Notes - Complex Analysis MT23, Rouché's theorem

Suppose:

• $f, g : U \to \mathbb C$ holomorphic
• $\overline B(a, r) \subseteq U$
• $ \vert f(z) \vert > \vert g(z) \vert $ for all $z \in \partial B(a, r)$

Then:

• $f$ and $f + g$ have the same change in argument around $\partial B(a, r)$…
• and hence the same number of zeroes, counted with multiplicity.

Parameterise the path $\partial B(a, r)$ by $\gamma(t)$. We aim to relate the zeroes of $f$ and zeroes of $f + g$, so consider the “helper function”

\[h(z) := \frac{f(z)+g(z)}{f(z)} = 1 + \frac{g(z)}{f(z)}\]

This is helpful because $h$ has a pole exactly when $f$ has a zero, and a zero exactly when $f + g$ has a zero.

Hence if we can show $N - P = 0$ for $h$ then we are done ($N$ is the number of zeroes, $P$ is the number of poles). The argument principle says that $N - P$ is the winding number of $h \circ \gamma$ or “the change in argument of $h$” (and in particular showing $N - P = 0$ means $f$ and $f + g$ have the same change in argument around $\partial B(a, r)$, since $N$ equals the change in argument of $f + g$ and $P$ equals the change in argument of $f$).

By the argument principle, we can do this by calculating

\[\frac{1}{2\pi i} \int_\gamma \frac{h'(z)}{h(z)} \text d z = \frac{1}{2\pi i} \int_{h \circ \gamma} \frac 1 z \text d z\]

The condition that $ \vert f(z) \vert > \vert g(z) \vert $ implies that $ \vert g(z)/f(z) \vert < 1$, so we actually have that $\text{Image}(h \circ \gamma) \subseteq B(1, 1)$ which is fully contained in $\{ z \in \mathbb C \mid \Re z \ge 0\}$. There’s a holomorphic branch of $\text{Log}$ here, so

\[\int_{h \circ \gamma} \frac 1 z \text d z = \text{Log}(h(\gamma(1))) - \text{Log}(h(\gamma(0))) = 0\]

(alternatively, $\frac 1 z$ is fully holomorphic in $B(1,1)$, so the integral around a closed loop will be zero).

Then we are done.

Note that on the circle $ \vert z \vert = 2$, we have $ \vert z^4 \vert = 16 > 5 \times 2 + 2 \ge \vert 5z + 2 \vert $, so if $g(z) = 5z + 2$, $ \vert P \vert > \vert g \vert $ on the boundary. Then $P - g$ and $z^4$ have the same number of roots in $B(0, 2)$, and since $z^4$ has 4 roots, it follows all roots of $z^4 + 5z + 2$ all have modulus less than $2$.