Cauchy Integral Formula

  • #1

Homework Statement


Using the Cauchy Integral Formula compute the following integrals,where C is a circle of radius 2a centered at z=o, where 2a<pi


Homework Equations



[tex]\oint\frac{(z-a)e^{z}}{(z+a)sinz}[/tex]

The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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Integrated around |z|=2a? If you write your integrand as f(z)/z (you figure out what f(z) would need to be) the Cauchy integral formula would tell you what the integral is in terms of f(0). Which would be very nice but the Cauchy integral formula doesn't apply because f(z)/z isn't holomorphic inside |z|=2a. Aside from a removable singularity at z=0 there's a pole at z=a. Are you sure you wrote the problem down correctly?
 
  • #3
Integrated around |z|=2a? If you write your integrand as f(z)/z (you figure out what f(z) would need to be) the Cauchy integral formula would tell you what the integral is in terms of f(0). Which would be very nice but the Cauchy integral formula doesn't apply because f(z)/z isn't holomorphic inside |z|=2a. Aside from a removable singularity at z=0 there's a pole at z=a. Are you sure you wrote the problem down correctly?
yes.
I have done the form like this:
[tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex] + [tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z+a)}[/tex]
however the first one is not the standard Cauchy Integral Formula
 
Last edited:
  • #4
Dick
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yes.
I have sperate the form like this:
[tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex]+[tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z-a)}[/tex]
I really don't understand what you are saying there. Did you alter the original problem statement? If so, what was the original?
 
  • #5
I really don't understand what you are saying there. Did you alter the original problem statement? If so, what was the original?
The original problem is this:
[tex]\oint\frac{(z-a)e^{z}}{(z+a)sinz}[/tex]dz c=2a centered at z=0 2a<pi

we can express the integral around the contour as the sum of the integral around z1 and z2 where the contour is a small circle around each pole. Call these contours C1 around z1 and C2 around z2.

So I am tying to express this original problem like:
[tex]\oint\frac{(z-a)e^{z}}{(z+a)}\frac{dz}{sinz}[/tex] + [tex]\oint\frac{(z-a)e^{z}}{sinz}\frac{dz}{(z+a)}[/tex]
 
  • #6
Dick
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Ok, so you are really doing as a residue theorem problem, not just a Cauchy integral problem. In one of those integrals you should multiply by (z-0) and let z approach 0 and in the other one multiply by (z+a) and let z approach -a, right?
 
  • #7
Ok, so you are really doing as a residue theorem problem, not just a Cauchy integral problem. In one of those integrals you should multiply by (z-0) and let z approach 0 and in the other one multiply by (z+a) and let z approach -a, right?
Well,thanks. Could you please give a quick explanation of Cauchy Residue Theorem?
 
  • #8
Dick
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You should probably look it up. I don't necessarily explain things that well. The Cauchy Integral Theorem just says f(a)=(1/(2*pi*i)) times the contour integral f(z)/(z-a) over a circle where f(z) is holomorphic. The residue theorem is the obvious generalization of that to the case where you have multiple poles in a single domain and you cut out a circle around each one and add them up. Which is what you are doing.
 
  • #9
You should probably look it up. I don't necessarily explain things that well. The Cauchy Integral Theorem just says f(a)=(1/(2*pi*i)) times the contour integral f(z)/(z-a) over a circle where f(z) is holomorphic. The residue theorem is the obvious generalization of that to the case where you have multiple poles in a single domain and you cut out a circle around each one and add them up. Which is what you are doing.
Thanks, I have worked it out
 

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