Cauchy Integral of Complex Function

  • #1
jaus tail
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Homework Statement


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Homework Equations


Using Cauchy Integration Formula
If function is analytic throughout the contour, then integraton = 0. If function is not analytic at point 'a' inside contour, then integration is 2*3.14*i* fn(a) divide by n!
f(a) is numerator.

The Attempt at a Solution


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But book answer is 2*3.14*i. They haven't included the negative sign. I got same error for similar question. They haven't included the negative sign there as well. They've used Laurentz series. Why is my method wrong?
 

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  • #2
Can you give the original problem? I can't read your hand writing.. Sorry :(
 
  • #3
Here you go:
The contour integral of
upload_2018-1-7_13-34-3.png
e1/zdz with c as the counter clockwise unit circle in the z-plane is equal to:
 

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  • #4
jaus tail said:
But book answer is 2*3.14*i. They haven't included the negative sign. I got same error for similar question. They haven't included the negative sign there as well. They've used Laurentz series. Why is my method wrong?

The contour is counter-clockwise in ##z##. What direction is the contour in ##t##? By the way, you should learn some latex. ##2 \pi i## looks much nicer than 2*3.14*i.
 
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  • #5
Another hint: You can do without any calculation by using the Laurent expansion of the integrand around ##z=0## and the theorem of residues.
 
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  • #6
Yeah i get the answer with laurent but i wanted to try without it. Contour in z is anti clockwise. So in 't' it'll be? How to find that? t = 1/z = 1/(x + iy) = re-iθ
I'm not able to understand as to how re-iθ is clockwise?
z can also be x - iy, so then t will be re
 
  • #7
Just parametrize the path along which to integrate. In your case it's ##z=\exp(\mathrm{i} \lambda)## then substitute ##t=1/z##. Where is the specific problem?
 
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  • #8
So I get t = exp(-iλ)
Okay so as angle for z varies from 0 to 90 and so on to 360
like this angle for t will vary opposite 0 to -90 and so on to -360.
Thanks.
 
  • #9
But you also need to transform ##\mathrm{d} z=\mathrm{d} t \frac{\mathrm{d} z}{\mathrm{d} t}##!
 
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  • #10
z = 1/t
So dz = (-1/t2) dt
 
  • #11
Then, what do you get for the integral?
 
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  • #12
Earlier I'd get -2πi But now I removed the minus sign so it matches answer by laurent.
 
  • #13
Sigh, why don't you present your calculation? Of course, you must get the same, no matter whether you use ##z## or ##t##, and you must not remove the - sign. It's crucial to get the sign right!
 
  • #14
Here it is:
##\oint e^{1/z}dz##
c is |z|=1
##\frac{1}{z} = t##
|t| = 1 angle is opposite in sign
##z = \frac{1}{t}##
##dz = \frac{-1}{t^2}dt##
##\oint \frac{e^t}{t^2}(-dt)##

Is analytic everywhere except at t = 0 which is within contour of |t| = 1
So using Cauchy Integral Formula

##2\pi i \frac{f'(a)}{1!}##
Where a = 0
f(a) is (-1)et
This gives 2*3.14*(-1)
and we remove the minus sign because of angle reverse so we get: 2*3.14*i
 
  • #15
jaus tail said:
Here it is:
##\oint e^{1/z}dz##
c is |z|=1
##\frac{1}{z} = t##
|t| = 1 angle is opposite in sign
##z = \frac{1}{t}##
##dz = \frac{-1}{t^2}dt##
##\oint \frac{e^t}{t^2}(-dt)##

Is analytic everywhere except at t = 0 which is within contour of |t| = 1
So using Cauchy Integral Formula

##2\pi i \frac{f'(a)}{1!}##
Where a = 0
f(a) is (-1)et
This gives 2*3.14*(-1)
and we remove the minus sign because of angle reverse so we get: 2*3.14*i

So it's clear now, right? The ##t## integration is clockwise which means you need an extra factor of ##(-1)## which cancels the minus sign from the ##dt##. So they both agree.
 
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  • #16
Thanks.
 
  • #17
jaus tail said:
Here it is:
##\oint e^{1/z}dz##
c is |z|=1
##\frac{1}{z} = t##
|t| = 1 angle is opposite in sign
##z = \frac{1}{t}##
##dz = \frac{-1}{t^2}dt##
##\oint \frac{e^t}{t^2}(-dt)##

Is analytic everywhere except at t = 0 which is within contour of |t| = 1
So using Cauchy Integral Formula

##2\pi i \frac{f'(a)}{1!}##
Where a = 0
f(a) is (-1)et
This gives 2*3.14*(-1)
and we remove the minus sign because of angle reverse so we get: 2*3.14*i

No, that is not what you get; you get ##2 \pi i##.

One approximation to that would be ##2 \times 3.14 \times i##, but that is by no means exact, or even particularly close to the true answer. A better approximation would be ##2 \times 3.141592653589793238462643383279502884197 \times i, ## but even that is not exact----it is just a much better approximation than what you gave.

The important point for you to grasp is that when the answer contains ##\pi## you should leave it like that; you plug in numbers only if you actually need a numerical answer, and even then, you tailor the number of decimals used to the context of the problem. Sometimes 3.14 will be good enough, but sometimes you will need more decimal places.
 
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