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Homework Help: Complex Analysis-Analytic Continuation

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A function f is analytic in the whole complex plane beside 4 poles.
    We know that -1,2,1+5i are poles of f and that f gets only real values in (-1,2).

    Find the fourth pole of f and show that f is a real-valued function for every real z which isn't a pole.

    2. Relevant equations
    3. The attempt at a solution

    I've tried using the symmetry principle but without any success..

    Help is needed

    Thanks !
     
  2. jcsd
  3. Aug 1, 2010 #2

    Dick

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    What symmetry principle did you try to apply and in what way didn't you succeed?
     
  4. Aug 2, 2010 #3
    The symmetry principle I've tried using is:
    Let [tex] R_1 [/tex] be a region located in one side of the real axis and let's assume that the open segment L of the axis is a part of the boundary of [tex] R_1 [/tex].
    Let [tex] f_1 [/tex] be an analytic function in [tex] R_1 [/tex] , continous in [tex] R_1 U L [/tex] and gets real values in L. Let [tex] R_2 [/tex] be the symmetric region of [tex] R_1 [/tex] according the real axis. If we'll look at the region [tex] R=R_1 U L U R_2 [/tex] and we'll define a function f by:
    f(z) = [tex] f_1(z) [/tex] when z is in R1UL and
    f(z) = [tex] \bar{ f_1(\bar{ z } ) } [/tex]
    then f is analytic in R.


    I've tried defining "a" to be the fourth pole and looking at L={(x,0) | -1<=x<=2 } ... I can't figure out how to continue from this point... It's obviously the goal of this question, but I'm pretty much stuck...


    Hope you'll be able to help me

    Thanks !
     
  5. Aug 2, 2010 #4

    Dick

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    Take your f in the symmetry principle to be the original f restricted to the upper half plane. What can you say about the behavior of the extended function near the complex conjugate of 1+5i?
     
  6. Aug 2, 2010 #5
    Well... By the symmetry principle I've quoted above we know that:
    [tex] 1-5i = \barf(1+5i) [/tex] , and for that reason, 1-5i is a pole of the new extanded function... But because f is analytic, we know that [tex] \barf(\barz) = f(z) [/tex] in the lower half plane, and for that reason-1-5i is also a pole of our f...

    As for the second part of the question-
    We can look at [tex] \Omega = C - (1-5i,1+5i,-1,2) [/tex] , then both f and
    [tex] \bar f (\bar z ) [/tex] are analytic in [tex] \Omega [/tex] and agree on (-1,2).
    Hence by the identity theorem we can deduce that they are the same in the whole plane, which means that f gets real-values in the real-axis...

    Am I right?

    Thanks !
     
  7. Aug 2, 2010 #6

    Dick

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    Some of your texing isn't coming out right, like /barf but yes, if 1+5i is a pole the 1-5i must also be a pole. To show f(a) must be real if a is real, let z approach a from the upper and lower half planes and consider the limits of your two functions. What does that tell you about f(a)?
     
  8. Aug 2, 2010 #7
    Thanks a lot :)
     
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