Open mapping theorem (complex analysis): Difference between revisions

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We know that ''g''(''z'') is not constant and holomorphic. The roots of ''g'' are isolated and by further decreasing the radius of the image disk ''d'', 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]], andon which |''g''(''z'')| is a positive [[continuous function]], so the [[extreme value theorem]] guarantees the existence of a positive minimum. Let''e'', that is, ''e'' beis the minimum of |''g''(''z'')| for ''z'' on the boundary of ''B'', aand positive''e'' > number0.

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'')−''w₁'' for any ''w₁'' within ain distance''D''. This is because
''eh''(''z'') = ''g''(''z'') + (''w''₀ - ''w''₁), and for ''z'' on the boundary of ''B'', |''g''(''z'')| ≥ ''e'' > |''w''₀ - ''w''₁|. Thus, for every ''w''₁ in ''D'', there exists at least one ''z''₁ in ''B'' sosuch that ''f''(''z''₁) = ''w₁''. 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 the image of an open set by a holomorphic function ''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 ==