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Thread: integrating in the complex plane

  1. #1
    Sep 2009

    integrating in the complex plane

    hi here is the question, by integrating explicitly in the complex plane, evaluate

    $\displaystyle \int_C (z-1)^6 dz $

    where C is the half-circle centered at z = 1 with radius 1, starting at z_0 = 2 and ending at z_1=0, and traveresed in the upper half-plane. Show that the fundamental theorem of calculus holds for this example.

    here's my working out:

    parametrise: $\displaystyle z(t) = 1 + e^{it} $
    $\displaystyle dz = ie^{it}, 0 <=t<= \pi $

    $\displaystyle \int^\pi_0 (e^{it} - 1 + 1)^6 i e^{it} dt $

    let $\displaystyle x = e^{it} $ and $\displaystyle dx = ie^{it} dt $

    $\displaystyle \frac{x^7}{7} ]^\pi_0 $
    $\displaystyle = -2/7 $

    my fundamental part doesnt equal, coz when i did it i got -128/7
    so can someone please help me where i got it wrong?

    coz when i FIRST did it i parametrise this way:
    $\displaystyle z(t) = e^{it} $

    $\displaystyle \int^\pi_0 (e^{it} - 1 )^6 i e^{it} dt $
    and i got the same answer as the fundamental part, but since the question says centered at z = 1 dont you have to add the 1 to z(t)?
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  2. #2
    Apr 2010
    Adelaide, Australia
    You got the first part right:

    A parametrisation of $\displaystyle C$ is $\displaystyle C\left(t\right)=1+e^{it}$.

    Now, for $\displaystyle f\left(z\right)=\left(z-1\right)^6$ the integral is

    $\displaystyle \int_C f=\int_C f\left(z\right)$ $\displaystyle =\int_a^b f\left(C\left(t\right)\right)C^\prime\left(t\right )~dt$ $\displaystyle =\int_0^\pi\left(1+e^{it}-1\right)^6\left(ie^{it}\right)~dt$ $\displaystyle =\int_0^\pi ie^{7it}$ $\displaystyle =\left[\frac{1}{7}e^{7it}\right]_0^\pi=-\frac{2}{7}$.

    You didn't show your working for the FTC part, but I believe you got that part wrong.

    For $\displaystyle f=\left(z-1\right)^6$, the anti-derivative $\displaystyle F=\frac{1}{7}\left(z-1\right)^7$.

    The fundamental theorem of calculus states that

    $\displaystyle \int_C f=F\left(C\left(b\right)\right)-F\left(C\left(a\right)\right)$ $\displaystyle =F\left(C\left(\pi\right)\right)-F\left(C\left(0\right)\right)$ $\displaystyle =F\left(0\right)-F\left(2\right)$ $\displaystyle =-\frac{1}{7}-\frac{1}{7}$ $\displaystyle =-\frac{2}{7}$.

    As expected, the result derived from the fundamental theorem and the result from explicit integration are the same.

    Maths help
    Last edited by lovek323; Apr 30th 2010 at 08:36 PM.
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