proof of argument principle
Since is meromorphic, is meromorphic, and hence is meromorphic. The singularities of can only occur at the zeros and the poles of .
I claim that all singularities of are simple poles. Furthermore, if has a zero at some point , then the residue of the pole at is positive and equals the multiplicity of the zero of at . If has a pole at some point , then the residue of the pole at is negative and equals minus the multiplicity of the pole of at .
To prove these assertions, write with . Then
Since , the only singularity of at comes from the first summand. Since is either the order of the zero of at if has a zero at or minus the order of the pole of at if has a pole at , the assertion is proven.
By the Cauchy residue theorem, the integral
equals the sum of the residues of . Combining this fact with the characterization of the poles of and their residues given above, one deduces that this integral equals the number of zeros of minus the number of poles of , counted with multiplicity.
|Title||proof of argument principle|
|Date of creation||2013-03-22 14:34:32|
|Last modified on||2013-03-22 14:34:32|
|Last modified by||rspuzio (6075)|
|Synonym||Cauchy’s argument principle|