Results 1 to 8 of 8

Math Help - complex integration

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    complex integration

    I am in the process of studying a little complex analysis when I have time.

    What is the best way to approach something like this?:

    \int\frac{z^{2}+sinh(2z)}{z^{3}-31z^{2}+4z-12i}

    First, I thought about using sinh(2z)=\frac{1}{2}(e^{2z}-e^{-2z})

    Do the roots of the denominator have a lot to do with it?.

    Just like some input from those in the know.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,903
    Thanks
    329
    Awards
    1
    Quote Originally Posted by galactus View Post
    I am in the process of studying a little complex analysis when I have time.

    What is the best way to approach something like this?:

    \int\frac{z^{2}+sinh(2z)}{z^{3}-31z^{2}+4z-12i}

    First, I thought about using sinh(2z)=\frac{1}{2}(e^{2z}-e^{-2z})

    Do the roots of the denominator have a lot to do with it?.

    Just like some input from those in the know.
    (I was wondering what happened to that little "dz" thingy in there. )

    But seriously, I can't be sure of what you are integrating over here. The real line, a curve, an area... The best method is going to depend heavily on what you are integrating over. (Warning: not that I am likely to be able to show you how once you specify it.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by galactus View Post
    I am in the process of studying a little complex analysis when I have time.

    What is the best way to approach something like this?:

    \int\frac{z^{2}+sinh(2z)}{z^{3}-31z^{2}+4z-12i}

    First, I thought about using sinh(2z)=\frac{1}{2}(e^{2z}-e^{-2z})

    Do the roots of the denominator have a lot to do with it?.

    Just like some input from those in the know.
    If the integral is around a well behaved closed curve then the residue
    theorem will give the integral.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Thanks fellas. Yes, it's a closed curve. And very well-behaved. Not a brat.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,903
    Thanks
    329
    Awards
    1
    Quote Originally Posted by galactus View Post
    Thanks fellas. Yes, it's a closed curve. And very well-behaved. Not a brat.
    The residue theorem is going to be a pain in the arse here. galactus: you are going to have to supply the curve as the three poles may or may not be included inside the contour.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I see what you mean. I knew there were three poles, but this is all that was given. I will try to gather more info regarding the curve. Thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    To galactus: Use \oint instead to signify around a closed curve.


    It is possible that Galactus is integrating in an open set where the function is holomorphic. In that case, Cauchy's Theorem will say the integral is zero.


    Where is this problem from?
    Last edited by ThePerfectHacker; September 8th 2007 at 05:13 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I seen it on another forum. I thought it was an unusually advanced and interesting problem. I have been studying complex analysis a bit and was curious also.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Integration II
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 12th 2010, 09:52 AM
  2. Complex Integration
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 9th 2009, 02:35 PM
  3. complex integration
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 28th 2009, 03:36 AM
  4. More Complex Integration Please Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 4th 2009, 09:45 PM
  5. Integration and complex nos
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 1st 2006, 05:45 AM

Search Tags


/mathhelpforum @mathhelpforum