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Title: complex analysis Post by Bo on Apr 9th, 2004, 5:45am 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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Title: Re: complex analysis Post by Icarus on Apr 9th, 2004, 3:25pm 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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Title: Re: complex analysis Post by Moon on Apr 14th, 2004, 9:54pm 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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Title: Re: complex analysis Post by Icarus on Apr 15th, 2004, 4:09pm 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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