Maximum modulus and open mapping

Try

The polynomial z²+1 has modulus at most two on the closed unit disk, and it reaches two at z=1 and z=−1. It does not achieve a larger value at a hidden interior point. This is an instance of the maximum-modulus principle.

The principle concerns modulus, not a nonexistent ordering of complex numbers. It also treats an interior maximum differently from an interior minimum: f(z)=z has a zero minimum at the center without being constant.

Predict

Predict whether a nonconstant holomorphic function can map an open disk into a line segment. The open-mapping theorem says no: its image must be open in the plane.

This conclusion fails for merely continuous or real-differentiable maps. Taking the real part maps a disk into a real interval, but it is not holomorphic. The complex derivative and its analytic consequences are doing essential work.

Experiment

Compare three maps on the open unit disk.

Map	Interior modulus minimum	Interior modulus maximum
z	Zero at the origin	None
z²	Zero at the origin	None
2	Two everywhere	Two everywhere

The constant map is the necessary exception. The open mapping theorem also excludes constant maps because a single point is not open.

Understand

Here is the local maximum argument. Suppose |f| has a local maximum M at a. If M=0, f vanishes nearby. Otherwise multiply f by a unit complex constant so f(a)=M is positive real. On each sufficiently small circle about a, Re f≤|f|≤M. Cauchy’s mean formula says the average of Re f is exactly M. A continuous function bounded above by M can have average M only by equaling M everywhere on that circle. Since |f|≤M, its imaginary part must also vanish. Varying the radius proves f constant on a disk. The identity theorem, proved shortly from power series, propagates that constant across a connected domain.

For a bounded domain with compact closure, a function continuous on the closure attains a modulus maximum. If it is holomorphic and nonconstant inside, that maximum must occur on the boundary. Boundary continuity and compactness are part of this version.

The open-mapping mechanism also uses the forthcoming Taylor theorem. Near a point where f is not locally constant, write f(z)−f(a)=(z−a)ᵐg(z), with g(a)≠0. On a small disk, g has a holomorphic mth root h. The map φ(z)=(z−a)h(z) has nonzero derivative at a, so the local inverse theorem applies. Its image contains a disk about zero; taking mth powers still contains a disk. Therefore f maps every sufficiently small neighborhood onto a set containing a neighborhood of f(a). This proves openness once the stated analytic factorization is available.

Apply

Repair “The minimum modulus principle says no interior minima.” At a zero, an interior minimum is entirely possible. At a nonzero point, a positive local minimum would give a local maximum for 1/f on a zero-free neighborhood, forcing constancy.

For f(z)=z³, describe what happens near zero. Its derivative vanishes there, but it still maps every small disk onto a disk. Open mapping does not require a nonzero derivative; local one-to-one conformality does.

Continue

Use maximum modulus for boundary estimates and open mapping for qualitative image constraints. The next two lessons supply the Taylor and identity arguments explicitly used here. Optional depth: compare the local inverse theorem with the power-map model, explaining how a critical point introduces multiplicity without destroying openness.