finite number of zeros of a complex functionentire function with only finitely many zerosprove an analytic function has at least n zerosI want to prove this resultFinite order function in the complex analysis.Finding Complex ZerosAnalytic function symmetric wrt inversion: $f(z^-1) = f(z)$Can a non-constant analytic function have infinitely many zeros on a closed disk?Analytic function on unit disk has finitely many zerosAnalytic, non- constant complex function has finitely many zeros inside the disk $D(0, R)$ for all $R > 0$Elliptic functions - proof non-constant function, finitely many zeros and poles.
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finite number of zeros of a complex function
entire function with only finitely many zerosprove an analytic function has at least n zerosI want to prove this resultFinite order function in the complex analysis.Finding Complex ZerosAnalytic function symmetric wrt inversion: $f(z^-1) = f(z)$Can a non-constant analytic function have infinitely many zeros on a closed disk?Analytic function on unit disk has finitely many zerosAnalytic, non- constant complex function has finitely many zeros inside the disk $D(0, R)$ for all $R > 0$Elliptic functions - proof non-constant function, finitely many zeros and poles.
$begingroup$
Suppose that $f$ is analytic on a region $D = z$, and the limit of $vert f vert$ at infinity exists. Show that $f$ is either the constant function $0$, or has only finitely many zeros.
I thought since the limit at infinity exists then Laurent seires of $f$ only contains the principal part plus a constant, but how do I proceed to show that this will only have finite zeros ?
complex-analysis
edited 2 days ago
Robert Lewis
47.9k23067
asked 2 days ago
Andrew Andrew
183
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Suppose that $f$ is analytic on a region $D = z$, and the limit of $vert f vert$ at infinity exists. Show that $f$ is either the constant function $0$, or has only finitely many zeros.
I thought since the limit at infinity exists then Laurent seires of $f$ only contains the principal part plus a constant, but how do I proceed to show that this will only have finite zeros ?
complex-analysis
edited 2 days ago
Robert Lewis
47.9k23067
asked 2 days ago
Andrew Andrew
183
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
By 'principal part' you mean the $z^-n $ terms?
$endgroup$
– Berci
2 days ago
$begingroup$
yes exactly I mean those negative power terms
$endgroup$
– Andrew
2 days ago
add a comment |
$begingroup$
Suppose that $f$ is analytic on a region $D = z$, and the limit of $vert f vert$ at infinity exists. Show that $f$ is either the constant function $0$, or has only finitely many zeros.
I thought since the limit at infinity exists then Laurent seires of $f$ only contains the principal part plus a constant, but how do I proceed to show that this will only have finite zeros ?
complex-analysis
edited 2 days ago
Robert Lewis
47.9k23067
asked 2 days ago
Andrew Andrew
183
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose that $f$ is analytic on a region $D = z$, and the limit of $vert f vert$ at infinity exists. Show that $f$ is either the constant function $0$, or has only finitely many zeros.
I thought since the limit at infinity exists then Laurent seires of $f$ only contains the principal part plus a constant, but how do I proceed to show that this will only have finite zeros ?
complex-analysis
complex-analysis
edited 2 days ago
Robert Lewis
47.9k23067
asked 2 days ago
Andrew Andrew
183
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Robert Lewis
47.9k23067
asked 2 days ago
Andrew Andrew
183
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Robert Lewis
47.9k23067
edited 2 days ago
Robert Lewis
47.9k23067
edited 2 days ago
Robert Lewis
47.9k23067
47.9k23067
asked 2 days ago
Andrew Andrew
183
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Andrew Andrew
183
asked 2 days ago
Andrew Andrew
183
183
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Andrew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
By 'principal part' you mean the $z^-n $ terms?
$endgroup$
– Berci
2 days ago
$begingroup$
yes exactly I mean those negative power terms
$endgroup$
– Andrew
2 days ago
add a comment |
$begingroup$
By 'principal part' you mean the $z^-n $ terms?
$endgroup$
– Berci
2 days ago
$begingroup$
yes exactly I mean those negative power terms
$endgroup$
– Andrew
2 days ago
$begingroup$
By 'principal part' you mean the $z^-n $ terms?
$endgroup$
– Berci
2 days ago
$begingroup$
By 'principal part' you mean the $z^-n $ terms?
$endgroup$
– Berci
2 days ago
$begingroup$
yes exactly I mean those negative power terms
$endgroup$
– Andrew
2 days ago
$begingroup$
yes exactly I mean those negative power terms
$endgroup$
– Andrew
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is false. Let $g$ be analytic in $z$ with in finitely many zeros but not identically zero. [There are many such functions. The zeros will necessarily accumulate near the boundary]. Let $f(z)=g(frac 1 z)$. Then the hypothesis is satisfied but the conclusion fails. An example of a function $g$ with stated properties is $g(z)=sin( frac pi frac 1 R -z)$. It vanishes at the points $z=frac 1 R -frac 1 n$.
However, if $f$ is assumed to be holomorphic in $$ (and $lim_z to infty f(z)$ exists) then it can have only finite number of zeros. Note that $f$ is holomorphic in some open set $U$ containing $ geq R$. Once again define $g(z)=f(frac 1 z)$ This is holomorphic in some open set containing $ leq frac 1 R$ . The hypothesis tells you that $g$ has a removable singularity at $0$. The extended function is holomorphic in an open set containing $ leq frac 1 R$ so it can have at most finite number of zeros (unless it is identically $0$) The same is true for $f$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
$begingroup$
Can you provide a more specific example?
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew I have now given an example in the answer.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Thanks for your answer! I understand your example might work, however I'm not rly sure if the function in your example has a limit at infinity? I was thinking about the function sin(z)/z but someone pointed out the limit doesn't exist at infinity: quora.com/…
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew The limit of $f$ at $infty$ is same as limit of $g$ at $0$. This limit exists and the value is $g(0)$. In the example I have give this limit is $sin (Rpi)$.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
@Andrew In that case the statement is true.
$endgroup$
– Kavi Rama Murthy
yesterday
|
show 7 more comments
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
This is false. Let $g$ be analytic in $z$ with in finitely many zeros but not identically zero. [There are many such functions. The zeros will necessarily accumulate near the boundary]. Let $f(z)=g(frac 1 z)$. Then the hypothesis is satisfied but the conclusion fails. An example of a function $g$ with stated properties is $g(z)=sin( frac pi frac 1 R -z)$. It vanishes at the points $z=frac 1 R -frac 1 n$.
However, if $f$ is assumed to be holomorphic in $$ (and $lim_z to infty f(z)$ exists) then it can have only finite number of zeros. Note that $f$ is holomorphic in some open set $U$ containing $ geq R$. Once again define $g(z)=f(frac 1 z)$ This is holomorphic in some open set containing $ leq frac 1 R$ . The hypothesis tells you that $g$ has a removable singularity at $0$. The extended function is holomorphic in an open set containing $ leq frac 1 R$ so it can have at most finite number of zeros (unless it is identically $0$) The same is true for $f$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
$begingroup$
Can you provide a more specific example?
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew I have now given an example in the answer.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Thanks for your answer! I understand your example might work, however I'm not rly sure if the function in your example has a limit at infinity? I was thinking about the function sin(z)/z but someone pointed out the limit doesn't exist at infinity: quora.com/…
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew The limit of $f$ at $infty$ is same as limit of $g$ at $0$. This limit exists and the value is $g(0)$. In the example I have give this limit is $sin (Rpi)$.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
@Andrew In that case the statement is true.
$endgroup$
– Kavi Rama Murthy
yesterday
|
show 7 more comments
$begingroup$
This is false. Let $g$ be analytic in $z$ with in finitely many zeros but not identically zero. [There are many such functions. The zeros will necessarily accumulate near the boundary]. Let $f(z)=g(frac 1 z)$. Then the hypothesis is satisfied but the conclusion fails. An example of a function $g$ with stated properties is $g(z)=sin( frac pi frac 1 R -z)$. It vanishes at the points $z=frac 1 R -frac 1 n$.
However, if $f$ is assumed to be holomorphic in $$ (and $lim_z to infty f(z)$ exists) then it can have only finite number of zeros. Note that $f$ is holomorphic in some open set $U$ containing $ geq R$. Once again define $g(z)=f(frac 1 z)$ This is holomorphic in some open set containing $ leq frac 1 R$ . The hypothesis tells you that $g$ has a removable singularity at $0$. The extended function is holomorphic in an open set containing $ leq frac 1 R$ so it can have at most finite number of zeros (unless it is identically $0$) The same is true for $f$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
$begingroup$
Can you provide a more specific example?
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew I have now given an example in the answer.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Thanks for your answer! I understand your example might work, however I'm not rly sure if the function in your example has a limit at infinity? I was thinking about the function sin(z)/z but someone pointed out the limit doesn't exist at infinity: quora.com/…
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew The limit of $f$ at $infty$ is same as limit of $g$ at $0$. This limit exists and the value is $g(0)$. In the example I have give this limit is $sin (Rpi)$.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
@Andrew In that case the statement is true.
$endgroup$
– Kavi Rama Murthy
yesterday
|
show 7 more comments
$begingroup$
This is false. Let $g$ be analytic in $z$ with in finitely many zeros but not identically zero. [There are many such functions. The zeros will necessarily accumulate near the boundary]. Let $f(z)=g(frac 1 z)$. Then the hypothesis is satisfied but the conclusion fails. An example of a function $g$ with stated properties is $g(z)=sin( frac pi frac 1 R -z)$. It vanishes at the points $z=frac 1 R -frac 1 n$.
However, if $f$ is assumed to be holomorphic in $$ (and $lim_z to infty f(z)$ exists) then it can have only finite number of zeros. Note that $f$ is holomorphic in some open set $U$ containing $ geq R$. Once again define $g(z)=f(frac 1 z)$ This is holomorphic in some open set containing $ leq frac 1 R$ . The hypothesis tells you that $g$ has a removable singularity at $0$. The extended function is holomorphic in an open set containing $ leq frac 1 R$ so it can have at most finite number of zeros (unless it is identically $0$) The same is true for $f$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
$endgroup$
This is false. Let $g$ be analytic in $z$ with in finitely many zeros but not identically zero. [There are many such functions. The zeros will necessarily accumulate near the boundary]. Let $f(z)=g(frac 1 z)$. Then the hypothesis is satisfied but the conclusion fails. An example of a function $g$ with stated properties is $g(z)=sin( frac pi frac 1 R -z)$. It vanishes at the points $z=frac 1 R -frac 1 n$.
However, if $f$ is assumed to be holomorphic in $$ (and $lim_z to infty f(z)$ exists) then it can have only finite number of zeros. Note that $f$ is holomorphic in some open set $U$ containing $ geq R$. Once again define $g(z)=f(frac 1 z)$ This is holomorphic in some open set containing $ leq frac 1 R$ . The hypothesis tells you that $g$ has a removable singularity at $0$. The extended function is holomorphic in an open set containing $ leq frac 1 R$ so it can have at most finite number of zeros (unless it is identically $0$) The same is true for $f$.
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
edited yesterday
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
answered 2 days ago
Kavi Rama MurthyKavi Rama Murthy
66.3k42867
66.3k42867
$begingroup$
Can you provide a more specific example?
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew I have now given an example in the answer.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Thanks for your answer! I understand your example might work, however I'm not rly sure if the function in your example has a limit at infinity? I was thinking about the function sin(z)/z but someone pointed out the limit doesn't exist at infinity: quora.com/…
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew The limit of $f$ at $infty$ is same as limit of $g$ at $0$. This limit exists and the value is $g(0)$. In the example I have give this limit is $sin (Rpi)$.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
@Andrew In that case the statement is true.
$endgroup$
– Kavi Rama Murthy
yesterday
|
show 7 more comments
$begingroup$
Can you provide a more specific example?
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew I have now given an example in the answer.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Thanks for your answer! I understand your example might work, however I'm not rly sure if the function in your example has a limit at infinity? I was thinking about the function sin(z)/z but someone pointed out the limit doesn't exist at infinity: quora.com/…
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew The limit of $f$ at $infty$ is same as limit of $g$ at $0$. This limit exists and the value is $g(0)$. In the example I have give this limit is $sin (Rpi)$.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
$begingroup$
@Andrew In that case the statement is true.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
Can you provide a more specific example?
$endgroup$
– Andrew
2 days ago
$begingroup$
Can you provide a more specific example?
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew I have now given an example in the answer.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@Andrew I have now given an example in the answer.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
Thanks for your answer! I understand your example might work, however I'm not rly sure if the function in your example has a limit at infinity? I was thinking about the function sin(z)/z but someone pointed out the limit doesn't exist at infinity: quora.com/…
$endgroup$
– Andrew
2 days ago
$begingroup$
Thanks for your answer! I understand your example might work, however I'm not rly sure if the function in your example has a limit at infinity? I was thinking about the function sin(z)/z but someone pointed out the limit doesn't exist at infinity: quora.com/…
$endgroup$
– Andrew
2 days ago
$begingroup$
@Andrew The limit of $f$ at $infty$ is same as limit of $g$ at $0$. This limit exists and the value is $g(0)$. In the example I have give this limit is $sin (Rpi)$.
$endgroup$
– Kavi Rama Murthy
2 days ago
$begingroup$
@Andrew The limit of $f$ at $infty$ is same as limit of $g$ at $0$. This limit exists and the value is $g(0)$. In the example I have give this limit is $sin (Rpi)$.
$endgroup$
– Kavi Rama Murthy
2 days ago
1
1
$begingroup$
@Andrew In that case the statement is true.
$endgroup$
– Kavi Rama Murthy
yesterday
$begingroup$
@Andrew In that case the statement is true.
$endgroup$
– Kavi Rama Murthy
yesterday
|
show 7 more comments
Andrew is a new contributor. Be nice, and check out our Code of Conduct.
Andrew is a new contributor. Be nice, and check out our Code of Conduct.
Andrew is a new contributor. Be nice, and check out our Code of Conduct.
Andrew is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
By 'principal part' you mean the $z^-n $ terms?
$endgroup$
– Berci
2 days ago
$begingroup$
yes exactly I mean those negative power terms
$endgroup$
– Andrew
2 days ago