Section 1.25 Rouché’s theorem and open mapping theorem
In this section we prove an important consequence of the argument principle: Rouché’s theorem.
Theorem 1.25.1. Rouché’s theorem.
Let \(f\) and \(g\) be analytic functions on the elementary region \(U\text{,}\) and let \(\gamma\) be a closed path in \(U\) elementary region, let \(\alpha\) be a closed path such that \(\chi_\gamma(z)=1\) for all \(z\in \Int \alpha\text{.}\) If \(\abs{g(z)}< \abs{f(z)}\) for all \(z\in \im \gamma\text{,}\) then \(f\) and \(g\) have the same number of zeros in \(\Int \gamma\) (counting multiplicity).
Proof A.
Observe first that the condition \(\abs{f(z)}> \abs{g(z)}\) for all \(z\in \im \gamma\) implies that \(f(z)\ne 0\) and \(f(z)+g(z)\ne 0\) for all \(z\in \im \gamma\text{.}\)
Let \(Z_f=\{z\in \Int \gamma\colon f(z)=0\}\) and \(Z_{f+g}=\{z\in \Int \gamma\colon (f+g)(z)=0\}\text{.}\) We wish to show that
\begin{equation}
\sum_{z\in Z_f}\ord_f(z)=\sum_{z\in Z_{f+g}}\ord_{f+g}(z)\text{.}\tag{1.96}
\end{equation}
This is the precise way of stating that the number of zeros of \(f\) in \(\Int \gamma\) is equal to the number of zeros of \(f+g\) in \(\Int \gamma\) (counting multipilicty).
Since \(f\) is nonzero on \(\im \gamma\) we have \(f+g=f(1+g/f)\) on \(\im \gamma\text{.}\) Letting \(h=1+g/f\text{,}\) we have
\begin{align*}
\sum_{z\in Z_{f+g}}\ord_{f+g}(z)\amp =\chi_{(f+g)\circ \gamma}(0) \\
\amp = \chi_{(fh)\circ \gamma}(0)\\
\amp =\chi_{(f\circ \gamma) \cdot (h\circ \gamma)}(0) \\
\amp = \chi_{f\circ\gamma}(0)+\chi_{h\circ \gamma}(0)\\
\amp = \sum_{z\in Z_{f}}\ord_{f}(z)+\chi_{h\circ \gamma}(0)\text{.}
\end{align*}
The penultimate equality is an instance of the more general property \(\chi_{\psi\cdot\phi}(0)=\chi_{\psi}(0)+\chi_{\phi}(0)\text{,}\) which is essentially a consequence of the product rule for derivatives. (We leave this elementary proof to the reader.)
Lastly, observe that since \(\abs{g(z)}< \abs{f(z)}\) for all \(z\in \im \gamma\text{,}\) we have
\begin{equation*}
\im h\circ \gamma\subseteq B_1(1)\text{.}
\end{equation*}
Since
\(B_1(1)\) is an elementary region, it follows from
Corollary 1.24.6 that
\(\Int h\circ \gamma\subseteq B_1(1)\text{,}\) and hence that
\(\chi_{h\circ\gamma}(0)=0\text{.}\) We conclude that
\begin{gather*}
\sum_{z\in Z_{f+g}}\ord_{f+g}(z)=\sum_{z\in Z_f}\ord_f(z)\text{,}
\end{gather*}
as desired.
Proof B (sketch).
As an alternative, somewhat slicker proof, consider the function \(h\colon [0,1]\rightarrow \Z\) defined as
\begin{align*}
h(t) \amp = \chi_{(f+tg)\circ \gamma}(0)\\
\amp = \frac{1}{2\pi i}\int_\gamma \frac{f'+tg'}{f+tg}\, dz\text{.}
\end{align*}
First show that \(h\) is continuous, then conclude that \(h\) is constant, since \([0,1]\) is connected and \(\Z\) is discrete. It follows that \(\chi_{f\circ\gamma}=h(0)=h(1)=\chi_{(f+g)\circ \gamma}\text{,}\) as desired.
Example 1.25.2. Roots of polynomials.
Let \(p(x)=x^7+6x^5+3x^2-14x+1\text{.}\) Prove the following statements.
\(p\) has exactly one root in \(B_1(0)\text{.}\)
\(p\) has exactly four roots in the open annulus \(1< \abs{z} < \abs{2}\text{.}\)
\(p\) has exactly 2 roots in the open annulus \(2< \abs{z} < 3\text{.}\)
