Vipul at 19:03, 12 September 2008

2008-09-12T19:03:40Z

← Older revision		Revision as of 19:03, 12 September 2008
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	==Statement==		==Statement==

	Suppose <math>c \in \mathbb{C}</math> is a complex number, and <math>U</math> is a domain in <math>\mathbb{C}</math>. Then, if <math>D</math> is a disk centered at <math>z_0~~</math> with radius <math>r~~</math>, and <math>z</math> is any point in the interion of the disk, we have:		Suppose <math>c \in \mathbb{C}</math> is a complex number, and <math>U</math> is a domain in <math>\mathbb{C}</math>. Then, if <math>D</math> is a disk centered at <math>z_0</math>, and <math>z</math> is any point in the interion of the disk, we have:

	<math>c = \frac{1}{2\pi i} \oint_{\partial D} \frac{c}{\xi - z} \, d\xi</math>.		<math>c = \frac{1}{2\pi i} \oint_{\partial D} \frac{c}{\xi - z} \, d\xi</math>.
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	==Proof==		==Proof==

	~~We use the parametrization of the circle~~ <math>\~~partial D~~</math> ~~by angle. In other words, we define~~ <math>~~\xi = re~~^{~~i\theta~~}</math>, with <math>\~~theta</math> moving from <math>~~0~~</math> to <math>~~2\pi</math>. ~~Thus~~:		Let us parametrize <math>\xi</math> as <math>z + r(t)e^{it}</math>, with <math>t \in [0,2\pi]</math>. Then, the integral becomes:

	<math>\oint_{\partial D} \frac{c}{\xi - z} \, d\xi = \int_0^{2\pi} \frac{~~cire~~^{i\~~theta~~}}{~~re^~~{i\~~theta~~} - z} \, d\~~theta~~</math>		<math>\oint_{\partial D} \frac{c}{\xi - z} \, d\xi = \int_0^{2\pi} \frac{c}{re^{it}} (rie^{it} + r'(t)e^{it}) \, dt = \int_0^{2\pi i} ci \, dt + \int_{0}^{2\pi} c\frac{r'(t)}{r(t)} \, dt</math>.

			The second integral is zero, because the expression being integrated is the differential of the function <math>\log(r(t))</math>, which has the same values at limits <math>0</math> and <math>2\pi</math>. Thus, we get:

			<math>\oint_{\partial D} \frac{c}{\xi - z} \, d\xi = ci \int_0^{2\pi} dt = 2\pi ic</math>.

			Rearranging this gives the statement of the Cauchy integral formula for constant functions.

Vipul: New page: ==Statement== Suppose $c \in \mathbb{C}$ is a complex number, and $U$ is a domain in $\mathbb{C}$ . Then, if $D$ is a disk centered at $z_0...$

2008-09-12T18:53:06Z

New page: ==Statement== Suppose <math>c \in \mathbb{C}</math> is a complex number, and <math>U</math> is a domain in <math>\mathbb{C}</math>. Then, if <math>D</math> is a disk centered at <math>z_0...

New page

==Statement==

Suppose <math>c \in \mathbb{C}</math> is a complex number, and <math>U</math> is a domain in <math>\mathbb{C}</math>. Then, if <math>D</math> is a disk centered at <math>z_0</math> with radius <math>r</math>, and <math>z</math> is any point in the interion of the disk, we have:

<math>c = \frac{1}{2\pi i} \oint_{\partial D} \frac{c}{\xi - z} \, d\xi</math>.

==Proof==

We use the parametrization of the circle <math>\partial D</math> by angle. In other words, we define <math>\xi = re^{i\theta}</math>, with <math>\theta</math> moving from <math>0</math> to <math>2\pi</math>. Thus:

<math>\oint_{\partial D} \frac{c}{\xi - z} \, d\xi = \int_0^{2\pi} \frac{cire^{i\theta}}{re^{i\theta} - z} \, d\theta</math>

Cauchy integral formula for constant functions - Revision history

Vipul at 19:03, 12 September 2008

Vipul: New page: ==Statement== Suppose c \in \mathbb{C} is a complex number, and U is a domain in \mathbb{C}. Then, if D is a disk centered at z_0...

Vipul: New page: ==Statement== Suppose $c \in \mathbb{C}$ is a complex number, and $U$ is a domain in $\mathbb{C}$ . Then, if $D$ is a disk centered at $z_0...$