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Homework Help: Integral of z^n on a closed contour

  1. Oct 14, 2007 #1
    I'm trying to show that
    [tex]\int_{c}z^{n}dz= \left\{\frac{0, n\neq-1}{2\pi i, n=-1}\right[/tex]

    I did a change of variables with [tex]z=e^{i\theta}[/tex] and [tex]dz=ire^{i\theta}d\theta[/tex]:

    [tex]=i\int^{2\pi}_{0}r^{n+1}e^{i(n+1)\theta}d\theta[/tex]

    [tex]=ir^{n+1}\int^{2\pi}_{0}e^{i(n+1)\theta}d\theta[/tex] Moving the constant out.

    [tex]=-(n+1)r^{n+1}\int^{2\pi}_{0}\frac{e^{i(n+1)\theta}d\theta}{i(n+1)}[/tex] Getting ready to integrate.

    [tex]=-r^{n+1}(n+1)\left[e^{i(n+1)\theta}\right]^{2\pi}_{0}[/tex]

    [tex]=-r^{n+1}(n+1)[/tex]???

    This is nothing like the answer... where am I going wrong?
     
  2. jcsd
  3. Oct 14, 2007 #2

    Hurkyl

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    Firstly, your work is only valid when n + 1 is nonzero.

    Secondly, your ultimate step doesn't follow from your penultimate step.

    Thirdly, you computed the antiderivative wrong when going from your antepenultimate step to your penultimate step.
     
  4. Oct 14, 2007 #3
    Okay thanks.
     
  5. Oct 14, 2007 #4

    HallsofIvy

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    If [itex]n+ 1\ne 0[/itex], what IS [tex]e^{i(n+1)(2\pi)}[/tex] and [tex]e^{i(n+1)(0)}[/tex]?

    (It's NOT 1!)
     
  6. Oct 14, 2007 #5
    Wait why isn't [tex]e^{i(n+1)(0)}=1?[/tex] or are you talking about the whole integral?

    [tex]e^{i(n+1)(2\pi)}[/tex]

    [tex]=e^{2\pi i}e^{2\pi in}[/tex]

    [tex]=e^{2\pi i}(e^{2\pi i})^n[/tex]

    [tex]=(1)(1)^n[/tex]

    [tex]=1[/tex]

    I thought that I fixed my errors since I had the correct solution after integrating in the right way. Then I looked at what happened when n=-1 and everything seemed fine.

    But, if this is wrong, then I'm still missing something.

    ****Nevermind****** I see what you mean now!
     
    Last edited: Oct 15, 2007
  7. Oct 14, 2007 #6
    [tex]=i\int^{2\pi}_{0}r^{n+1}e^{i(n+1)\theta}d\theta[/tex]

    [tex]=ir^{n+1}\int^{2\pi}_{0}e^{i(n+1)\theta}d\theta[/tex] Moving the constant out.

    [tex]=\frac{r^{n+1}}{n+1}\int^{2\pi}_{0}i(n+1)e^{i(n+1)\theta}d\theta[/tex] Getting ready to integrate, the right way

    [tex]=\frac{r^{n+1}}{n+1}\left[e^{i(n+1)\theta}\right]^{2\pi}_{0}[/tex]

    [tex]=\left[e^{i(n+1)2\pi}-e^{0}\right][/tex]

    [tex]=0[/tex]

    But this whole process only makes sense if [tex]n\neq-1[/tex], so for that case we just plug in -1 for n to:

    [tex]i\int^{2\pi}_{0}r^{0}e^{i(0)\theta}d\theta=2\pi i[/tex]

    So,

    [tex]\int_{c}z^{n}dz= \left\{\frac{0, n\neq-1}{2\pi i, n=-1}\right[/tex]
     
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