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Topic: complex analysis (Read 3120 times) |
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Bo
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Dear Respected Sir:
Could please have a look at this question:
If the complex function g(z) has an isolated singularity at z, then the
function exp(g(z))does not have a pole at z. (given hint: f has a pole of
order n at z, then f'/f has a simple pole with residue -n)
Thanks a lot.
Bo
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Icarus
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Re: complex analysis
« Reply #1 on: Apr 9th, 2004, 3:25pm » |
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Let f(z) = exp(g(z)), and apply the result in your hint.
The result you get is the contrapositive of the question.
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| « Last Edit: Apr 9th, 2004, 3:27pm by Icarus » |
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Moon
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Is my solution of this problem correct ?
Suppose that f has a pole at z0 of order n, then f'/f=g' has a simple pole at z0 of order -n, i.e. g'(z)=-n*(z-z0)-1 +a0+a1*(z-z0)+...
1) If g(z) has a removable singularity at z0, then g(z)=a0+a1*(z-z0)+..., hence g'(z)=a1+a2(z-z0)+...
So we have that g(z) doesn't have removable singularity at z0.
2) If g(z) has a essential singularity then g(z)= [sum]-[infty]+[infty]an(z-z0), and we have that g'(z)=...+a-1(z-z0)-2+a1+a2 (z - z0)+...
So g'(z) doesn't have an essential singularity.
3) If g(z) has a pole at z0, then g(z)=a-1(z-z0)-1+a0+a1(z-z 0)+... so that g'(z)=-a-1(z-z0)-2+a1+a2(z-z 0).
Hence all type of singularities contradicts to condition that g'(z) has a simple pole with residue -n. Hence f(z) doesn't have pole at z0.
Correct me please if I did something wrong.
Thank you.
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Icarus
wu::riddles Moderator
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Re: complex analysis
« Reply #3 on: Apr 15th, 2004, 4:09pm » |
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That is it.
If g has a singularity, then g' cannot have a simple pole, and so exp(g(z)) cannot have any pole.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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