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Complex circuation problem.

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-prn2/1395894_10201699046814591_1790916582_n.jpg

    Assignment 3.

    2. Relevant equations
    Cauchy's differentiation formula
    http://en.wikipedia.org/wiki/Cauchy's_integral_formula
    3. The attempt at a solution

    I checked if ##f(z) = e^{-z}## is analytic, and it is not. So it seems I have to do the circulation the hard way: parametrize z into ##z(t) = \cos(t) +i \sin(t)##

    But that sounds like a bad way to do it. So you guys got any better ideas? Maybe I am missing something?

    thanks for all help! :)
     
  2. jcsd
  3. Oct 30, 2013 #2

    Office_Shredder

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    Why is e-z not analytic?
     
  4. Oct 30, 2013 #3
    Oh right. Now that I have rested a bit at home I see it is obviously analytic. how stupid of me.

    Anyway, how do I solve it?

    I guess the strategy should be something like this?

    1) Apply Cauchy's differentiation formula, with ##z_0 = 0## and ##f(z_0) = e^{-z_0}##
    2) Rearrange so the integral in assignment 3 is a function of ##f^{(n)} (z_0)##.
    3) Find ##f^{(n)}##.

    Err, how do I do step 3? Won't ##f^{(n)}## vary between being zero and 1 after inserting ##z_0 = 0##, depending on the value of n?
     
    Last edited: Oct 30, 2013
  5. Oct 30, 2013 #4

    Dick

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    You are making this sound a lot more elaborate than it is. Just do it. Try a few examples, like n=1 and n=2 first. Sure ##f^{(n)}(0)## will depend on n. But it doesn't vary between zero and one.
     
  6. Oct 31, 2013 #5
    Yes, you're right.

    But I end up with

    [tex] \frac{2 \pi}{(n-1)!} \cdot (-1)^{n-1} i [/tex]

    while the correct answer is just

    [tex]\frac{2 \pi}{(n-1)!}[/tex]
     
  7. Oct 31, 2013 #6

    Dick

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    That 'correct answer' doesn't look correct to me. I agree with your answer.
     
  8. Oct 31, 2013 #7
    OK, thanks! :) I guess my book's solution-appendix is wrong on this assignment.
     
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