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Math Help - Check solution of integral

  1. #1
    Junior Member piglet's Avatar
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    Check solution of integral

    I'm preparing for an exam and would like someone to check this if possible

     \int_{C}(z^{2}+z)dz where  C is any path from  -i to  2+i

    I figured out that  (z^{2}+z) was not analytic using cauchy riemann equations

    So i parametrized   -i to  2+i using the formula
     z(t) = a + (b-a)t where  0<=t<=1

    Filling into the above formula i got  z(t) = 2t + i(t-1)
    and  \frac{dz}{dt} = 2+i

    So i said  \int_{C}(z^{2}+z)dz = \int^{1}_{0}[(2t + i(t-1))^{2} + (2t +i(t-1))](2+i)dt

    i proceed to work this out an got an answer of  -\frac{5}{3} - \frac{4}{3}i

    The answer isn't really that important to me, i just would like to know if the line above my answer is correct

    Thanks,
    Piglet
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  2. #2
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    Shouldn't the parametrisation be 2t + i(2t - 1) instead, because yours never attains the point 2+i for t in [0,1]. Otherwise you have the right idea with regards to computing the integral.
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  3. #3
    Junior Member piglet's Avatar
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    Quote Originally Posted by Magus01 View Post
    Shouldn't the parametrisation be 2t + i(2t - 1) instead, because yours never attains the point 2+i for t in [0,1]. Otherwise you have the right idea with regards to computing the integral.
    Yeah your spot on there, silly mistake

    note to self: practice simple multiplication
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  4. #4
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    suppose z=t+(t+1)i and t blong to [-1,1]
    then your question can be transfered into
    \int_{C}(z^{2}+z)dz = \int^{-1}_{1}[-(t+1)+(2t^{2}+3t+1)i]dt=\int^{-1}_{1}(2t^{2}+3t+1)idt-2=(4/3+2)i-2

    i trust you can solve the problem above!
    Last edited by nicklus; April 21st 2010 at 07:54 AM.
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  5. #5
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    my result is (4/3+2)i-2
    Last edited by nicklus; April 21st 2010 at 07:55 AM.
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