Determination of entire functions given with a removable singularity. [closed]
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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 Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
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closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56
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." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
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 Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
$endgroup$
closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56
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." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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Constants, by Liouville's theorem.
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– metamorphy
Nov 30 '18 at 18:23
add a comment |
$begingroup$
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 Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
$endgroup$
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 Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
asked Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
asked Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
asked Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
asked Nov 30 '18 at 18:14
Mittal GMittal G
1,193515
1,193515
closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56
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." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician Dec 1 '18 at 15:56
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." – Saad, Chinnapparaj R, Brahadeesh, rtybase, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Constants, by Liouville's theorem.
$endgroup$
– metamorphy
Nov 30 '18 at 18:23
add a comment |
2
$begingroup$
Constants, by Liouville's theorem.
$endgroup$
– metamorphy
Nov 30 '18 at 18:23
2
2
$begingroup$
Constants, by Liouville's theorem.
$endgroup$
– metamorphy
Nov 30 '18 at 18:23
$begingroup$
Constants, by Liouville's theorem.
$endgroup$
– metamorphy
Nov 30 '18 at 18:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
$endgroup$
add a comment |
$begingroup$
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 Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
$endgroup$
add a comment |
$begingroup$
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 Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
$endgroup$
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 Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
answered Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
answered Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
answered Dec 1 '18 at 13:53
ShubhamShubham
1,5951519
1,5951519
add a comment |
add a comment |
2
$begingroup$
Constants, by Liouville's theorem.
$endgroup$
– metamorphy
Nov 30 '18 at 18:23