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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 $k$.

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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 $|k|$.

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  • $\begingroup$ Thanks! Could you, please, explain to me how to conclude continuity from the Hurwitz theorem? $\endgroup$ – Arastas Feb 14 at 20:18
  • $\begingroup$ 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. $\endgroup$ – Conrad Feb 14 at 21:12
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Set $A(s)=-1$, $B(s)=1$, and $k=2\pi$. Then you obtain the equation $$e^{2 \pi s}=1,$$ which has solutions $s = 0, \pm i, \pm 2i, \pm 3i, \ldots$.

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  • $\begingroup$ 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? $\endgroup$ – Arastas Feb 14 at 17:31

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