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Complex Analysis - Contour integral

  1. Jun 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I have the following problem:
    Compute
    [itex] \operatorname{Re} \int _\gamma \frac{\sqrt{z}}{z+1} dz, [/itex]
    where [itex] \gamma [/itex] is the quarter-circle [itex] \{ z: |z|=1, \operatorname{Re}z \geq 0 , \operatorname{Im} z \geq 0 \} [/itex] oriented from 1 to [itex]i[/itex], and [itex] \sqrt{z} [/itex] denotes the principal branch.

    2. Relevant equations



    3. The attempt at a solution
    I've been trying to solve this using the complex analog of the 2nd Fundamental Theorem of Calculus. Substituting [itex] u = \sqrt{z} [/itex] and using partial fractions, I get
    [itex] \int_\delta 2 - \frac{i}{u+i} + \frac{i}{u-i} du [/itex]
    where delta is the eighth-circle from 1 to [itex] e^{i\pi /4} [/itex]
    This is equal to
    [itex] [2u - i \log(u+i) + i\log (u-i)]_{u=1} ^{u=e^{i\pi/4}} [/itex],

    and the real part is then

    [itex] [\operatorname{Re} u + \operatorname{Arg}(u+i) -\operatorname{Arg}(u-i)]_{u=1} ^{u = e^{i \pi /4}} [/itex]

    However, the arguments at [itex] u=e^{i\pi /4} [/itex] do not come out to nice forms. I am wondering if my approach is even correct, and if there's a better way to solve problems of this type.
    Thanks.

    [Mathematica says the value is approximately -0.584].
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 16, 2012 #2

    vela

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    Combine the logs:
    $$-i \log (u+i) + i\log(u-i) = i\log\frac{u-i}{u+i}$$ For ##u=e^{i\pi/4}##, you should be able to show that ##\frac{u-i}{u+i} = -i \tan \pi/8##.
     
  4. Jun 17, 2012 #3


    Yes. You can approach it entirely in terms of an analytically-continuous antiderivative. But I realize that analytically-continuous thing is what, not easy to understand? First, what is any-old antiderivative? That's easy right:

    [tex]
    \begin{align}
    \int_1^i \frac{\sqrt{z}}{z+1}dz&=\int_1^{\sqrt{i}} \frac{2u^2}{u^2+1}du\\
    &=2(u-\arctan(u)\biggr|_1^{\sqrt{i}}
    \end{align}
    [/tex]

    Ok then, as long as you evaluate that antiderivative along an analytically-continuous path from 1 to [itex]e^{\pi i/4}[/itex] then you can simply evaluate the antiderivative at it's end-points. But the multi-valued function [itex]\arctan(z)[/itex] has branch-points at [itex]\pm i[/itex] so my path from 1 to [itex]\sqrt{i}[/itex] doesn't pass near these points or go around them, so I can use the prinicpal-valued branch of arctan(z) and simply write:

    [tex]\int_{1}^{i}\frac{\sqrt{z}}{z+1}=2\left[(\sqrt{i}-\arctan(\sqrt{i}))-(1-\arctan(1))\right][/tex]
     
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