Results 1 to 3 of 3

Math Help - Complex function of complexvariable

  1. #1
    Member Ruun's Avatar
    Joined
    Mar 2009
    From
    North of Spain
    Posts
    129
    Thanks
    13

    Complex function of complexvariable

    If I am given the function u(x,y) wich could be the real part of an analytic complex function of complex variable, and z=x+iy, how can I compute a function f(z)=u(x,y)+iv(x,y)? I have to apply the Cauchy Riemann conditions ( u_{x}=v_{y} and u_{y}=-v_{x}) and integrate both equations?

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Assume u(x,y) is harmonic, then it has a harmonic conjugate v(x,y) such that f(x,y)=u(x,y)+i v(x,y) is analytic and satisfies the partials:

    \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\quad \quad \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}

    so yes, you can partially integrate them like:

    \int \partial v=\int \frac{\partial u}{\partial x}\partial y

    and after the integration with respect to y with x held constant, we obtain something like:

    v(x,y)=\int \frac{\partial u}{\partial x}\partial y+\phi(x)

    where the constant of integration is now a function of x right? Now we use the second CR equations to solve for \phi(x) by writing:

    \frac{\partial u}{\partial y}=-\frac{\partial}{\partial x}\left\{\int \frac{\partial u}{\partial x}\partial y+\phi(x)\right\}

    Now, solve for \phi'(x) and integrate.
    Do it manually a few times first to get the hang of it then explain how that's the same as:

    v(x,y)=\int\frac{\partial u}{\partial x}\partial y-\int\left(\frac{\partial}{\partial x}\int \frac{\partial u}{\partial x}\partial y+\frac{\partial u}{\partial y}\right)dx

    where all the integrations are without the arbitrary function, and it looks intimidating at first sight but becomes a piece of cake after you solve a few of them. For example u(x,y)=y^3-3x^2y, then:

    \int \frac{\partial u}{\partial x}\partial y=\int (-6xy)\partial y=-3xy^2

    Now take the derivative of that with respect to x:

    \frac{\partial}{\partial x}\left(\int \frac{\partial u}{\partial x}\partial y\right)=\frac{\partial}{\partial x}(-3xy^2)=-3y^2

    and continue working the partials and integrals until you get the expression for v(x,y). Then the most general harmonic conjugate is v(x,y)+k.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Ruun's Avatar
    Joined
    Mar 2009
    From
    North of Spain
    Posts
    129
    Thanks
    13
    Yes, by "integrating the equations" I mean what you have done, taking in count that function \phi (x) with the partial respect to y, not just integrating both.

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: October 4th 2011, 05:30 AM
  2. [SOLVED] complex analysis: differentiability of a complex function?
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: February 16th 2011, 08:33 AM
  3. Complex function help
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: September 17th 2010, 08:59 AM
  4. complex function
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 19th 2010, 05:45 AM
  5. Replies: 1
    Last Post: March 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum