Rouché's theorem

If $f,g$ are holomorphic near the closure of a bounded domain and $|g|<|f|$ on its boundary, then $f$ and $f+g$ have the same number of interior zeros, counted with multiplicity. Apply the argument principle to the zero-free boundary homotopy $f+tg$.

Let $P(z)={\prod }_{\ell }(z-r_{\ell })$ have distinct roots. Choose a disc around $r_j$ with radius $\delta$ smaller than every separation from $r_j$. On its boundary, $|P(z)|\ge \delta {\prod }_{\ell \ne j}(|r_j-r_{\ell }|-\delta )$. If a holomorphic perturbation $H$ satisfies the displayed inequality, Rouché's theorem shows that $P+H$ has exactly one root in that disc, counted with multiplicity. Discs confined to specified half-planes or quadrants give direct geometric root-location proofs without needing a global root-continuation argument.

A locally uniform limit of nonvanishing holomorphic functions on a connected open set is either nonvanishing or identically zero. More generally, isolated zeros persist nearby with their multiplicity.

A nonconstant holomorphic function on a domain maps open sets to open sets. Around any point, isolate its zero relative to the value there and use Rouché's theorem to show that every sufficiently nearby value has a preimage.

If a nonconstant holomorphic function had a local maximum of its modulus, its open image near that point would contain values of larger modulus. Hence no such local maximum exists.

If $f$ is holomorphic near the closed unit disc, $|f|\ge 1$ on its boundary, and $|f(z_0)|<1$ at one interior point, then $f$ maps the open unit disc onto a set containing the open unit disc. Rouché's theorem first shows that $f$ has a zero and then that every $f-w$ with $|w|<1$ has one.