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Simple complex integral

  1. Oct 9, 2007 #1
    Evaluate [tex] \int_C f(z)dz [/tex] by theorem 1 and check your result by theorem 2 where f(z) = z^4 and C is the semicircle |z|=2 from -2i to 2i in the right half-plane.
    Theorem 1 : [tex] \int_{z_0} ^{z_1} f(z)dz = F(z_1) - F(z_0) \ \ \frac{dF}{dz}=f(z) \\ [/tex]
    [tex] \int_{-2\iota} ^{2\iota}z^4dz = \frac{z^5}{5} = \frac{2^6 \iota}{5} \\[/tex]
    Theorem 2: [tex] \int_C f(z)dz = \int_a ^bf(z(t)) \frac{dz}{dt}dt\\ [/tex]. The path C can be represented by [tex] z(t)=2\exp{\iota t} \ f(z)=(2\exp{\iota t})^4 \ = \ 2^4\exp{4\iota t}\ \mbox{ therefore }\ \frac{dz}{dt}= 2 \iota\exp{\iota t} \\ [/tex] [tex] \ \mbox{ therefore, the integral}\ = 2^5\iota\int_{\frac{-\pi}{2}} ^{\frac{\pi}{2}} \exp{5\iota t}dt \ = \ 2^5\iota\frac{\exp{6it}}{6\iota} \ = \ \frac{2^5}{3}(\frac{\exp{3\pi \iota} - \exp{-3\pi \iota} }{2}) =0 \\ [/tex]. Both theorems should give the same answer, but I don't get the same answer. Where have I gone wrong. Help please. Thanks.
     
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  3. Oct 9, 2007 #2

    Dick

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    In the last line, how did exp(5it) become exp(6it) after the integration?
     
  4. Oct 9, 2007 #3
    What is it then?
     
  5. Oct 9, 2007 #4

    Dick

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    What's the integral of exp(k*x)dx? I don't think it has a k+1 in it.
     
  6. Oct 9, 2007 #5
    Thanks.
     
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