Open mapping theorem (complex analysis)
Theorem on holomorphic functions
title: "Open mapping theorem (complex analysis)" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["theorems-in-complex-analysis", "articles-containing-proofs"] description: "Theorem on holomorphic functions" topic_path: "general/theorems-in-complex-analysis" source: "https://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0
::summary Theorem on holomorphic functions ::
In complex analysis, the open mapping theorem states that if U is a domain of the complex plane \mathbb{C} and f: U\to \mathbb{C} is a non-constant holomorphic function, then f is an open map (i.e. it sends open subsets of U to open subsets of \mathbb{C}, and we have invariance of domain.).
The open mapping theorem points to the sharp difference between holomorphy and real-differentiability. On the real line, for example, the differentiable function f(x)=x^2 is not an open map, as the image of the open interval (-1, 1) is the half-open interval [0, 1).
The theorem for example implies that a non-constant holomorphic function cannot map an open disk onto a portion of any line embedded in the complex plane. Images of holomorphic functions can be of real dimension zero (if constant) or two (if non-constant) but never of dimension 1.
Proof
::figure[src="https://upload.wikimedia.org/wikipedia/commons/e/ef/Openmappingtheorem.png" caption="Black dots represent zeros of g(z). Black annuli represent poles. The boundary of the open set U is given by the dashed line. Note that all poles are exterior to the open set. The smaller red disk is B, centered at z_0."] ::
Assume f: U\to \mathbb{C} is a non-constant holomorphic function and U is a domain of the complex plane. We have to show that every point in f(U) is an interior point of f(U), i.e. that every point in f(U) has a neighborhood (open disk) which is also in f(U).
Consider an arbitrary w_0 in f(U). Then there exists a point z_0 in U such that w_0 = f(z_0). Since U is open, we can find d 0 such that the closed disk B around z_0 with radius d is fully contained in U. Consider the function g(z)=f(z)-w_0. Note that z_0 is a root of the function.
We know that g(z) is non-constant and holomorphic. The roots of g are isolated by the identity theorem, and by further decreasing the radius of the disk B, we can assure that g(z) has only a single root in B (although this single root may have multiplicity greater than 1).
The boundary of B is a circle and hence a compact set, on which |g(z)| is a positive continuous function, so the extreme value theorem guarantees the existence of a positive minimum e, that is, e is the minimum of |g(z)| for z on the boundary of B and e0.
Denote by D the open disk around w_0 with radius e. By Rouché's theorem, the function g(z)=f(z)-w_0 will have the same number of roots (counted with multiplicity) in B as h(z) := f(z)-w_1 for any w_1 in D. This is because h(z) = g(z) + (w_0-w_1), and for z on the boundary of B, |g(z)| \geq e |w_0-w_1|. Thus, for every w_1 in D, there exists at least one z_1 in B such that f(z_1) = w_1. This means that the disk D is contained in f(B).
The image of the ball B, f(B) is a subset of the image of U, f(U). Thus w_0 is an interior point of f(U). Since w_0 was arbitrary in f(U) we know that f(U) is open. Since U was arbitrary, the function f is open.
Applications
References
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