - 1}}$, which proves the claim.