Complex Analysis-Analytic Continuation

[WannaBe22]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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

Help is needed

Thanks !

 

[Dick]

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

 

[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 , continuous 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 !

 

[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?

 

[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 !

 

[Dick]

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 !

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)?

 

[WannaBe22]

Thanks a lot :)