Rouché's Theorem

Theorem
Let $f$ and $g$ be complex-valued functions which are holomorphic in the interior of some simply connected region $D$.

Let $\left\vert{g \left({z}\right)}\right\vert < \left\vert{f \left({z}\right)}\right\vert$ on the boundary of $D$.

Then $f$ and $f + g$ have the same number of zeroes in the interior of $D$ 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:


 * $\displaystyle N_f = \frac 1 {2 \pi i} \oint_D \frac {f' \left({z}\right)} {f \left({z}\right)} \rd z$

Similarly:


 * $\displaystyle N_{f + g} = \frac 1 {2 \pi i} \oint_D \frac {\left({f + g}\right)' \left({z}\right)} {\left({f + g}\right) \left({z}\right)} \rd z$

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

From $\left\vert{g \left({z}\right)}\right\vert < \left\vert{f \left({z}\right)}\right\vert$ we have that $f$ is non-zero on $D$, as $\left\vert{z}\right\vert \ge 0$ for all $z \in \C$.

From the fact that $\left\vert{g \left({z}\right)}\right\vert \ne \left\vert{f \left({z}\right)}\right\vert$ we also have that $g \left({z}\right) \ne - f \left({z}\right)$, so $f + g$ is also non-zero on $D$.

We have:

As $g \left({z}\right) \ne - f \left({z}\right)$, we have $\dfrac {g \left({z}\right)} {f \left({z}\right)} \ne -1$, so $1 + \dfrac g f$ has no zeroes on $D$.

So:

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