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        general >> complex analysis >> About the pole
        
(Message started by: immanuel78 on Dec 9^th, 2006, 5:05am)

Title: About the pole
Post by immanuel78 on Dec 9^th, 2006, 5:05am

Let f be analytic in G={z:0<|z-a|<r} except that there is a sequence of poles {a_n} in G with a_n -> a.
Show that for any w in C there is a sequence {z_n} in G with a=lim z_n and w=lim f(z_n)

Title: Re: About the pole
Post by Icarus on Dec 9^th, 2006, 7:00am

If there is a sequence of poles converging to a, then a is an essential singularity of f. By Weierstrass, the image of any neighborhood of an essential singularity is dense in C. Hence such a sequence must exist.

Title: Re: About the pole
Post by immanuel78 on Dec 9^th, 2006, 7:29am

But z=a is not an isolated singularity of f.
Hence z=a is not an essential singularity of f.

Title: Re: About the pole
Post by Icarus on Dec 9^th, 2006, 12:57pm

Consider g(z) = 1/(f(z) - w). In many ways, poles are not really a failure of the function to be analytic.

Title: Re: About the pole
Post by immanuel78 on Dec 11^th, 2006, 10:50pm

Icarus, may you tell me the way in detail?
I think this problem by usuing your hint but I can't solve it perfectly.
But I can understand this problem more than before because of your hint.

Title: Re: About the pole
Post by Icarus on Dec 12^th, 2006, 4:15pm

A pole is really just a place where an analytic function takes on the value http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif. Around poles, the function is as well-behaved as it is everywhere else. In particular, if you take the inverse, then you get a function that is analytic at that point (there is no such thing as a "removable singularity" - if you ever encounter one, it just means someone has been lazy and not got around to removing it yet).

If f(b) = w for some b in G, then we could just take z_n to be any sequence converging to b. So assume that f does not take on the value w. Then g(z) = 1/(f(z) - w) is analytic everywhere in G. the point a is also an essential singularity of g (If it was a pole or removable, then f(z) would have had a worst at pole at a.) Therefore by Weierstrass's theorem, g(G) is dense in C. In particular, for every n, there must be z_n in G such that |g(z_n)| > n. From this it follows that f(z_n) --> w.

(We differ on the definition of essential singularity, by the way - my definition is that any place where analycity fails that is not removable or a pole is an essential singularity, regardless of whether it is isolated. However, Weierstrass's theorem does require an isolated essential singularity - though it can be extended to allow poles by exactly the means I've used here.)