Argument principle: Difference between revisions

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<math>n(f \circ c; 0) = \sum ord(z_j)n(c;z_j)</math>
<math>n(f \circ c; 0) = \sum ord(z_j)n(c;z_j)</math>


Where the sum is taken over all zeros/poles <math>z_j</math> for <math>f</math> and <math>ord(z_j)</math> is the unique integer <math>n</math> such that <math>(z - z_j)^nf(z)</math> has neither a zero nor a pole at <math>z_j</math>.
Where the sum is taken over all zeros/poles <math>z_j</math> for <math>f</math> and <math>ord(z_j)</math> is the [[order of zero for function at point|order]] of <math>f</math> at </math>z_j</math>: the unique integer <math>n</math> such that <math>(z - z_j)^{-n}f(z)</math> has neither a zero nor a pole at <math>z_j</math>.


Equivalently, we can write this as:
Equivalently, we can write this as:

Revision as of 22:35, 26 April 2008

This article gives the statement, and possibly proof, of a basic fact in complex analysis.
View a complete list of basic facts in complex analysis

This fact is an application of the following pivotal fact/result/idea: residue theorem
View other applications of residue theorem OR Read a survey article on applying residue theorem

Statement

Suppose is an open subset and is a 0-homologous cycle in . Suppose is a meromorphic function on such that no zero or pole of lies in . Then we have:

Where the sum is taken over all zeros/poles for and is the order of at </math>z_j</math>: the unique integer such that has neither a zero nor a pole at .

Equivalently, we can write this as:

where the first summation is for zeros, counted with multiplicity, and the second summation is for poles, counted with multiplicity.