# How many roots does an exponential polynomial have?

Let $$s$$ be a complex variable and consider two polynomials with real coefficients: $$A(s) = s^n + a_{n-1}s^{n-1}+\ldots+a_1s+a_0,$$ $$B(s) = s^m + b_{m-1}s^{m-1}+\ldots+b_1s+b_0,$$ where $$n \ge m$$.

Let $$k$$ be a real constant. I am looking for roots of the function $$A(s)+e^{sk}B(s)=0.$$

Obviously, when $$k=0$$ I have just a polynomial of degree $$n$$ and I have $$n$$ roots. Then I have two questions:

1. Is it true that this function always has $$n$$ roots for any fixed $$k$$? (Negative, see UPD2).
2. Do these roots depend continuously on $$k$$?

UPD: We also assume that $$n\ge 1$$.

UPD2: Let us take $$A(s)=s$$ and $$B(s)=1$$. Then we have $$s+e^{ks} = 0.$$ For $$k=0$$ the only roots is $$s_1=-1$$. However, for $$k=-1$$ we have $$se^{s}=-1$$ and we have multiple solutions.

So the answer to the first question is negative. However, I do not know if the roots are continuous with respect to $$k$$.

It always has infinitely many roots if $$A$$ or $$B$$ are not the zero polynomial and $$k>0$$ since then the function $$f(s)$$ = $$A(s)+e^{sk}B(s)$$ is integral of order 1 and finite type $$k$$ and such have infinitely many zeros unless they are of the type $$e^{P(s)}C(s)$$, where $$P$$ has degree 1 and $$C$$ is a non-zero polynomial; if so, it follows $$P(s)$$=$$ks+r$$ and $$A(s)$$ identically zero. As for continuos dependency, it should follow on any compact set by Hurwitz's theorem.
For $$k<0$$ same applies just that the type is now $$|k|$$.
• Fix $A, B, k$, so $f$ as above and pick a large enough $R$ s.t the disk of radius $R$ contains the root you want, say $z(k)$ (or a finite number of roots, same argument works); if the root is of order 1, the uniform convergence of $A(s)+e^{sl}B(s)$ to $f$ on the disk of radius $R$ as $l$ goes to $k$, and the fact that roots of $f$ are isolated ensures $A(s)+e^{sl}B(s)$ has a root $z(l)$ close to $z(k)$ for $l$ near $k$ and $z(l)$ converges to $z(k)$, so you can locally define a continuous function of the roots; if the root is multiple, you may get branches, but still some local continuity. – Conrad Feb 14 at 21:12
Set $$A(s)=-1$$, $$B(s)=1$$, and $$k=2\pi$$. Then you obtain the equation $$e^{2 \pi s}=1,$$ which has solutions $$s = 0, \pm i, \pm 2i, \pm 3i, \ldots$$.
• Thanks! But let us assume that $n\ge 1$ (I have updated the question). Then this example does not work, can we find another one? – Arastas Feb 14 at 17:31