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Topic: About the pole (Read 4032 times) 

immanuel78
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About the pole
« on: Dec 9^{th}, 2006, 5:05am » 
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Let f be analytic in G={z:0<za<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)

« Last Edit: Dec 9^{th}, 2006, 5:06am by immanuel78 » 
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Icarus
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Re: About the pole
« Reply #1 on: Dec 9^{th}, 2006, 7:00am » 
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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.


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immanuel78
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Re: About the pole
« Reply #2 on: Dec 9^{th}, 2006, 7:29am » 
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But z=a is not an isolated singularity of f. Hence z=a is not an essential singularity of f.

« Last Edit: Dec 9^{th}, 2006, 7:30am by immanuel78 » 
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Icarus
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Re: About the pole
« Reply #3 on: Dec 9^{th}, 2006, 12:57pm » 
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Consider g(z) = 1/(f(z)  w). In many ways, poles are not really a failure of the function to be analytic.

« Last Edit: Dec 9^{th}, 2006, 1:27pm by Icarus » 
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immanuel78
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Re: About the pole
« Reply #4 on: Dec 11^{th}, 2006, 10:50pm » 
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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.

« Last Edit: Dec 11^{th}, 2006, 10:52pm by immanuel78 » 
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Icarus
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Re: About the pole
« Reply #5 on: Dec 12^{th}, 2006, 4:15pm » 
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A pole is really just a place where an analytic function takes on the value . Around poles, the function is as wellbehaved 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.)

« Last Edit: Dec 12^{th}, 2006, 4:16pm by Icarus » 
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