# Solving Contour Integration Homework

• claralou_
In summary: You need to be more careful. In summary, the conversation discusses the continuity of the function f(z) = 1/sin^(2)(z) on a circle ΓR with radius R > 0 in the complex plane. The main question is for which values of R is the function continuous on ΓR, and the answer may depend on R. The conversation also mentions the use of the function sin(x) = (e^(ix) - e^(-ix)) / 2i and the method of finding residues to evaluate the contour integral of f(z) over ΓR. The student attempts to find the poles of the function and calculate the corresponding residues, but there are some errors and confusion along the way.
claralou_

## Homework Statement

For R > 0,
assume ΓR is a circle {z ∈ C : |z| = R} with anticlockwise direction.
For which R>0, does the the function f(z) = 1/sin^(2)(z) be continuous on ΓR
and evaluate ∫_{ΓR} dz/sin^(2)(z) for each R (the answer may be dependent on R).

## Homework Equations

sinx= (e^(ix) - e^(-ix)) / 2i (possibly)
Resf(z)= lim (pole x f(z))

## The Attempt at a Solution

used 1/e^(iz) = sin(x)
so found
1/sin^{2}x = 1 / (e^{2iz}

Began using the

closed integral over C of f(z)dz = Integral from -R to +R of f(x)dx + integral f(z) dz

Found that there was a pole at z = 1/2i and found the residue at that point to be 1/(2ie)

The way you use x and z as the same thing (?) is confusing.

Your first equation in part 3 is wrong. If it would be right there wouldn’t be any pole (and it would imply the sine has no zeros). It would also mean the sine is just the exponential function rotated in the complex plane. It is not.

I don’t understand the pole you calculated, it is not a pole of the original function and not a pole of the other one.

mfb said:
The way you use x and z as the same thing (?) is confusing.

Your first equation in part 3 is wrong. If it would be right there wouldn’t be any pole (and it would imply the sine has no zeros). It would also mean the sine is just the exponential function rotated in the complex plane. It is not.

I don’t understand the pole you calculated, it is not a pole of the original function and not a pole of the other one.

I tried following a method I found on a website for contour integration. Feel this is where I have gone wrong.
Should I be using the first equation in part 2?

claralou_ said:
Should I be using the first equation in part 2?
That will work.
Alternatively, find the zeros first without using any exponentials.

mfb said:
That will work.
Alternatively, find the zeros first without using any exponentials.

Do you mean to find where 1/sin^2(x) would be zero?

claralou_ said:
Do you mean to find where 1/sin^2(x) would be zero?
No, obviouslsy not! You want points where ##1/ \sin^2(z)## is singular. What are the only points where a ratio gives singular results?

Ray Vickson said:
No, obviouslsy not! You want points where ##1/ \sin^2(z)## is singular. What are the only points where a ratio gives singular results?

I thought the isolated singularities were when the bottom line can equal 0 ie where f in this case would have poles of pi*k for some k in the complex numbers?

claralou_ said:
I thought the isolated singularities were when the bottom line can equal 0 ie where f in this case would have poles of pi*k for some k in the complex numbers?

You are getting close, but those are not the points where ##1/\sin(z) = 0##, which is what you said!

Also, you need more details: exactly what values of ##k \in C## should you use?

Ray Vickson said:
You are getting close, but those are not the points where ##1/\sin(z) = 0##, which is what you said!

Also, you need more details: exactly what values of ##k \in C## should you use?

I meant k in integers sorry! Following an alternative method I've learnt, if i just sub in sin^(z) as (z - 1/z) / (2i) and multiply it out I get z^2 over (z^4 - 2z^2 +1).
After solving I found z^2 to be equal to plus/minus 1 so could z be equal to plus/minus i?

claralou_ said:
I meant k in integers sorry! Following an alternative method I've learnt, if i just sub in sin^(z) as (z - 1/z) / (2i) and multiply it out I get z^2 over (z^4 - 2z^2 +1).
After solving I found z^2 to be equal to plus/minus 1 so could z be equal to plus/minus i?
I have absolutely no idea what you are trying to do or say. You have a function ##f(z) = 1/\sin^2(z)## with known poles in ##C##.

You can compute the residues of ##f## at these poles, then use the residue theorem to finish the job.

That's all there is to it!

claralou_ said:
sin^(z) as (z - 1/z) / (2i)
These two things are not the same. There are some exponentials missing.

## 1. What is contour integration?

Contour integration is a mathematical technique used to evaluate integrals along a specified path or contour in the complex plane. It is commonly used in complex analysis and has applications in physics, engineering, and other fields.

## 2. How do I solve contour integration problems?

The first step in solving a contour integration problem is to identify the contour or path along which the integral is to be evaluated. Then, you can use techniques such as the Cauchy-Goursat theorem, residue theorem, or Cauchy's integral formula to evaluate the integral.

## 3. What are some common mistakes made when solving contour integration problems?

Some common mistakes in contour integration include not considering the direction of the contour, not correctly applying the Cauchy integral formula, and not accounting for poles or branch points in the contour.

## 4. How can I improve my skills in solving contour integration problems?

Practice and familiarity with the techniques used in contour integration are key to improving your skills. It is also helpful to have a strong understanding of complex analysis and its underlying principles.

## 5. Can contour integration be used to solve real-world problems?

Yes, contour integration has many applications in physics, engineering, and other fields. It can be used to solve problems involving electric fields, fluid mechanics, and signal processing, among others.

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