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 $|k|$.

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.