📂Complex Anaylsis

Proof of Rouché's Theorem

Theorem ¹

If $f$ and $g$ are analytic on a simple closed contour $\mathscr{C}$ and its interior, and satisfy $|g(z)| < |f(z)|$ on $\mathscr{C}$, then $f$ and $f + g$ have the same number of zeros in the interior of $\mathscr{C}$.

Explanation

This is a theorem in which the originally given function is regarded as $h = f + g$ and is written by neatly separating it into $f$ and $g$. In particular, for polynomial functions, such manipulation is very easy, so it can be applied usefully. Moreover, combined with numerical methods, one can determine quite accurately where exactly the solutions of the equation $h(z) = 0$ are located.

Proof

Strategy: Define $Z$ so that $Z(0)$ denotes the number of zeros of $f$ and $Z(1)$ denotes the number of zeros of $g$, and show that $Z(0) = Z(1)$. The values of $Z$ are integers, and if $Z$ is a continuous function, then such a $Z$ can only be a constant function, so we obtain $Z(0) = Z(1)$.

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For $t \in [0,1]$, consider $h(z,t) := f(z) + t g(z)$. Since $f$ and $g$ are analytic on $\mathscr{C}$ and its interior, $h(z,t)$ has no poles on $\mathscr{C}$ and its interior for a fixed $t$. If $f(z) + t g(z) = 0$, then $$ |f(z)| = |-tg(z)| = |tg(z)| $$ but since $t \in [0,1]$, we have $|f(z)| \le |g(z)|$. This contradicts the assumption, so $f(z) + t g(z) \ne 0$ must hold.

Since $f(z) + t g(z) \ne 0$, we can define a new function $\displaystyle Z(t) := {{1} \over {2 \pi i}} \int_{\mathscr{C}} {{f ' (z) + t g ' (z) } \over {f(z) + t g (z) }} dz$.

By the definition of $Z: [0,1] \to \mathbb{Z}$, $Z(t)$ is the number of zeros of the meromorphic function $f(z) + t g (z)$. Here, if $t=0$ then $Z(0)$ is the number of zeros of $f$, and if $t=1$ then $Z(1)$ is the number of zeros of $f+g$. That is, the proof is complete once we show $Z(0) = Z(1)$.

Meanwhile, since the codomain of $Z: [0,1] \to \mathbb{Z}$ is the set of integers $\mathbb{Z}$, if $Z$ is a continuous function, then $Z$ can only be a constant function. Computing $| Z(t) - Z(s)|$, upon taking a common denominator we obtain $$ |Z(t) - Z(s)| = {{ |t-s| } \over {2 \pi}} \left| \int_{\mathscr{C}} {{f(z) g ' (z) - f '(z) g(z)} \over { (f(z) + t g(z))(f(z) + s g(z)) }} dz \right| $$ Meanwhile, $$ | f(z) + t g(z) | \ge |f(z)| - t |g(z)| \ge |f(z)| - |g(z)| > 0 $$

$$ | f(z) + s g(z) | \ge |f(z)| - |g(z)| > 0 $$ Here, since $\mathscr{C}$ is compact, by the extreme value theorem there exists $M>0$ satisfying $\displaystyle \left| {{f(z) g ' (z) - f '(z) g(z)} \over { (|f(z)| - |g(z)|)^2 }} \right| \le M$.

ML Lemma: For a positive number $M$ satisfying $|f(z)| \le M$ and the length $L$ of $\mathscr{C}$, $$ \left| \int_{\mathscr{C}} f(z) dz \right| \le ML $$

Letting the length of $\mathscr{C}$ be $L$, $$ |Z(t) - Z(s)| \le |t - s|{{ML} \over {2 \pi}} $$ Therefore, as $| t - s | \to 0$, we have $ | Z(t) - Z(s) | \to 0$, which means that $Z$ is (uniformly) continuous. The values of $Z(t)$ are integers, and to have continuity, $Z$ can only be a constant function, so we obtain $Z(0) = Z(1)$.

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