Results 1 to 5 of 5

Math Help - line integral

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    45

    line integral

    \int_C (z*^2dz + z^2dz*), where C is the curve defined by z^2+2|z|^2+z*^2=2(1-i)z+2(1+i)z*
    z is a complex number x+iy, z* is his complex conjugate
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by hurz View Post
    \int_C (z*^2dz + z^2dz*), where C is the curve defined by z^2+2|z|^2+z*^2=2(1-i)z+2(1+i)z*
    z is a complex number x+iy, z* is his complex conjugate
    Thanks
    I think that there is a problem with your curve.

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

    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

    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

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

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

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

    \oint (2x^2-2y^2)dx+4xydy
    but the curve of integration (the parabola) is unbounded so the integral will be as well.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2011
    Posts
    45
    Quote Originally Posted by TheEmptySet View Post
    I think that there is a problem with your curve.

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

    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

    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

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

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

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

    \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 x^2-x and dy for (2x-1)dx in the integral and evaluate it from x=1 to x=2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    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 x^2-x and dy for (2x-1)dx in the integral and evaluate it from x=1 to x=2?
    Yes that is correct
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2011
    Posts
    45
    Quote Originally Posted by TheEmptySet View Post
    Yes that is correct
    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 12th 2011, 12:30 AM
  2. [SOLVED] Line Integral along a straight line.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 11th 2011, 08:18 PM
  3. Replies: 0
    Last Post: May 9th 2010, 02:52 PM
  4. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: September 16th 2009, 12:50 PM
  5. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 15th 2009, 02:28 PM

Search Tags


/mathhelpforum @mathhelpforum