# Path integral

• Sep 10th 2009, 10:03 AM
synclastica_86
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?
• Sep 10th 2009, 11:53 AM
shawsend
$\displaystyle z=x+iy$
• Sep 10th 2009, 05:26 PM
synclastica_86
Yes but $\displaystyle z^*=x-iy$ right?