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

Real- and Complex- Analytic Functions.

  1. Jul 23, 2010 #1
    Hi Again:

    I don't know if this is obvious or not: An analytic complex function
    f(z)=u(x,y)+iv(x,y) , can be made into an analytic function
    f: R^2 -->R^2, since each of u(x,y) and v(x,y) is itself a real-analytic
    function, i.e., we can use a standard argument by component function.

    How about in the opposite direction, i.e., we have a real-analytic
    function f(x), analytic in an interval (a,b). When can we extend
    f(x) into a complex-analytic function.?. I suspect , thinking of power series,
    that we can use the radius of convergence to construct an analytic function, i.e.,
    if f(x) is analytic in (a-r,a+r), then f(x) can be extended to a complex-analytic
    function in |z-a|<r .

    Is this correct.?
  2. jcsd
  3. Jul 24, 2010 #2
    This can be seen as follows: if [tex]f(z)=u(x,y)+\mathrm{i} v(x,y)[/tex] is complex analytic, then [tex]u[/tex] and [tex]v[/tex] are harmonic hence (real) analytic.

    Yes, given a real analytic function on an (a,b), it can be extended to a complex analytic function.
  4. Jul 24, 2010 #3
    While Anthony's statement is true, the answer to Bacle's question is "no". Consider

    [tex]f(x) = \frac{1}{1 + x^2}[/tex]

    in the interval (-1, 3), Using Bacle's notation, a = 1, r = 2. Then f is real-analytic in the interval (-1, 3), but f is not complex-analytic in |z - 1| < 2 due to the poles at +-i. The function f would be complex-analytic in some disk that didn't include the poles, however.
  5. Jul 24, 2010 #4
    Apologies, I only read the first bit! That will teach me.
  6. Jul 29, 2010 #5
    PeteK, Anthony: Thnaks for your posts. Please check on this; I would appreciate

    your comments:

    Just a question: I guess we could look at 1/(1+x^2) on the real line as agreeing

    with 1-x^2+x^4 -... =Sum_(x=0...n) x^2n on (-1,1). We can also see it as

    the restriction of 1/(1+z^2) , to the real axis, so that, by the identity theorem

    (using the fact that (-1,1) has limit points) , 1/(1+z^2) is the analytic extension

    of ( 1/(1+x^2)) to the complex plane, and :

    1/(1+z^2) = 1-z^2+z^4 -...... on |z|=1

    But then it seems like 1/(1+x^2) on (-1,1) can be extended to the disk |z|=1.

    And it seems like the above could be generalized to analytic functions defined on

    the real line; by the comparison criterion, if f(x) were analytic in (-r,r) (we can

    translate the function if necessary so that it converges in that interval) , then

    we would have -1< |(a_nx^n)/ (a_(n-1)x^n-1)|<1 in (-r,r), which I believe

    would imply that the analytic extension of f(x) to the complex plane would converge

    in that same interval.?
    real line
  7. Jul 29, 2010 #6
    Sorry, I forgot to ask another followup:

    Anthony: I understand that if f(z)=u+iv , then both u,v are harmonic, but
    how does it follow from u,v being harmonic, that each is real-analytic.?.

    I understand that from harmonicity, the second derivatives would be killed,
    but it seems many other mixed partials would not be killed by u (equiv. v)
    being harmonic.

    Would you please expand.?

  8. Jul 29, 2010 #7
    What's a real analytic function?
  9. Jul 29, 2010 #8
    First, let's consider Anthony's statement that

    It would be more precise to say that "Any real analytic function on some open set on the real line can be extended to a complex analytic function on some open set of the complex plane." The function [itex]f(x) = 1/(1 + x^2)[/itex] shows that a real-valued function can be analytic on all of R, but not have a continuation that is analytic on all of C.

    Next, in your above statement you probably mean the disk {z: |z| < 1} since the complex power series [itex]1 - x^2 + x^4 - ...[/itex] doesn't converge, for example, if z = 1.

    Finally, in your last paragraph, do you mean something like "If f(x) is analytic in (r,r), then it has a continuation to the complex numbers that is analytic in the disk {z: |z| < r}"? If so, it not true as shown by the example of f(x) as given above and with any r > 1. If you meant something else, please restate.
  10. Jul 29, 2010 #9
    See the Wikipedia article on analytic functions.
  11. Jul 29, 2010 #10
    Thanks, PeteK.

    Would you please give a hint for why 1/(1+x^2) is real-analytic in
    the interval (-1,3).? I can't see it. I can see how it agrees with:

    Sum_i=0...oo x^(2n) =1-x^2+x^4-........ in (-1,1)

    But I don't see how it is real-analytic beyond (-1,1).

    Would you please explain.?

    P.S: you were right, the radius of convergence of f(z)=1-z^2+z^4-....

    is 1 , and, yes, there are poles at +i, -i' my bad. Tho I think

    f(z) can be continued beyond the disk.

  12. Jul 29, 2010 #11


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2.
  13. Jul 29, 2010 #12
    "If you wanted to see whether or not it's real analytic at, say, 2, then you should look to see whether or not there is a power series defined on an interval containing 2. "

    Yes, but that is precisely where I am stuck; I can see how :


    is defined on (-1,1), and it agrees with 1/(1+x^2) , but I cannot

    think of any other series defined outside of (-1,1), which agrees

    with 1/(1+x^2). Can you spare a paradigm.?
  14. Jul 29, 2010 #13
    Do you know Taylor's Theorem and Taylor Series?
  15. Jul 29, 2010 #14
    Yes, I do know both, but I have not seen them for a few years now;
    I don't know if it is more accurate to say that I knew them.

    Since the functions we are working with are real C<sup>oo</sup> ,
    I guess the part of the theorem that you may be referring to is that,

    I do know, e.g., that, in the real case, if the radius of convergence
    of the Taylor series T(f(x)) of f(x) ) is r (let's standardize to (-r,r)
    again, by translation) , then T(f(x)) does not converge to f(x)
    either at r, or at -r , or both. Same goes for the complex case; there
    must be a point of non-convergence in the boundary of the circle of
    convergence, like in the case of 1/(1+z^2) , where you pointed out
    that there are poles at +i and at -i . (tho we may be able to continue
    f(z) analytically in C-{i,-i} . )

    This is the property I was using for 1/(1+x^2) , which has radius of
    convergence one, i.e., it converges in (-1,1), and diverges at x=1 and
    at x=-1 .
  16. Jul 29, 2010 #15


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The point is that a disk centered on 0 is probably not the best choice if you're looking for a disk that contains 2.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook