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Question integrating |f(z)||dz| over a contour C

  1. Feb 28, 2009 #1
    1. The problem statement, all variables and given/known data

    I want to compute [tex]\int_{C}^{}{|f(z)||dz|}[/tex] along the contour C given by the curve [tex]y=x^2[/tex] using endpoints (0,0) and (1,1). I am to use [tex]f(z)=e^{i\cdot \texrm{arg}(z)}[/tex]


    2. Relevant equations

    3. The attempt at a solution
    I know that for all complex numbers z, [tex]|e^{i\cdot \texrm{arg}(z)}}|=1[/tex]. So now I am looking at the integral [tex]\int_{C}^{}{1|dz|}[/tex]

    Is the approach I take below correct?

    A complex representation of [tex]C[/tex] can be given by [tex]\gamma (t)=t+it^2[/tex] for [tex]0 \leq t \leq 1[/tex]. Then [tex]\gamma^{'}(t)=1+i2t[/tex]. We have

    [tex]\int_{C}^{}{|f(z)||dz|}[/tex]

    [tex]=\int_{0}^{1}{\gamma^{'}(t)\bigg|\frac{dz}{dt}\bigg||dt|}[/tex]

    [tex]=\int_{0}^{1}{(1+i2t)|dt|}[/tex]

    [tex]=[t+it^2]_{t=0}^{t=1}[/tex]

    [tex]=1+i[/tex]
     
  2. jcsd
  3. Feb 28, 2009 #2

    Dick

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    It's a real integral. The result can't have an i in it. dz=d(t+it^2)=dt*(1+i2t). |dz|=dt*|1+i2t|.
     
  4. Feb 28, 2009 #3
    OK, I should have had [tex]|\gamma^{'}(t)|[/tex], instead of [tex]\gamma^{'}(t)[/tex].

    So the original integral becomes [tex]\int_{0}^{1}{\sqrt{1+4t^2}}dt [/tex] or would I have
    [tex]\int_{0}^{1}{\sqrt{1+4t^2} \cdot \sqrt{1+4t^2}dt}[/tex] ?
     
    Last edited: Feb 28, 2009
  5. Feb 28, 2009 #4

    Dick

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    The first choice looks just fine.
     
  6. Feb 28, 2009 #5
    Ok, I see why. Thank you very much!
     
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