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:


*

*Is it true that this function always has $n$ roots for any fixed $k$? (Negative, see UPD2).

*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 $k$.

Set $A(s)=-1$, $B(s)=1$, and $k=2\pi$. Then you obtain the equation $$e^{2 \pi s}=1,$$ which has solutions $s = 0, \pm i, \pm 2i, \pm 3i, \ldots$.
