Residue theorem - Revision history

Vipul: 5 revisions

2008-05-18T19:18:07Z

5 revisions

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

Vipul: /* Statement */

2008-04-27T22:32:06Z

Statement

← Older revision		Revision as of 22:32, 27 April 2008
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	<math>\oint_c f(z) \, dz = \sum_{j=1}^r n(c;z_j)\operatorname{res}(f; z_j)</math>		<math>\oint_c f(z) \, dz = \sum_{j=1}^r n(c;z_j)\operatorname{res}(f; z_j)</math>

	The residue <math>res</math> here denotes the coefficient of <math>1/(z - z_j)</math> in the laurent expansion about <math>z_j</math>.		The residue <math>\operatorname{res}</math> here denotes the coefficient of <math>1/(z - z_j)</math> in the laurent expansion about <math>z_j</math>.

	==Related facts==		==Related facts==

	* [[Winding number version of Cauchy integral formula]] is a special case of this, where the function has precisely one pole of order one. However, we typically ''use'' the winding number version of Cauchy integral formula to establish the residue theorem.		* [[Winding number version of Cauchy integral formula]] is a special case of this, where the function has precisely one pole of order one. However, we typically ''use'' the winding number version of Cauchy integral formula to establish the residue theorem.

Vipul at 22:31, 27 April 2008

2008-04-27T22:31:48Z

← Older revision		Revision as of 22:31, 27 April 2008
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	Suppose <math>U</math> is an open subset of <math>\mathbb{C}</math>. Let <math>z_1, z_2, \ldots, z_r</math> be points in <math>U</math> and <math>f:U \setminus \{ z_j\}_{1 \le j \le r} \to \mathbb{C}</math> be a [[holomorphic function]]. Let <math>c</math> be a 0-homologous sum of loops in <math>U \setminus \{ z_j\}_{1 \le j \le r}</math> such that <math>c</math> is zero-homologous. Then, we have:		Suppose <math>U</math> is an open subset of <math>\mathbb{C}</math>. Let <math>z_1, z_2, \ldots, z_r</math> be points in <math>U</math> and <math>f:U \setminus \{ z_j\}_{1 \le j \le r} \to \mathbb{C}</math> be a [[holomorphic function]]. Let <math>c</math> be a 0-homologous sum of loops in <math>U \setminus \{ z_j\}_{1 \le j \le r}</math> such that <math>c</math> is zero-homologous. Then, we have:

	<math>\oint_c f(z) \, dz = \sum_{j=1}^r n(c;z_j)res(f; z_j)</math>		<math>\oint_c f(z) \, dz = \sum_{j=1}^r n(c;z_j)\operatorname{res}(f; z_j)</math>

	The residue <math>res</math> here denotes the coefficient of <math>1/(z - z_j)</math> in the laurent expansion about <math>z_j</math>.		The residue <math>res</math> here denotes the coefficient of <math>1/(z - z_j)</math> in the laurent expansion about <math>z_j</math>.

Vipul at 19:33, 26 April 2008

2008-04-26T19:33:55Z

← Older revision		Revision as of 19:33, 26 April 2008
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			{{basic fact}}
	==Statement==		==Statement==

Vipul: /* Related facts */

2008-04-19T15:00:57Z

Related facts

← Older revision		Revision as of 15:00, 19 April 2008
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	==Related facts==		==Related facts==

	* [[Winding number version of Cauchy integral formula]] is a special case of this, where the function has precisely one pole of order one.		* [[Winding number version of Cauchy integral formula]] is a special case of this, where the function has precisely one pole of order one. However, we typically ''use'' the winding number version of Cauchy integral formula to establish the residue theorem.

Vipul: New page: ==Statement== Suppose $U$ is an open subset of $\mathbb{C}$ . Let $z_1, z_2, \ldots, z_r$ be points in $U$ and $f:U \setminus \{ z_j\}_{1 \...$

2008-04-19T14:58:37Z

New page: ==Statement== Suppose <math>U</math> is an open subset of <math>\mathbb{C}</math>. Let <math>z_1, z_2, \ldots, z_r</math> be points in <math>U</math> and <math>f:U \setminus \{ z_j\}_{1 \...

New page

==Statement==

Suppose <math>U</math> is an open subset of <math>\mathbb{C}</math>. Let <math>z_1, z_2, \ldots, z_r</math> be points in <math>U</math> and <math>f:U \setminus \{ z_j\}_{1 \le j \le r} \to \mathbb{C}</math> be a [[holomorphic function]]. Let <math>c</math> be a 0-homologous sum of loops in <math>U \setminus \{ z_j\}_{1 \le j \le r}</math> such that <math>c</math> is zero-homologous. Then, we have:

<math>\oint_c f(z) \, dz = \sum_{j=1}^r n(c;z_j)res(f; z_j)</math>

The residue <math>res</math> here denotes the coefficient of <math>1/(z - z_j)</math> in the laurent expansion about <math>z_j</math>.

==Related facts==

* [[Winding number version of Cauchy integral formula]] is a special case of this, where the function has precisely one pole of order one.