Argument principle: Difference between revisions

From Companal
No edit summary
Line 7: Line 7:
<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 UC is an open subset and c is a 0-homologous cycle in U. Suppose f is a meromorphic function on U such that no zero or pole of f lies in U. Then we have:

n(fc;0)=ord(zj)n(c;zj)

Where the sum is taken over all zeros/poles zj for f and ord(zj) is the order of f at </math>z_j</math>: the unique integer n such that (zzj)nf(z) has neither a zero nor a pole at zj.

Equivalently, we can write this as:

n(fc;0)=n(c;zj)n(c;zj)

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