# Cauchy's Integral Formula

• May 11th 2011, 08:47 AM
Alexrey
Cauchy's Integral Formula
Hey guys, I was just wondering how I would apply Cauchy's Integral Formula to something like this:

http://img813.imageshack.us/img813/7181/integral.jpg

I would be able to do it if there wasn't a constant of 1/2i there but that constant makes me uncertain.
• May 11th 2011, 08:54 AM
TheEmptySet
Quote:

Originally Posted by Alexrey
Hey guys, I was just wondering how I would apply Cauchy's Integral Formula to something like this:

http://img813.imageshack.us/img813/7181/integral.jpg

I would be able to do it if there wasn't a constant of 1/2i there but that constant makes me uncertain.

I have a question for you. What do you mean by this notation

$\gamma(0,1)$

Is this a circle of radius 1 centered at the origin?

If so the Cauchy integral formula does not apply because the integrand is discontinous at z=1.
• May 11th 2011, 09:22 AM
Alexrey
My bad, yes, gamma (0,1) is an open disk of radius 1 centered at the origin. Surely you could apply Cauchy's integral formula:
http://img38.imageshack.us/img38/506...ralformula.png

with f(w) = z and w-a = z-1, therefore since f(w) is holomorphic in gamma(0, 1) we can apply Cauchy's Integral Formula, with a=1?
• May 11th 2011, 09:41 AM
TheEmptySet
Quote:

Originally Posted by Alexrey
My bad, yes, gamma (0,1) is an open disk of radius 1 centered at the origin. Surely you could apply Cauchy's integral formula:
http://img38.imageshack.us/img38/506...ralformula.png

with f(w) = z and w-a = z-1, therefore since f(w) is holomorphic in gamma(0, 1) we can apply Cauchy's Integral Formula, with a=1?

a must be an interior point of

$\gamma$

Cauchy's integral formula - Wikipedia, the free encyclopedia

The problem is that there is a pole on the countour. Also something else is fishy you sure that it is an open disk?
• May 11th 2011, 09:54 AM
Alexrey
Ah I see, I screwed that one up. Regarding gamma being an open disk, I thought that it was open since I read in my notes somewhere that the interior of gamma was open and therefore presumed that the whole disk was open as well. What can I do if the singularity is ON the contour? Will residues work (I haven't gone over this section yet).

Okay lets rephrase the question to gamma(0,2). I would be able to use Cauchy's Integral Formula then right, since a is inside the contour.
• May 11th 2011, 10:01 AM
TheEmptySet
Quote:

Originally Posted by Alexrey
Ah I see, I screwed that one up. Regarding gamma being an open disk, I thought that it was open since I read in my notes somewhere that the interior of gamma was open and therefore presumed that the whole disk was open as well. What can I do if the singularity is ON the contour? Will residues work (I haven't gone over this section yet).

Okay lets rephrase the question to gamma(0,2). I would be able to use Cauchy's Integral Formula then right, since a is inside the contour.

Yes, this would give

$f(1)=\frac{1}{2\pi i}\int_{\gamma (0,2)}\frac{f(z)}{z-1}dz, \quad f(z)=z$ so we get

$1=\frac{1}{2\pi i}\int_{\gamma (0,2)}\frac{z}{z-1}dz \iff \int_{\gamma (0,2)}\frac{z}{z-1}dz=2\pi i$

So finaly

$\frac{1}{2i}\int_{\gamma (0,2)}\frac{z}{z-1}dz=\frac{1}{2i}(2\pi i)=\pi$
• May 11th 2011, 10:54 AM
Alexrey
Awesome, thanks so much.