Rouché's Theorem

Theorem
Let $\gamma$ be a closed contour.

Let $D$ be the region enclosed by $\gamma$.

Let $f$ and $g$ be complex-valued functions which are holomorphic in $D$.

Let $\size {\map g z} < \size {\map f z}$ on $\gamma$.

Then $f$ and $f + g$ have the same number of zeroes in $D$ counted up to multiplicity.

Proof
Let $N_f$ and $N_{f + g}$ be the number of zeroes of $f$ and $f + g$ in $D$ respectively.

By the Argument Principle:


 * $\ds N_f = \frac 1 {2 \pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z$

Similarly:


 * $\ds N_{f + g} = \frac 1 {2 \pi i} \oint_\gamma \frac {\map {\paren {f + g}'} z} {\map {\paren {f + g} } z} \rd z$

We aim to show that $N_f = N_{f + g}$.

From $\size {\map g z} < \size {\map f z}$ we have that $f$ is non-zero on $\gamma$, otherwise we would have $\size {\map g z} < 0$.

From the fact that $\size {\map g z} \ne \size {\map f z}$ we also have that $\map g z \ne - \map f z$, so $f + g$ is also non-zero on $\gamma$.

We have:

As $\map g z \ne - \map f z$, we have $\dfrac {\map g z} {\map f z} \ne -1$, so $1 + \dfrac g f$ has no zeroes on $\gamma$.

So:

Hence $N_{f + g} = N_f$.