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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>z_0</math> in <math>U</math>. Since ''U'' is open, we can find <math>d>0</math> such that the closed disk <math>B</math> around ''z''<sub>0</sub> with radius ''d'' is fully contained in ''U''. Since ''U'' is connected and ''f'' is not constant on ''U'', we then know that ''f'' is not constant on ''B''. Consider the [[Image (mathematics) | image]] point, <math>w_0 = f(z_0)</math>. Then <math>f(z_0)-w_0 = 0</math>, making <math>z_0</math> a [[
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>.
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