The title says it all: a now deleted question on the Mathematics Stackexchange asked more or less the same thing, and I answered by citing the work [1] of Wolfgang Tutschke, whose version of Rouche's theorem gives some information on the localization of the zeros of $f+g$ assuming only a priori knowledge of the $n$ (distinct) zeros $z_{o1}, \ldots, z_{on}$ of $f$ contained in $D$. However, this refinement is meaningful only if we know that $$ {\sup_{z\in{\partial D}}|g(z)|}<{\inf_{z\in{\partial D}}|f(z)|} $$ (see the development below) and thus it does require further informations on $f$ and $g$. This triggered my curiosity and thus, since no other answered to the question on Math.SE, I decided to re-post it here also my answer as a reference to "prior art": in sum, I'm asking if
is it possible, without assuming nothing more than the classical hypotheses of Rouche's theorem, to prove that the zeros of $f+g$ are indeed contained in an explicitly defined region strictly (compactly) contained in $D$?
Classical Rouche's theorem: if $f$ and $g$ are functions analytic inside a region $G\in\Bbb C$ such that $0<|g(z)|<|f(z)|$ on the boundary $\partial D$ of a given subregion $D\Subset G$, then $f$ and $f+g$ have the same number of zeros inside $D$.
Description of Tutschke's refinement: let's start by assuming that $f$ and $g$ are two functions holomorphic in a region $G\Supset D$ for which the standard hypotheses of Rouche's theorem holds in the subregion $D$, i.e. $$ |g(z)|< |f(z)| \quad \forall z\in\partial D. $$ Define
- $q_1,\ldots, q_n$ as the respective multiplicities of the zeros $z_{o1}, \ldots, z_{on}$, and call $N=\sum_{k=1}^{n} q_k$ their sum (the total number of zeros of $f$ in $D$),
- two strictly positive numbers $M$ and $m$ as $$ \begin{align} M &= \frac{\sup_{z\in{\partial D}}|g(z)|}{\inf_{z\in{\partial D}}\dfrac{|f(z)|}{\prod_{k=1}^n|z-z_{ok}|^{q_k}}}, \label{1}\tag{1}\\ \\ m &= \frac{\sup_{z\in{\partial D}}|g(z)|}{\inf_{z\in{\partial D}}|f(z)|}, \label{2}\tag{2} \end{align} $$
- two finite families of open disks indexed by $k= 1, \ldots, n$, $$ \begin{align} \Bbb D_M(z_{ok}) & =\bigg\{z\in\Bbb C : |z-z_{ok}|\le M^{1\over N}\bigg\} \\ \Bbb D_m(z_{ok}) & =\bigg\{z\in\Bbb C : |z-z_{ok}|\le m^{1\over N}\sup_{z\in{\partial D}}|z-z_{ok}|\bigg\}\\ \end{align} $$ and their unions $$ \Bbb D_M \triangleq \bigcup_{k=1}^n \Bbb D_M(z_{ok})\qquad \Bbb D_m \triangleq \bigcup_{k=1}^n \Bbb D_m(z_{ok}). $$
Then the following theorem holds: all the zeros of $f(z)+g(z)$ in $D$ are also contained in the domain $$ \Bbb D_M \cap \Bbb D_m. $$ Proof. By the minimum modulus principle for analytic functions, we have that $$ \dfrac{|f(z)|}{\prod_{k=1}^n|z-z_{ok}|^{q_k}} \ge \inf_{z\in{\partial D}} \dfrac{|f(z)|}{\prod_{k=1}^n|z-z_{ok}|^{q_k}}. $$ By the standard version of Rouché's theorem, $f(z) + g(z)$ has $N$ zeros inside $D$, therefore if ${\bar z}$ is one of these then $$ {\prod_{k=1}^n|{\bar z}-z_{ok}|^{q_k}} \le \left(\inf_{z\in{\partial D}} \dfrac{|f(z)|}{\prod_{k=1}^n|z-z_{ok}|^{q_k}}\right)^{-1}|f({\bar z})|. $$ Since $|f({\bar z})|=|g({\bar z})|\le \sup_{z\in\partial D} |g(z)|$ the zeros of $f(z) + g(z)$ lie inside the set defined by the following inequality $$ {\prod_{k=1}^n|z-z_{ok}|^{q_k}} \le M \quad z\in D. \label{3}\tag{3} $$ From \eqref{3}, putting $\Delta=\sup_{z\in{\partial D}} \prod_{k=1}^n|z-z_{ok}|^{q_k}$ and considering $m$ defined by \eqref{2} the following is deduced $$ M\le \Delta \cdot m. \label{4}\tag{4} $$ If $z\in D$ is a point satisfying \eqref{3} then there's at least an index $k$ for which $$ |z-z_{ok}|\le M^{1\over N}, \label{5}\tag{5} $$ (see the below for a simple proof of this fact) and this implies that all the zeros of $f(z)+g(z)$ are included in $\Bbb D_M$. In a completely analogous way, noting that $\Delta\le\prod_{k=1}^n\sup_{z\in{\partial D}}|z-z_{ok}|^{q_k}$, \eqref{3} and \eqref{4} imply that every zero of $f(z)+g(z)$ satisfies the following inequality $$ \prod_{k=1}^n \left(\frac{|z-z_{ok}|}{\sup_{z\in{\partial D}}|z-z_{ok}|}\right)^{q_k} \le m. $$ For any $z$ satisfying the preceding inequality there exists at least an index $k$ such that $$ \frac{|z-z_{ok}|}{\sup_{z\in{\partial D}}|z-z_{ok}|} \le m^{1\over N} \iff |z-z_{ok}| \le m^{1\over N} \sup_{z\in{\partial D}}|z-z_{ok}| $$ thus all the zeros of $f(z)+g(z)$ are included in $\Bbb D_m$. $\blacksquare$
Notes
- Relation \eqref{4} is proved by a simple reductio ad absurdum argument: assuming its falsity, then it must be $$ |z-z_{ok}|>M^{1\over N} $$ for every point $z\in D$ and every $k=1,\ldots, n$ and thus no $z$ can satisfy \eqref{4}, clearly a contradiction.
- Tutschke's theorem is really a refinement of the usual Rouché's theorem only when $m$ as defined by \eqref{2} is $<1$ (as noted by Tutschke himself in [1], p.433 footnote 3). Indeed, in this case we have that $$ \prod_{k=1}^n|z-z_{ok}|^{q_k}\le M < \sup_{z\in{\partial D}}\prod_{k=1}^n|z-z_{ok}|^{q_k} $$ i.e. the set defined by \eqref{5} is compactly contained in $D$. And obviously the lower $m$ is, the smaller is the set $\Bbb D_M \cap \Bbb D_m$ including all zeros of $f(z)+g(z)$.
Reference
[1] Wolfgang Tutschke, "Über eine Verschärfung der Aussage des Satzes von Rouche" (German), Archiv der Mathematik 17, 432-434 (1966), MR0199353, Zbl 0152.26804.
