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Complex analysis - Cauchy Theorem

  1. Apr 13, 2006 #1
    Hi again. Can somebody help me out with this question?

    "[tex]\int_{C_1(0)} \frac {e^{z^n + z^{n-1}+.....+ z + 1}} {e^{z^2}} \,dz[/tex]

    Where [itex]C_r(p)[/itex] is a circle with centre p and radius r, traced anticlockwise."

    I'd be guessing that you have to compare this integral with the Cauchy integral formula. I have no idea how to go about doing this though!

    Thanks in advance.
     
    Last edited: Apr 13, 2006
  2. jcsd
  3. Apr 13, 2006 #2

    HallsofIvy

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    I'm afraid you are going to have to explain what those "zn", "zn-1", etc. mean. Are those powers of z: "zn", "zn-1", etc.?
    Oh, and is that "z" supposed to be ez?
    That is, do you mean:
    [tex]\int_{C_1(0)} \frac {e^{z^n} + e^{z^{n-1}}+.....+ e^z + 1} {e^{z^2}} \,dz[/tex]
    (Click on that to see the code and note, especially my use of "{ }".)

    Since, looking at your LaTex code, it is clear that is what you mean, here is how I would do it: Go ahead and divide each term by [itex]e^{z^2}[/itex]:
    [tex]\int_{C_1(0)} \left({e^{z^{n-1}} + e^z^{n-3}+.....+ e^{z^{-1}} + e^{-2}}\right)dz[/tex]

    Now, which of those, if any, are singular inside C1(0)?
     
  4. Apr 16, 2006 #3
    I assure you, this is what it says in the question:

    [tex]\int_{C_1(0)} \frac {e^{z^n + z^{n-1}+.....+ z + 1}} {e^{z^2}} \,dz[/tex]

    Unless my lecturer made a mistake in printing the question. I doubt it though.

    I think you're supposed to look at it as [itex]e^w[/itex] where [itex] w = z^n + z^{n-1}+.....+ z + 1[/itex]

    I suppose you could use the rule [itex] e^{a+b} = e^a * e^b [/itex] to try and start sorting this out.
     
    Last edited: Apr 16, 2006
  5. Apr 16, 2006 #4

    HallsofIvy

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    Well, okay, then, do what I suggested with that integrand and you get
    [tex]\int_{C_1(0)}\left(e^{z^{n-2}}+ e^{z^{n-3}}+ ...+ ze^{-2z}+ e^{-2z}\right)dz[/tex]
    Once again, which of those, if any, are singular inside the unit circle? Those that are not singular (the exponent of e is greater not negative) will give 0 integral. For those that are, you might try writing [itex]z= e^{i\theta}[/itex] and integrating from [itex]\theta= 0 to 2\pi[/itex].
     
  6. Apr 16, 2006 #5

    Hurkyl

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    No, he gets

    [tex]e^{z^n} \cdot e^{z^{n-1}} \cdot \ldots
    \cdot e^{z^3} \cdot e^{z^1} \cdot e^{z^0}[/tex]
     
    Last edited: Apr 16, 2006
  7. Apr 17, 2006 #6

    HallsofIvy

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    Hey, just because I can't do arithmetic!
     
  8. Apr 17, 2006 #7
    Would I be right in saying that the answer is just zero?

    I'm guessing this is because all the exponential powers are positive and so Cauchy's Theorem applies?
     
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