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Math Help - compute the integral of a complex-valued polynomial along a curve

  1. #1
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    compute the integral of a complex-valued polynomial along a curve

     \gamma \text{ is any curve joining } -1+i \text{ to }1.\\ \text{ Letting }\gamma(b)=1, \, \gamma(a)=-1+i  \text{ and know that } (-1+i)^2 = -2i,  \begin{align*} \int_\gamma (z^3-6z^2 +4)dz  &= \frac{\gamma^4(b)-\gamma^4(a)}{4}-6\frac{\gamma^3(b)-\gamma^3(a)}{3}+4(\gamma(b)-\gamma(a)) \\&= \frac{1^4- (-1+i)^4}{4}-6\frac{1^3- (-1+i)^3}{3}+4(1 - (-1+i)) \\ &= \frac{1^4- (-2i)^2}{4}-2(1^3- (-1+i)^2(-1+i))+4(2- i)) \\ &= \frac{1-(-4i)}{4} -2(1- (-2i)(-1+i)) +8 - 4i   \\ &= \frac{1+ 4i}{4} -2(-1 -2i) +8 - 4i\\ &= \frac{41}{4} +i \end{align*}

    This question is taken from Fisher's "Complex Variables 2nd Edition" on page 74.

    Is my computation correct? Let me know.


    Thanks a lot.
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  2. #2
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    Re: compute the integral of a complex-valued polynomial along a curve

    I think there is a much easier parameterisation. If you are going from the point \displaystyle \begin{align*} z = 1 + 0i \end{align*} and going to \displaystyle \begin{align*} z = -1 + i \end{align*}, that means you are trying to find the line between the points (1, 0) and (-1, 1).

    The direction vector is \displaystyle \begin{align*} (-1 - 1, 1 - 0) = (-2, 1) \end{align*}, and so your line is \displaystyle \begin{align*} (-2, 1)t + (1, 0) \end{align*}.

    Therefore your parameterisation is \displaystyle \begin{align*} x = -2t + 1 \end{align*} and \displaystyle \begin{align*} y = t \end{align*} with \displaystyle \begin{align*} 0 \leq t \leq 1 \end{align*}. And therefore your line \displaystyle \begin{align*} \gamma \end{align*} is given by the function \displaystyle \begin{align*} z = x + i\,y = -2t + 1 + i\,t \end{align*}.

    So evaluating your integral:

    \displaystyle \begin{align*} \int_{\gamma}{z^3 - 6z^2 + 4\,dz} &= \int_0^1{\left( -2t + 1 + i\,t \right) ^3 - 6 \left( -2t + 1 + i\,t \right) ^2 + 4 \, dt} \end{align*}

    Go from here.
    Thanks from chen09 and topsquark
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  3. #3
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    Re: compute the integral of a complex-valued polynomial along a curve

    As I was using your method, it gave me the right answer and also, showed me where I did wrong on my previous solution (I had (-2i)^2 = -4i, which should be just -4).

    Using the result of a theorem to evaluate this is more efficient, but I appreciate all the help man!
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    Re: compute the integral of a complex-valued polynomial along a curve

    Just out of interest, which theorem is more efficient in this case?
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