Argument principle - Revision history

Vipul: 4 revisions

2008-05-18T19:09:43Z

4 revisions

← Older revision	Revision as of 19:09, 18 May 2008
(No difference)

Vipul at 22:36, 26 April 2008

2008-04-26T22:36:13Z

← Older revision		Revision as of 22:36, 26 April 2008
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 [[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>.		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:
Line 14:		Line 14:

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

			==Facts used==

			* [[Residue theorem]]
			* [[Order of zero equals residue of logarithmic derivative]]

Vipul: /* Statement */

2008-04-26T22:35:39Z

Statement

← Older revision		Revision as of 22:35, 26 April 2008
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:

Vipul at 19:30, 26 April 2008

2008-04-26T19:30:37Z

← Older revision		Revision as of 19:30, 26 April 2008
Line 1:		Line 1:
			{{basic fact}}
			{{application of\|residue theorem}}
	==Statement==		==Statement==

Vipul: New page: ==Statement== Suppose $U \subset \mathbb{C}$ is an open subset and $c$ is a 0-homologous cycle in $U$ . Suppose $f$ is a [[meromorphic function]...

2008-04-19T20:34:37Z

New page: ==Statement== Suppose <math>U \subset \mathbb{C}</math> is an open subset and <math>c</math> is a 0-homologous cycle in <math>U</math>. Suppose <math>f</math> is a [[meromorphic function]...

New page

==Statement==

Suppose <math>U \subset \mathbb{C}</math> is an open subset and <math>c</math> is a 0-homologous cycle in <math>U</math>. Suppose <math>f</math> is a [[meromorphic function]] on <math>U</math> such that no zero or pole of <math>f</math> lies in <math>U</math>. Then we have:

<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>.

Equivalently, we can write this as:

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

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

Argument principle - Revision history

Vipul: 4 revisions

Vipul at 22:36, 26 April 2008

Vipul: /* Statement */

Vipul at 19:30, 26 April 2008

Vipul: New page: ==Statement== Suppose U \subset \mathbb{C} is an open subset and c is a 0-homologous cycle in U. Suppose f is a [[meromorphic function]...

Vipul: New page: ==Statement== Suppose $U \subset \mathbb{C}$ is an open subset and $c$ is a 0-homologous cycle in $U$ . Suppose $f$ is a [[meromorphic function]...