Seeking Help: Showing Cauchy's Integral with Non-Standard Denominator

  • Thread starter garyman
  • Start date
  • Tags
    Integral
In summary, you are trying to use the Cauchy's integral to show that \ointcos(z) / [ z (z2+9) ] dz = 2\pii/9. However, you are not sure how to get the denominator in the form (z-z0) to apply the theorem.
  • #1
garyman
19
0
Hi,

I'm stuck on this question:

Use the Cauchy's integral to show [tex]\oint[/tex]cos(z) / [ z (z2 +9) ] dz = 2[tex]\pi[/tex]i/9

C is within the square defined by x= +/- 2 , y = +/- 2

I can't see how to get the denominator in the form (z-z0) to apply the theorem. Any help would be much appreciated!
[
 
Physics news on Phys.org
  • #2
can you write z^2 + 9 as something of the form (z-z0) ?
 
  • #3
You must specify what the path of integration is. There are singularities at z = 0, and z = ±3i.

The answer you gave is correct if the contour encloses only z = 0, and is traversed in the counter clockwise direction.

To find the residue at z = 0, you multiply the integrand by (z - 0) = z and evaluate the result at z = 0. If you do this, you will get 1/9.
 
  • #4
malawi_glenn said:
can you write z^2 + 9 as something of the form (z-z0) ?

Not sure, but if you split the fraction up into partial fractions its possible to get an integral of that form. Just not sure how to get the coefficents or indeed if its even the correct method.
 
  • #5
dx said:
You must specify what the path of integration is. There are singularities at z = 0, and z = ±3i.

The answer you gave is correct if the contour encloses only z = 0, and is traversed in the counter clockwise direction.

To find the residue at z = 0, you multiply the integrand by (z - 0) = z and evaluate the result at z = 0. If you do this, you will get 1/9.

Sorry, the path is around the square with sides at x= +/- 2 and y= +/-2 .
 
  • #6
garyman said:
Not sure, but if you split the fraction up into partial fractions its possible to get an integral of that form. Just not sure how to get the coefficents or indeed if its even the correct method.

hint: Yes you can do it, read dx's post - you also need to specify what the integration contour is
 
  • #7
malawi_glenn said:
hint: Yes you can do it, read dx's post - you also need to specify what the integration contour is

Sorry, I don't understand. How does that get it in the form f(z)dz / [z-zo] ?
 
  • #8
First of all, what is z0? And why are you trying to get it into that form?
 
  • #9
the strategy is to find all points in the complex plane where your integrand is singular, then you find out which one are lying inside your contour of integration.

Can you please write down the entire Cauchy's integral Theorem for us so we see that you are trying.
 
  • #10
dx said:
First of all, what is z0? And why are you trying to get it into that form?

cauchys integral is : [tex]\int[/tex]f(z)dz / [z-z0] = 2[tex]\pi[/tex]i f(z0) . I am trying to use this result to prove the statement.
 
  • #11
garyman said:
cauchys integral is : [tex]\int[/tex]f(z)dz / [z-z0] = 2[tex]\pi[/tex]i f(z0) . I am trying to use this result to prove the statement.

you have still not said anything what z0 is ... is it something special about it?
 
  • #12
malawi_glenn said:
you have still not said anything what z0 is ... is it something special about it?

z0 is the value of the singularity.
 
  • #13
garyman said:
z0 is the value of the singularity.

so which possible singularities do your integrand posses (hint: there are three such) and which one is closed within your contour??
 
  • #14
dx said:
That's not correct. The RHS should be

[tex] 2 \pi i \sum Res(f, z_i) [/tex].

In this case there is only one singularity in the contour, so it becomes just [tex] 2 \pi i Res(f, z_0) [/tex]. What is z0 here? I.e. where is the singularity? Do you know what a residue is? What methods of evaluating residues do you know? Do you know what a simple pole is?

One does not need to know the Residue theorem to evaluate this for single poles, in many textbooks Cauchy's formula is introduced earlier than the more general Residue theorem...
 
  • #15
Oh, right. Deleted my prev. post.
 
  • #16
malawi_glenn said:
so which possible singularities do your integrand posses (hint: there are three such) and which one is closed within your contour??

only the singularity at z=0 is enclosed, with the others being at +/- 3i. But i still can't see how to get it inot the correct form... If i could split up into multiple integrals, one with the singularity in the cauchy form and one without a singularity then i could make use of the fact that the integral of fz is zero if it doesn't enclose a singularity. But I'm not sure how to split it in this way...
 
  • #17
As hinted previously, to find all singularities write [itex]z^2+9=(z-a)(z+a)[/itex]. The value a should be easy to find.
 
  • #18
garyman said:
only the singularity at z=0 is enclosed, with the others being at +/- 3i. But i still can't see how to get it inot the correct form...

oh my, then you are really in trouble, i don't even know how to explain such simple thing xD

if you have cos(z)/[ z(z^2 + 9) ] and you want the the form f(z)/(z-0), can you GUESS??
 
  • #19
malawi_glenn said:
oh my, then you are really in trouble, i don't even know how to explain such simple thing xD

if you have cos(z)/[ z(z^2 + 9) ] and you want the the form f(z)/(z-0), can you GUESS??

No. Has it got anything to do with partial fractions?
 
  • #20
There's already a z in the denominator. Just replace it with (z - 0).
 
  • #21
garyman said:
No. Has it got anything to do with partial fractions?

you are kidding me, why not try with f(z) = cos(z)/(z^2 + 9) ??
 
  • #22
dx said:
There's already a z in the denominator. Just replace it with (z - 0).

lol, finally I understand. I blame tiredness!

Cheers for your help!
 

What is Cauchy's Integral with Non-Standard Denominator?

Cauchy's Integral with Non-Standard Denominator is a mathematical concept in complex analysis that allows for the integration of functions with a non-standard denominator, such as square roots or logarithms.

Why is Cauchy's Integral with Non-Standard Denominator important?

This concept is important because it allows for the evaluation of integrals that cannot be solved using traditional methods. It is also a fundamental tool in the study of complex analysis and has many applications in physics and engineering.

How is Cauchy's Integral with Non-Standard Denominator shown?

To show Cauchy's Integral with Non-Standard Denominator, one must use the Cauchy's Integral Formula and manipulate the integrand to have a standard denominator. This can be done through techniques such as partial fraction decomposition or using Cauchy's Residue Theorem.

What are some common mistakes made when showing Cauchy's Integral with Non-Standard Denominator?

Some common mistakes include not properly manipulating the integrand to have a standard denominator, forgetting to include the appropriate singularity in the integration path, or making algebraic errors when evaluating the integral.

Are there any tips for successfully showing Cauchy's Integral with Non-Standard Denominator?

Some tips for success include thoroughly understanding the concept of Cauchy's Integral Formula, practicing with different types of non-standard denominators, and carefully checking for mistakes when evaluating the integral. It can also be helpful to break the problem down into smaller steps and use visual aids, such as diagrams, to better understand the integration path.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
688
Back
Top