Difference between revisions of "Cauchy Integral theorem"

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(Created page with "The Cauchy Integral theorem states that for a function <math> f(z) </math> which is analytic inside and on a simple closed curve <math> C </math> in some region <math> {\m...")
 
 
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For <math> f(a) </math> a constant, we may rewrite <math> dz = d(z-a) </math>, and noting that <math> z - a = | z - a | e^{i \phi} </math>
 
For <math> f(a) </math> a constant, we may rewrite <math> dz = d(z-a) </math>, and noting that <math> z - a = | z - a | e^{i \phi} </math>
 
and <math> d(z-a) = i |z -a | e^{i\phi} \; d\phi </math>, and
 
and <math> d(z-a) = i |z -a | e^{i\phi} \; d\phi </math>, and
taking the contour of integration to be a circle of unit radius <math> C : |z -a= 1 </math>, we may write
+
taking the contour of integration to be a circle of unit radius <math> C : |z -a |= 1 </math>, we may write
  
  
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Because the derivative of an analytic function is also analytic, the integral vanishes identically within a neighborhood of <math> z = a </math>.
 
Because the derivative of an analytic function is also analytic, the integral vanishes identically within a neighborhood of <math> z = a </math>.
By Cauchy's theorem, the contour of integration may be expanded to any closed curve within {\mathcal R} that contains the point <math> z = a</math>
+
By Cauchy's theorem, the contour of integration may be expanded to any closed curve within <math> {\mathcal R} </math> that contains the point <math> z = a</math>
 
thus showing that the integral is identically zero.
 
thus showing that the integral is identically zero.

Latest revision as of 15:59, 17 August 2021

The Cauchy Integral theorem states that for a function which is analytic inside and on a simple closed curve in some region of the complex plane, for a complex number inside

Proof:

For a constant, we may rewrite , and noting that and , and taking the contour of integration to be a circle of unit radius , we may write


.

By Cauchy's Theorem, we may deform the contour into any closed curve that contains the point and the result holds.


For the case when is not constant we may write


.

We must show that the second term on the left is identically zero.

In a vanishingly small neighborhood

.


Because the derivative of an analytic function is also analytic, the integral vanishes identically within a neighborhood of . By Cauchy's theorem, the contour of integration may be expanded to any closed curve within that contains the point thus showing that the integral is identically zero.