Cauchy integral formula for constant functions: Difference between revisions

Latest revision as of 19:03, 12 September 2008

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 $z_{0}$ , and $z$ is any point in the interion of the disk, we have:

$c={\frac {1}{2\pi i}}\oint _{\partial D}{\frac {c}{\xi -z}}\,d\xi$ .

Proof

Let us parametrize $\xi$ as $z+r(t)e^{it}$ , with $t\in [0,2\pi ]$ . Then, the integral becomes:

$\oint _{\partial D}{\frac {c}{\xi -z}}\,d\xi =\int _{0}^{2\pi }{\frac {c}{re^{it}}}(rie^{it}+r'(t)e^{it})\,dt=\int _{0}^{2\pi i}ci\,dt+\int _{0}^{2\pi }c{\frac {r'(t)}{r(t)}}\,dt$ .

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

$\oint _{\partial D}{\frac {c}{\xi -z}}\,d\xi =ci\int _{0}^{2\pi }dt=2\pi ic$ .

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

@@ Line 1: / Line 1: @@
 ==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>.
@@ Line 7: / Line 7: @@
 ==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.