Contour Integration

  • Thread starter claralou_
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  • #1
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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)
 

Answers and Replies

  • #2
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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.
 
  • #3
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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?
 
  • #4
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Should I be using the first equation in part 2?
That will work.
Alternatively, find the zeros first without using any exponentials.
 
  • #5
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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?
 
  • #6
Ray Vickson
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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?
 
  • #7
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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?
 
  • #8
Ray Vickson
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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?
 
  • #9
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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?
V grateful your help!
 
  • #10
Ray Vickson
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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?
V grateful your help!
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!
 
  • #11
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sin^(z) as (z - 1/z) / (2i)
These two things are not the same. There are some exponentials missing.
 

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