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==Proof==
[[Image:
Assume ''f'':''U''
Consider an arbitrary <math>w_0</math> in <math>f(U)</math>. Then there exists a point <math>z_0</math> in ''U'' such that <math>w_0 = f(z_0)</math>. Since ''U'' is open, we can find <math>d>0</math> such that the closed disk <math>B</math> around <math>z_0</math> 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'' because of [[Analytic continuation]]. Consider the function <math>g(z)=f(z)-w_0</math>. Note that <math>z_0</math> is a [[root of a function|
We know that ''g''(''z'') is not constant and holomorphic. The reciprocal of any holomorphic ''g''(''z'') is [[meromorphic]] and has isolated poles. Thus the roots of holomorphic non-constant functions are isolated. In particular, 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''.
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Denote by <math>D</math> the disk around <math>w_0</math> with [[radius]] ''e''. 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 <math>f(B)</math>.
The image of the ball ''B'', <math>f(B)</math> is a subset of the image of ''U'', <math>f(U)</math>. Thus <math>w_0</math> is an
== Applications ==
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