Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditions for regularity

  1. Jul 11, 2012 #1
    Are there other ways of determining whether or not a function of a complex variable is analytic without using the Cauchy Riemann conditions? It seems for more complicated functions it's too difficult to decompose an arbitrary function into its real and imaginary parts, so it would be nice if there was another way to determine if the function possesses this property.
  2. jcsd
  3. Jul 11, 2012 #2

    Google "Morera's Theorem": if a function is continuous in a domain and if its line integral over any simple closed piecewise smooth path contained fully in the domain is zero, then the function is analytin in that domain.

  4. Jul 13, 2012 #3
    Couldn't we also say that if a complex function f(z)=f(x+iy) obeys laplace's equation for x and y, i.e [itex]\nabla^{2}=(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}})[/itex] then the function f(z) is analytic over the specified domain?
  5. Jul 13, 2012 #4

    Well, yes: but this is just the Cauchy-Riemann equations in disguise...:)

  6. Jul 13, 2012 #5
    I understand that. It seems that the latter condition is easier to verify for an arbitrary complex function, assuming the function doesn't have to be written in f(z)=u(x,y)+iv(x,y) form to satisfy laplace's equation. Is this true? Could you have a particular analytic function that satisfies laplace's equation without being written in u+iv form?
  7. Jul 13, 2012 #6


    User Avatar
    Science Advisor

    The basic definition of "f is an analytic function" in a given region is "At every point in the region the Taylor's series of f exists and, in some neighborhood of that point converges to the value of f". Of course, the whole point of the Cauchy-Riemann conditions is that they are much easier to show that finding the Taylor's series at every point!

    Oh, and a function is analytic on a region if and only if it is differentiable at every point in the region. But, again, it is typically easier to show the Cauchy-Riemann conditions.
  8. Jul 13, 2012 #7
    I'm just curious though as to whether or not there are conditions that are easier to verify for some functions. It's not always easy to decompose a function in the form u+iv and verify the Cauchy Riemann conditions so I was wondering if instead we could just compute the laplacian of the function without writing it as f(z)=u+iv and see whether or not it equals zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook