Complex analysis: Integration About Singularity

In summary, the conversation discusses evaluating the integral ∫sin(z)/(z^2-4) dz around the contour C, a circle of radius 2 centered at z = 2. The Cauchy Integral Formula is used to split the integral into two parts, one around the outer contour C1 and one around the inner contour C2. By cancelling out the bridges between the two contours, it is determined that the summation of these two integrals is equal to 0. The integral around the entire contour C is then evaluated using the Cauchy Integral Formula and the final answer is determined to be πi sin(2)/2.
  • #1
903
5

Homework Statement



Evaluate the integral ∫sin(z)/(z^2-4) dz about the contour C such that it is a circle of radius 2 centered at z = 2.

Homework Equations



All theorems of complex analysis except residue theorem.

The Attempt at a Solution



There is a singularity at z = 2, so we avoid it by drawing a bridge going from the outer contour now renamed C1, to another circular contour C2 with opposite orientation going around z = 2. The bridges have opposite orientation but same direction, so they cancel out, and the distance between them can be made arbitrarily small.

Therefore the summation of these 2 integrals is C, within which the function is analytic.

Then we recognize that ∫ C1 + ∫ C2 = ∫C = 0.

However I don't know how to apply this directly to solve the problem.

I tried a different way. Let's take the analytic contour C.

Note that f(z) = sin(z)/(z^2-4) = sin(z)/{ (z-2)(z+2) } = sin(z)(z+2)^-1/(z-2)

∫f(z)dz is now in the form ∫f(z)/(z-a) dz = 2πi f(a). Here a = 2, f(z) is now sin(z)(z+2)^-1

Substitute 2 into the equation and we have the integral around C is equal to
2πi*sin(2)/4 = πi sin(2)/2.

However this is for the entire contour C, while what I'm looking for is C1, the outer contour.

Can I just say that since C2 can be made arbitrarily small, the integral around it must be zero, and the answer I got for C is the final answer? If not what should my next step be?
 
Physics news on Phys.org
  • #2
You have the Cauchy Integral Formula, yes? If so then just split 1/(x^2-4) using partial fractions and go from there.
 
  • #3
thank you greatly. I already split it up with the Cauchy integral formula and got the answer. I was just wondering if that's my final answer. I think it is now.
 

1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with functions of complex numbers. It studies the properties and behavior of complex-valued functions, including differentiation, integration, and series representation.

2. What is a singularity in complex analysis?

A singularity in complex analysis refers to a point at which a function is not defined or behaves in an unusual way. This can include poles, essential singularities, and branch points.

3. How is integration around a singularity different from regular integration?

Integration around a singularity involves taking into account the behavior of a function at that singularity, such as how it approaches the singularity and any singularities contained within the integration path. This can result in different integration techniques and non-standard integration paths compared to regular integration.

4. What are some applications of complex analysis in real life?

Complex analysis has applications in various fields, including physics, engineering, and economics. Some examples include the study of fluid dynamics, electrical circuits, and the pricing of financial derivatives.

5. How can I learn more about complex analysis?

There are many resources available to learn more about complex analysis, including textbooks, online courses, and lectures. It is recommended to have a strong foundation in calculus and real analysis before delving into complex analysis. Additionally, practicing with problems and examples can help solidify understanding of the subject.

Suggested for: Complex analysis: Integration About Singularity

Replies
1
Views
448
Replies
3
Views
763
Replies
1
Views
641
Replies
7
Views
919
Replies
7
Views
1K
Replies
2
Views
2K
Replies
3
Views
476
Back
Top