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Topic: singularity at punctured disk (Read 3540 times) |
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Michael
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Let f:D*(z_0,r)->C (function from punctured disk around z_0 with radius r to complex plane)
and let f(z) not real for all z in the domain
I need to proove that z_0 is removable singularity.
please help me!
Michael
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| « Last Edit: Mar 26th, 2005, 9:06am by Icarus » |
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Michael
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Re: singularity at punctured disk
« Reply #1 on: Jan 5th, 2005, 6:54am » |
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may be it's because zeros of 1/f imply real values of 1/f and than f have too real values,contradiction.
[z_0 is sure not essential(picard thm), and if z_0 is pole this imply zero of 1/f extention]
are it's true?
Michael
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Icarus
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Re: singularity at punctured disk
« Reply #2 on: Jan 5th, 2005, 5:00pm » |
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zeros of 1/f do indeed imply real values of 1/f in the neighborhood, but this is not likely to impress your instructor, since it is effectively what you are being asked to prove. (The zero itself does not imply a real value, since 1/0 is not real.)
Basically the reason z0 has to be removable is that if f had a pole there, it would need to approximate, as z[to]z0, A(z - z0)-n for some A, n. But all such functions have both values with positive and with negative imaginary parts. Since f is trapped to have either all positive or all negative imaginary parts, it cannot approximate these functions. Therefore f cannot have a pole, and as you have already pointed out, f cannot have an essential singularity by Picard's theorem.
But you are going to have to clean that argument up before a teacher will accept it.
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Michael Dagg
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Re: singularity at punctured disk
« Reply #3 on: Apr 17th, 2005, 10:34am » |
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I guess we will never know if the original Michael was able to take this reasonsing and make a proof from it, as a proof undoubtedly follows.
One of your ideas can be exploited to give a shorter proof. As you note, connectedness considerations mean that f maps the punctured disk into either the open upper half-plane or the open lower half-plane. Either of these is mapped conformally to a disk by a Moebius transformation, say, M, and then M(f(z)) is bounded near the (putative) singularity.
Regards,
MD
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