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Math Help - Complex Integration

  1. #1
    Super Member Deadstar's Avatar
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    Complex Integration

    Right isn't so much a question as an 'am i doing this right'.

    Evaluate \int_\delta \frac{z^2}{z - 1} dz where \delta is the circle with centre 0 and radius 2, traversed in the usual anti-clockwise sense.

    And my answer...

    The function \frac{z^2}{z - 1} is holomorphic on the complex plane except at z=1.

    So, using Cauchys integral formula we get f(1) = \frac{1}{2 \pi i} \int_\delta \frac{f(z)}{z - 1} dz with f(z) = z^2.

    Hence, \int_\delta \frac{f(z)}{z - 1} = 2 \pi i f(1) = 2 \pi i 1^2 = 2 \pi i.

    This right?

    If so, how come the answer doesnt depend on \delta? Does it not matter what the radius of the circle is?
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Deadstar View Post
    Right isn't so much a question as an 'am i doing this right'.

    Evaluate \int_\delta \frac{z^2}{z - 1} dz where \delta is the circle with centre 0 and radius 2, traversed in the usual anti-clockwise sense.

    And my answer...

    The function \frac{z^2}{z - 1} is holomorphic on the complex plane except at z=1.

    So, using Cauchys integral formula we get f(1) = \frac{1}{2 \pi i} \int_\delta \frac{f(z)}{z - 1} dz with f(z) = z^2.

    Hence, \int_\delta \frac{f(z)}{z - 1} = 2 \pi i f(1) = 2 \pi i 1^2 = 2 \pi i.

    This right?
    Yes

    If so, how come the answer doesnt depend on \delta? Does it not matter what the radius of the circle is?
    It does depend on it.
    Because the pole z=1 is within the circle, you get the result.
    If it wasn't (say you considered \delta as the circle of center 5 and radius 1), then the integral would have been 0.
    Look here : Cauchy's integral formula - Wikipedia, the free encyclopedia (they say 'Then for every a in the interior of D')
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