Determination of entire functions given with a removable singularity. [on hold]
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Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.
I have no idea how to start with.
complex-analysis singularity entire-functions
asked 2 days ago
Mittal G
1,182515
put on hold as off-topic by user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.
I have no idea how to start with.
complex-analysis singularity entire-functions
asked 2 days ago
Mittal G
1,182515
put on hold as off-topic by user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Constants, by Liouville's theorem.
– metamorphy
2 days ago
add a comment |
up vote
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up vote
0
down vote
favorite
Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.
I have no idea how to start with.
complex-analysis singularity entire-functions
asked 2 days ago
Mittal G
1,182515
Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $fbig(frac{1}{z}big)$.
I have no idea how to start with.
complex-analysis singularity entire-functions
complex-analysis singularity entire-functions
asked 2 days ago
Mittal G
1,182515
asked 2 days ago
Mittal G
1,182515
asked 2 days ago
Mittal G
1,182515
asked 2 days ago
Mittal G
1,182515
asked 2 days ago
Mittal G
1,182515
1,182515
put on hold as off-topic by user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user302797, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
2
Constants, by Liouville's theorem.
– metamorphy
2 days ago
add a comment |
2
Constants, by Liouville's theorem.
– metamorphy
2 days ago
2
2
Constants, by Liouville's theorem.
– metamorphy
2 days ago
Constants, by Liouville's theorem.
– metamorphy
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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accepted
By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood
SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so
By Liouvillies theorem
f is constant
answered yesterday
Shubham
1,5721519
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood
SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so
By Liouvillies theorem
f is constant
answered yesterday
Shubham
1,5721519
add a comment |
up vote
0
down vote
accepted
By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood
SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so
By Liouvillies theorem
f is constant
answered yesterday
Shubham
1,5721519
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood
SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so
By Liouvillies theorem
f is constant
answered yesterday
Shubham
1,5721519
By Reimann Theorem for removable singularity, If $ f$ has a removable singularity at $z_0$ iff it is bounded and holomorphic in a neighbourhood
SO as $g(z)=f(1/z)$ has removable singularity at 0 so at infinity $f(z)$ is bounded so
By Liouvillies theorem
f is constant
answered yesterday
Shubham
1,5721519
answered yesterday
Shubham
1,5721519
answered yesterday
Shubham
1,5721519
answered yesterday
Shubham
1,5721519
1,5721519
add a comment |
add a comment |
2
Constants, by Liouville's theorem.
– metamorphy
2 days ago