Isolated and non isolated essential singularity at same point?
I need to find the singularities of
$$f(z) = frac{1-e^z}{2+e^z}$$
My effort: Poles of function are given by
$$2+e^z=0implies e^z = -2 implies z = log 2+i(2k+1)pi$$ for k integer.
All these are singularities termed as simple poles. By definition, limit point of these which is $infty$ is a non-isolated singularity.
Further, limit point of zeros is again infinity which is a isolated-essential singularity.
But if both of isolated and non isolated coincides we take it as a non-isolated singularity. Am i correct? These are the only singularities?
complex-analysis singularity
asked Nov 25 '18 at 8:46
Mittal G
1,188515
add a comment |
I need to find the singularities of
$$f(z) = frac{1-e^z}{2+e^z}$$
My effort: Poles of function are given by
$$2+e^z=0implies e^z = -2 implies z = log 2+i(2k+1)pi$$ for k integer.
All these are singularities termed as simple poles. By definition, limit point of these which is $infty$ is a non-isolated singularity.
Further, limit point of zeros is again infinity which is a isolated-essential singularity.
But if both of isolated and non isolated coincides we take it as a non-isolated singularity. Am i correct? These are the only singularities?
complex-analysis singularity
asked Nov 25 '18 at 8:46
Mittal G
1,188515
add a comment |
I need to find the singularities of
$$f(z) = frac{1-e^z}{2+e^z}$$
My effort: Poles of function are given by
$$2+e^z=0implies e^z = -2 implies z = log 2+i(2k+1)pi$$ for k integer.
All these are singularities termed as simple poles. By definition, limit point of these which is $infty$ is a non-isolated singularity.
Further, limit point of zeros is again infinity which is a isolated-essential singularity.
But if both of isolated and non isolated coincides we take it as a non-isolated singularity. Am i correct? These are the only singularities?
complex-analysis singularity
asked Nov 25 '18 at 8:46
Mittal G
1,188515
I need to find the singularities of
$$f(z) = frac{1-e^z}{2+e^z}$$
My effort: Poles of function are given by
$$2+e^z=0implies e^z = -2 implies z = log 2+i(2k+1)pi$$ for k integer.
All these are singularities termed as simple poles. By definition, limit point of these which is $infty$ is a non-isolated singularity.
Further, limit point of zeros is again infinity which is a isolated-essential singularity.
But if both of isolated and non isolated coincides we take it as a non-isolated singularity. Am i correct? These are the only singularities?
complex-analysis singularity
complex-analysis singularity
asked Nov 25 '18 at 8:46
Mittal G
1,188515
asked Nov 25 '18 at 8:46
Mittal G
1,188515
asked Nov 25 '18 at 8:46
Mittal G
1,188515
asked Nov 25 '18 at 8:46
Mittal G
1,188515
asked Nov 25 '18 at 8:46
Mittal G
1,188515
1,188515
add a comment |
add a comment |
1 Answer
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Every neighborhood of $infty$ contains a pole, implying, as you correctly state, that $infty$ is not an isolated singularity. What makes you believe that it is also an isolated singularity?
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
I have read somewhere else that limit point of zeros is such a singularity.
– Mittal G
Nov 26 '18 at 13:10
Probably you misremember what you read.
– Julián Aguirre
Nov 26 '18 at 15:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Every neighborhood of $infty$ contains a pole, implying, as you correctly state, that $infty$ is not an isolated singularity. What makes you believe that it is also an isolated singularity?
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
I have read somewhere else that limit point of zeros is such a singularity.
– Mittal G
Nov 26 '18 at 13:10
Probably you misremember what you read.
– Julián Aguirre
Nov 26 '18 at 15:31
add a comment |
Every neighborhood of $infty$ contains a pole, implying, as you correctly state, that $infty$ is not an isolated singularity. What makes you believe that it is also an isolated singularity?
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
I have read somewhere else that limit point of zeros is such a singularity.
– Mittal G
Nov 26 '18 at 13:10
Probably you misremember what you read.
– Julián Aguirre
Nov 26 '18 at 15:31
add a comment |
Every neighborhood of $infty$ contains a pole, implying, as you correctly state, that $infty$ is not an isolated singularity. What makes you believe that it is also an isolated singularity?
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
Every neighborhood of $infty$ contains a pole, implying, as you correctly state, that $infty$ is not an isolated singularity. What makes you believe that it is also an isolated singularity?
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
answered Nov 25 '18 at 19:40
Julián Aguirre
67.6k24094
67.6k24094
I have read somewhere else that limit point of zeros is such a singularity.
– Mittal G
Nov 26 '18 at 13:10
Probably you misremember what you read.
– Julián Aguirre
Nov 26 '18 at 15:31
add a comment |
I have read somewhere else that limit point of zeros is such a singularity.
– Mittal G
Nov 26 '18 at 13:10
Probably you misremember what you read.
– Julián Aguirre
Nov 26 '18 at 15:31
I have read somewhere else that limit point of zeros is such a singularity.
– Mittal G
Nov 26 '18 at 13:10
I have read somewhere else that limit point of zeros is such a singularity.
– Mittal G
Nov 26 '18 at 13:10
Probably you misremember what you read.
– Julián Aguirre
Nov 26 '18 at 15:31
Probably you misremember what you read.
– Julián Aguirre
Nov 26 '18 at 15:31
add a comment |
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