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Shawnarchy (talk | contribs) →Proof: changed "radius of image disk d" to "radius of the disk B" |
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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.
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
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 ''e'' > 0.
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