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Thread: Path integral

  1. #1
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    Path integral

    I need to show that $\displaystyle \int_{0,0}^{1,1}z^*dz$ is path dependent by choosing 2 paths. $\displaystyle C_1$ passes through (1,0) and $\displaystyle C_2$ passes through(0,1)

    Here's my attempt:

    path 1:
    $\displaystyle \int_{0,0}^{1,1}z^*dz=\int_{0,0}^{1,1}(x-iy)(dx+idy)=\int_{0,0}^{1,1}(xdx+ydy)+i\int_{0,0}^ {1,1}(xdy-ydx)$
    $\displaystyle =\int_{0,0}^{1,0}(xdx+ydy)+\int_{1,0}^{1,1}(xdx+yd y)+i\int_{0,0}^{1,0}(xdy-ydx)+i\int_{1,0}^{1,1}(xdy-ydx)$
    $\displaystyle =\left (\frac{x^2}{2}+\frac{y^2}{2} \right )^{0,1}_{0,0}+\left (\frac{x^2}{2}+\frac{y^2}{2} \right )^{1,1}_{0,1}+i\left (xy-yx \right )^{0,1}_{0,0}+i\left (xy-yx \right )^{1,1}_{0,1}$
    $\displaystyle =\frac{1}{2}-0+1-\frac{1}{2}$
    the imaginary part cancels since $\displaystyle xy-yx=0$
    $\displaystyle =1$

    Following the same procedure for path 2:
    $\displaystyle \int_{0,0}^{1,1}z^*dz=\int_{0,0}^{1,1}(x-iy)(dx+idy)=\int_{0,0}^{1,1}(xdx+ydy)+i\int_{0,0}^ {1,1}(xdy-ydx)$
    $\displaystyle =\int_{0,0}^{0,1}(xdx+ydy)+\int_{0,1}^{1,1}(xdx+yd y)+i\int_{0,0}^{0,1}(xdy-ydx)+i\int_{0,1}^{1,1}(xdy-ydx)$
    $\displaystyle =\left (\frac{x^2}{2}+\frac{y^2}{2} \right )^{0,1}_{0,0}+\left (\frac{x^2}{2}+\frac{y^2}{2} \right )^{1,1}_{0,1}+i\left (xy-yx \right )^{0,1}_{0,0}+i\left (xy-yx \right )^{1,1}_{0,1}$
    $\displaystyle =\frac{1}{2}-0+1-\frac{1}{2}$
    the imaginary part cancels since $\displaystyle xy-yx=0$
    $\displaystyle =1$

    I recover the same answer, what am I doing wrong?
    Last edited by synclastica_86; Sep 10th 2009 at 05:25 PM.
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  2. #2
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    $\displaystyle z=x+iy$
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  3. #3
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    Yes but $\displaystyle z^*=x-iy$ right?
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