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Math Help - Integrating complex numbers

  1. #1
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    Integrating complex numbers

    I was asked to evaluate the following:
    \oint _C \frac{dz}{z^2-1}
    where C is the circle \left | z \right |=2
    Well I figure that is the function f(z)= \frac{1}{z^2-1} is analytical then according to Cauchy's intergral theorem:
    \oint _C f(z)=0
    This is somewhat complicated since it's difficult to separate the function into the real and imaginary part to take their derivatives. I pluged it into the calculator and it seems like it's not analytical. So, I went about trying to do it.
    z=r e^{it} and  dz=ire^{it}dt
    then it becomes:
    \oint _C \frac{dz}{z^2-1}=\int_0^{2\pi} \frac{ire^{it}dt}{r^2 e^{2it}-1}=\frac{i}{r}\int_0^{2\pi} \frac{e^{it}dt}{e^{2it}-\frac{1}{r^2}}
    let:
    u=e^{2it}-\frac{1}{r^2} and dv=e^{it}dt
    then:
    du=2ie^{2it} and v=\frac{e^{it}}{i}
    And using by parts:
    =(e^{2it}-\frac{1}{r^2})\frac{e^{it}}{i}-\int\frac{e^{it}}{i}2ie^{2it}dt
    =\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\int e^{3it}dt
    =\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\frac{e^{3it}}{3i}
    =\frac{3e^{3it}-2e^{3it}-\frac{3e^{it}}{r^2}}{3i}
    since \left | z \right |=2 r=2, and we have:
    =\frac{e^{3it}-\frac{3e^{it}}{4}}{3i}
    Now pluging in the limits 2\pi and 0 we have:
    =\frac{1-\frac{3}{4}}{3i}-\frac{1-\frac{3}{4}}{3i}=0
    How can this be? It shouldn't be 0 since it's not analytical. What am I missing here?
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  2. #2
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    Quote Originally Posted by synclastica_86 View Post
    I was asked to evaluate the following:
    \oint _C \frac{dz}{z^2-1}
    where C is the circle \left | z \right |=2
    Well I figure that is the function f(z)= \frac{1}{z^2-1} is analytical then according to Cauchy's intergral theorem:
    \oint _C f(z)=0
    This is somewhat complicated since it's difficult to separate the function into the real and imaginary part to take their derivatives. I pluged it into the calculator and it seems like it's not analytical. So, I went about trying to do it.
    z=r e^{it} and  dz=ire^{it}dt
    then it becomes:
    \oint _C \frac{dz}{z^2-1}=\int_0^{2\pi} \frac{ire^{it}dt}{r^2 e^{2it}-1}=\frac{i}{r}\int_0^{2\pi} \frac{e^{it}dt}{e^{2it}-\frac{1}{r^2}}
    let:
    u=e^{2it}-\frac{1}{r^2} and dv=e^{it}dt
    then:
    du=2ie^{2it} and v=\frac{e^{it}}{i}
    And using by parts:
    =(e^{2it}-\frac{1}{r^2})\frac{e^{it}}{i}-\int\frac{e^{it}}{i}2ie^{2it}dt
    =\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\int e^{3it}dt
    =\frac{e^{3it}-\frac{e^{it}}{r^2}}{i}-2\frac{e^{3it}}{3i}
    =\frac{3e^{3it}-2e^{3it}-\frac{3e^{it}}{r^2}}{3i}
    since \left | z \right |=2 r=2, and we have:
    =\frac{e^{3it}-\frac{3e^{it}}{4}}{3i}
    Now pluging in the limits 2\pi and 0 we have:
    =\frac{1-\frac{3}{4}}{3i}-\frac{1-\frac{3}{4}}{3i}=0
    How can this be? It shouldn't be 0 since it's not analytical. What am I missing here?
    Just because f(z) is not analytic at all points in the region enclosed by the contour C doesn't mean that \oint_{C} f(z) \, dz can't equal zero. In fact, if you deform the contour in the given question into a 'dumb-bell' contour you get:

    \oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz = - \frac{2 \pi i}{2} + \frac{2 \pi i}{2} = 0

    where C_1 and C_2 are circles inside C centred at z = -1 and z = 1.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Just because f(z) is not analytic at all points in the region enclosed by the contour C doesn't mean that \oint_{C} f(z) \, dz can't equal zero. In fact, if you deform the contour in the given question into a 'dumb-bell' contour you get:

    \oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz = - \frac{2 \pi i}{2} + \frac{2 \pi i}{2} = 0

    where C_1 and C_2 are circles inside C centred at z = -1 and z = 1.
    Forgive me, my math background isn't that solid. I'm a physics major and tend t o learn math on need to know basis, but what do you mean by dumbell contour? Why does:
    \oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz
    How do know to redefine the new paths?
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  4. #4
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    Quote Originally Posted by synclastica_86 View Post
    Forgive me, my math background isn't that solid. I'm a physics major and tend t o learn math on need to know basis, but what do you mean by dumbell contour? Why does:
    \oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz
    How do know to redefine the new paths?
    http://www.mathhelpforum.com/math-he...tegration.html might help.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Just because f(z) is not analytic at all points in the region enclosed by the contour C doesn't mean that \oint_{C} f(z) \, dz can't equal zero. In fact, if you deform the contour in the given question into a 'dumb-bell' contour you get:

    \oint _C \frac{dz}{z^2-1} = \oint_{C_1} \frac{\frac{1}{z-1}}{z + 1} \, dz + \oint_{C_2} \frac{\frac{1}{z+1}}{z - 1} \, dz = - \frac{2 \pi i}{2} + \frac{2 \pi i}{2} = 0

    where C_1 and C_2 are circles inside C centred at z = -1 and z = 1.
    Ok, so did you use Cauchy's integral theorem here? and assume f(z)=1\frac{1}{z-1} for C_1
    and
    f(z)=1\frac{1}{z+1} for C_2
    But doesn't this required f(z) to be analytical?
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  6. #6
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    I worked on it a bit more. Please tell me if I am on the right track.
    Using partial fraction I constructed 2 circles around z=1 and -1.
    \oint_C\frac{dz}{z^2-1}=\oint_{C_1}\frac{dz}{2(z-1)}-\oint_{C_2}\frac{dz}{2(z+1)} where C_1 is the circle around z=1 and C_2 is a circle around z=-1
    Letting  f(z)=\frac{1}{2} and using the Cauchy's Integral Formula \oint_C\frac{f(z)dz}{z-z_0}=2\pi if(z_0)
    Here, f(z) must be analytical because all the partial derivatives are indeed 0. Is this right?
    This gives:
    \oint_{C_1}\frac{dz}{2(z-1)}-\oint_{C_2}\frac{dz}{2(z+1)}=\frac{2\pi i}{2}-\frac{2\pi i}{2}=0
    Is this correct?
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