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Calculate integral using Cauchy's formula

  1. Mar 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##K## be a circle ##|z-2|=2##. Using Cauchy's formula calculate ##\int _K\frac{(z-1)\sin(z)}{z^2-2z-3}dz##


    2. Relevant equations

    Cauchy's formula: ##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##


    3. The attempt at a solution

    It's my first time dealing with this kind of problems so I am not really sure that what I am doing is in fact the right way to do it. I would be really happy if one could check my solution.

    ##\frac{(z-1)\sin(z)}{z^2-2z-3}=\frac{(z-1)\sin(z)}{(z-3)(z+1)}## therefore ##a_1=3## and ##a_2=-1##.

    For ##a_1=3##:

    ##f(z)=\frac{(z-1)\sin(z)}{z+1}##

    ##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

    ##\frac{2\sin(3)}{4}2\pi i=\int _K\frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}dz##

    For ##a_2=-1##:

    ##f(z)=\frac{(z-1)\sin(z)}{z-3}##

    ##f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}##

    ##\frac{-2\sin(-1)}{-4}2\pi i=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##

    ##-\pi i\sin(1)=\int _K\frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}dz##


    Since ##K## is a circle with center in 2 and radius 2, point ##a_2=-1## is not inside ##K##. Which means that only integral around ##a_1=3## counts and therefore

    ##\int _K\frac{(z-1)\sin(z)}{z^2-2z-3}dz=\int _K\frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}dz=\frac{2\sin(3)}{4}2\pi i##.

    I HOPE so. Thanks in advance!
     
  2. jcsd
  3. Mar 26, 2014 #2

    vela

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    You have the basic idea down, but what you wrote below is incorrect.

    For the given K and the way you chose ##a## and f, you can't say that
    $$f(a)2\pi i=\oint _k\frac{f(z)dz}{z-a}$$ because function f isn't differentiable everywhere inside the region bounded by K. Therefore, the stuff you wrote below isn't correct:
    Note also that when you considered the other pole, you said
    $$\int _K \frac{\frac{(z-1)\sin(z)}{z+1}}{z-3}\,dz = \frac{2 \sin 3}{4}2\pi i.$$ You also said
    $$\int _K \frac{\frac{(z-1)\sin(z)}{z-3}}{z+1}\,dz = -\pi i \sin 1.$$ You have the same integral equal to two different quantities!

    When you calculated what the integral equaled by considering the pole at z=3, you were done. The pole at z=-1 is outside the contour, so you don't need to consider it at all.
     
  4. Mar 26, 2014 #3
    I completely agree with that. I only wanted to check if my idea is right, where I completely forgot that ##f## may not be Holomorphic outside ##K##.

    So in case I would have two singular points inside given ##K##, than the integral would be a sum of two, or not?
     
  5. Mar 26, 2014 #4

    vela

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    Yes, that's how it will turn out, as you will learn later.
     
  6. Mar 26, 2014 #5
    Thank you!
     
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