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Assume ''f'':''U'' → '''C''' is a non-constant holomorphic function and <math>U</math> is a connected [[open subset]] of the complex plane. We have to show that every [[Point (geometry) | point]] in <math>f(U)</math> is an [[interior point]] of <math>f(U)</math>, i.e. that every point in <math>f(U)</math> is contained in a disk which is contained in <math>f(U)</math>.
Consider an arbitrary <math>
We know that ''g''(''z'') is not constant, and by further decreasing ''d'', we can assure that ''g''(''z'') has only a single root in ''B''. (The roots of holomorphic non-constant functions are isolated.) Let ''e'' be the minimum of |''g''(''z'')| for ''z'' on the boundary of ''B'', a positive number. (The boundary of ''B'' is a circle and hence a [[compact set]], and |''g''(''z'')| is a [[continuous function]], so the [[extreme value theorem]] guarantees the existence of this minimum.) Denote by <math>D</math> the disk around <math>w_0</math> with [[radius]] <math>e</math>. By [[Rouché's theorem]], the function <math>g(z)=f(z)-w_0</math> will have the same number of roots in ''B'' as <math>f(z)-w</math> for any <math>w</math> within a distance <math>e</math> of <math>w_0</math>. Thus, for every <math>w</math> in <math>D</math>, there exists one (and only one) <math>z_1</math> in <math>B</math> so that <math>f(z_1) = w</math>. This means that the disk ''D'' is contained in ''f''(''B''), which is a subset of <math>f(U)</math>. Thus <math>w_0</math> is an [[interior point]] of <math>f(U)</math>.
== References ==
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