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Osmium2356 (talk | contribs) m Clarified some wording in the proof. |
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[[Image:OpenMapping1.png|frame|right|Blue dots represent zeros of ''g''(''z''). Black spikes 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 circle is the boundary of the disk ''B'' centered at ''z''<sub>0</sub>.]]
Assume ''f'' : ''U'' → '''C''' is a non-constant holomorphic function and ''U'' is a [[Domain (mathematical analysis)|___domain]] of the complex plane. We have to show that every [[Point (geometry)|point]] in ''f''(''U'') is an [[interior point]] of ''f''(''U''), i.e. that every point in ''f''(''U'')
Consider an arbitrary ''w''<sub>0</sub> in ''f''(''U''). Then there exists a point ''z''<sub>0</sub> in ''U'' such that ''w''<sub>0</sub> = ''f''(''z''<sub>0</sub>). Since ''U'' is open, we can find ''d'' > 0 such that the closed disk ''B'' around ''z''<sub>0</sub> with radius ''d'' is fully contained in ''U''. Consider the function ''g''(''z'') = ''f''(''z'')−''w''<sub>0</sub>. Note that ''z''<sub>0</sub> is a [[root of a function|root]] of the function.
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