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Math Help - Complex Integration

  1. #1
    Member Haven's Avatar
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    Complex Integration

    Let  f(z) = \bar{z} where  \bar{z} is the conjugate of z.

    Let  g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{f(s)}{s-z}\\ds

    Where C is the boundary of the unit circle. Show  g(z) is analytic in the interior of C by evaluating the integral.

    By multiplying the top and bottom of the integrand by s, I get

     g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{|s|^{2}}{s(s-z)}\\ds

    and since we are integrating on C and  C = \{ z = e^{i\theta} : 0\leq\theta\leq\\2\pi \} , then  |s|^{2} = 1

    which brings us to:

     g(z) = \frac{1}{2\pi\\i}\int_{C}\frac{1}{s(s-z)}\\ds

    How ever I cannot figure out what allows me to integrate here.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The residue theorem extablishes that...

    \int_{C} f(s)\cdot ds= 2 \pi i \sum_{n} R_{n} (1)

    ... where...

    R_{n} = \lim_{s \rightarrow s_{n}} (s-s_{n})\cdot f(s)_{s=s_{n}} (2)

    ... being s_{n} is every pole of f(*) inside the path C. In your case is...

    f(s)= \frac{1}{2 \pi i}\cdot \frac{1}{s\cdot (s-z)} (3)

    ... so that is...

    g(z)= \frac{1}{z} - \frac{1}{z} =0 , |z|<1

    g(z)= -\frac{1}{z} , |z|>1 (4)

    If |z|=1 there is a pole is on the path C and the residue theorem can't be applied...

    Kind regards

    \chi \sigma
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  3. #3
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    As nice of an argument that is, we're only on integration in our complex variables course. So we're expected to give an argument that relies completely on integration methods. I love the residue theorem, but sadly I cannot use it.
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  4. #4
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    How about thiis:

    First note if z=0 then g(0)=\oint\frac{ds}{s^2}=0 via s=e^{it} and direct integration.
    If z\ne 0 then g(z)=\frac{1}{2\pi i z}\mathop\oint\limits_{|s|=1}\left(\frac{1}{s-z}-\frac{1}{s}\right)ds.
    Now, you can directly integrate those. For the first, we can deform the path:

    \mathop\oint\limits_{|s|=1}\frac{ds}{s-z}dz=\mathop\oint\limits_{s=z+e^{it}} \frac{ds}{s-z}=2\pi i
    and:
    \mathop\oint\limits_{|s|=1}\frac{ds}{s}=2\pi i
    as well by the same substitution method. Note I could deform the path for the first one only because the deformation was over a region in which the function was analytic. I could not do this if |z|>1, since then I would be deforming it across the singular point at z. So then g(z)=0 in the unit disc and certainly the zero function is analytic.
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