I really got stuck with this question...guess it is not difficult, but i haven't done enough complex analysis...could anyone help me out? I am gonna take sub in 21, April!!!
If f is a function defined by a complex power series expansion in za which converges for za<1 and diverges for za>1, which of the following must be true?
A. f(z) is analytic in the open unit disk with center at a
B. The power series for f(z+a) converges for z+a<1
C. f'(a)=0
D. Integral f(z)dz over c=0 for any circle C in the plane
E. f(z) has a pole of order 1 at z=a
Answer: A
My thought is to apply the ratio test for convergence, and got lim f(n+1)/f(n)=1, but how can I get the answer and exclude other choices? Plzzzzzzzz help!
SOS: 9367 Q65
Re: SOS: 9367 Q65
A satisfactory answer is dependent on your choice of definitions; using wikipedia's, the answer is trivial. http://en.wikipedia.org/wiki/Analytic_f ... efinitions
Some authors take $$f$$ analytic at $$x$$ to mean it is differentiable in an open neighborhood of $$x$$. It turns out the choice of definitions doesn't matter much. http://en.wikipedia.org/wiki/Analyticit ... _functions
Some authors take $$f$$ analytic at $$x$$ to mean it is differentiable in an open neighborhood of $$x$$. It turns out the choice of definitions doesn't matter much. http://en.wikipedia.org/wiki/Analyticit ... _functions

 Posts: 2
 Joined: Thu Sep 25, 2014 8:08 pm
Re: SOS: 9367 Q65
I was badly struggling with Complex Analysis part but after reading through a textbook it became very clear to me.
You might have also got the similar confusion about analyticity and hope this helps:
If f is analytic at a point 'a', it means that f is analytic in an open disc containing 'a'. (so it's not necessarily differentiable at x=a in terms of the conventional differentiablity in the Real).
Also, here is a useful theorem:
Let f be analytic in an open disc D except for a finite number of exceptional points in D then for any closed curve, say gamma, in D that is not passing through any of the exceptional points the integral over gamma of f is 0.
You might have also got the similar confusion about analyticity and hope this helps:
If f is analytic at a point 'a', it means that f is analytic in an open disc containing 'a'. (so it's not necessarily differentiable at x=a in terms of the conventional differentiablity in the Real).
Also, here is a useful theorem:
Let f be analytic in an open disc D except for a finite number of exceptional points in D then for any closed curve, say gamma, in D that is not passing through any of the exceptional points the integral over gamma of f is 0.

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: SOS: 9367 Q65
That isn't true. You need the region to be simply connected, or at least the curve has to be able to homotopy retract to a point.