Homework Help: Complex Analysis-Analytic Continuation

1. Aug 1, 2010

WannaBe22

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. Aug 1, 2010

Dick

What symmetry principle did you try to apply and in what way didn't you succeed?

3. Aug 2, 2010

WannaBe22

The symmetry principle I've tried using is:
Let $$R_1$$ 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 $$R_1$$.
Let $$f_1$$ be an analytic function in $$R_1$$ , continous in $$R_1 U L$$ and gets real values in L. Let $$R_2$$ be the symmetric region of $$R_1$$ according the real axis. If we'll look at the region $$R=R_1 U L U R_2$$ and we'll define a function f by:
f(z) = $$f_1(z)$$ when z is in R1UL and
f(z) = $$\bar{ f_1(\bar{ z } ) }$$
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 !

4. Aug 2, 2010

Dick

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?

5. Aug 2, 2010

WannaBe22

Well... By the symmetry principle I've quoted above we know that:
$$1-5i = \barf(1+5i)$$ , and for that reason, 1-5i is a pole of the new extanded function... But because f is analytic, we know that $$\barf(\barz) = f(z)$$ 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 $$\Omega = C - (1-5i,1+5i,-1,2)$$ , then both f and
$$\bar f (\bar z )$$ are analytic in $$\Omega$$ 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 !

6. Aug 2, 2010

Dick

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)?

7. Aug 2, 2010

WannaBe22

Thanks a lot :)