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Thread: line integral

  1. #1
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    line integral

    $\displaystyle \int_C (z*^2dz + z^2dz*)$, where C is the curve defined by $\displaystyle z^2+2|z|^2+z*^2=2(1-i)z+2(1+i)z*$
    $\displaystyle z$ is a complex number x+iy, $\displaystyle z*$ is his complex conjugate
    Thanks
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  2. #2
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    Quote Originally Posted by hurz View Post
    $\displaystyle \int_C (z*^2dz + z^2dz*)$, where C is the curve defined by $\displaystyle z^2+2|z|^2+z*^2=2(1-i)z+2(1+i)z*$
    $\displaystyle z$ is a complex number x+iy, $\displaystyle z*$ is his complex conjugate
    Thanks
    I think that there is a problem with your curve.

    first I will use $\displaystyle z*=\bar{z}$ for complex conjugation
    recall that
    $\displaystyle |z|^2=z\bar{z}$ so we can factor the left hand side as follows

    $\displaystyle z^2+z\bar{z}+z\bar{z}+\bar{z}^2=z(z+\bar{z})+\bar{ z}(z+\bar{z})=(z+\bar{z})^2=(2\text{Re}(z))^2=4x^2$

    Now for the right hand side we get

    $\displaystyle 2(1-i)(x+iy)+2(1+i)(x-iy)=2[(x+iy-ix+y)+(x-iy+ix+y)]=4x+4y$

    This gives the curve as

    $\displaystyle 4x^2=4x+4y \iff y=x^2-x$

    If you expand out the integrand in a similar fashion you will get

    $\displaystyle (2x^2-2y^2)dx+4xydy$
    remember that
    $\displaystyle dz=dx+idy; \quad d\bar{z}=dx-idy$
    So this gives the real line integral

    $\displaystyle \oint (2x^2-2y^2)dx+4xydy$
    but the curve of integration (the parabola) is unbounded so the integral will be as well.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    I think that there is a problem with your curve.

    first I will use $\displaystyle z*=\bar{z}$ for complex conjugation
    recall that
    $\displaystyle |z|^2=z\bar{z}$ so we can factor the left hand side as follows

    $\displaystyle z^2+z\bar{z}+z\bar{z}+\bar{z}^2=z(z+\bar{z})+\bar{ z}(z+\bar{z})=(z+\bar{z})^2=(2\text{Re}(z))^2=4x^2$

    Now for the right hand side we get

    $\displaystyle 2(1-i)(x+iy)+2(1+i)(x-iy)=2[(x+iy-ix+y)+(x-iy+ix+y)]=4x+4y$

    This gives the curve as

    $\displaystyle 4x^2=4x+4y \iff y=x^2-x$

    If you expand out the integrand in a similar fashion you will get

    $\displaystyle (2x^2-2y^2)dx+4xydy$
    remember that
    $\displaystyle dz=dx+idy; \quad d\bar{z}=dx-idy$
    So this gives the real line integral

    $\displaystyle \oint (2x^2-2y^2)dx+4xydy$
    but the curve of integration (the parabola) is unbounded so the integral will be as well.
    If we evaluate the integral from z=1 to z=2(1+i), can i replace y for $\displaystyle x^2-x$ and dy for $\displaystyle (2x-1)dx$ in the integral and evaluate it from x=1 to x=2?
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  4. #4
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    Quote Originally Posted by hurz View Post
    If we evaluate the integral from z=1 to z=2(1+i), can i replace y for $\displaystyle x^2-x$ and dy for $\displaystyle (2x-1)dx$ in the integral and evaluate it from x=1 to x=2?
    Yes that is correct
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    Yes that is correct
    Thanks!
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