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Why does a meromorphic function in the (extended) complex plane have finitely many poles?
Let $f$ be meromorphic in $mathbb{C} cup {infty}$. Why must $f$ have only finitely many poles?
Edit: Renamed question following the comments.
complex-analysis
asked Oct 5 '11 at 3:44
user1205
14828
add a comment |
Let $f$ be meromorphic in $mathbb{C} cup {infty}$. Why must $f$ have only finitely many poles?
Edit: Renamed question following the comments.
complex-analysis
asked Oct 5 '11 at 3:44
user1205
14828
10
The title of the post asks a different question than the body and the question in the title is inaccurate: a meromorphic function in the complex plane MAY have infinitely many poles. The space $mathbb Ccup{infty}$ is usually called the extended complex plane or the Riemann sphere.
– Did
Oct 5 '11 at 9:12
1
As an explicit counterexample to the title question, consider $frac{1}{2 - e^z}$.
– Qiaochu Yuan
Oct 5 '11 at 15:22
add a comment |
Let $f$ be meromorphic in $mathbb{C} cup {infty}$. Why must $f$ have only finitely many poles?
Edit: Renamed question following the comments.
complex-analysis
asked Oct 5 '11 at 3:44
user1205
14828
Let $f$ be meromorphic in $mathbb{C} cup {infty}$. Why must $f$ have only finitely many poles?
Edit: Renamed question following the comments.
complex-analysis
complex-analysis
asked Oct 5 '11 at 3:44
user1205
14828
asked Oct 5 '11 at 3:44
user1205
14828
edited Oct 8 '11 at 6:20
asked Oct 5 '11 at 3:44
user1205
14828
asked Oct 5 '11 at 3:44
user1205
14828
asked Oct 5 '11 at 3:44
user1205
14828
14828
10
The title of the post asks a different question than the body and the question in the title is inaccurate: a meromorphic function in the complex plane MAY have infinitely many poles. The space $mathbb Ccup{infty}$ is usually called the extended complex plane or the Riemann sphere.
– Did
Oct 5 '11 at 9:12
1
As an explicit counterexample to the title question, consider $frac{1}{2 - e^z}$.
– Qiaochu Yuan
Oct 5 '11 at 15:22
add a comment |
10
The title of the post asks a different question than the body and the question in the title is inaccurate: a meromorphic function in the complex plane MAY have infinitely many poles. The space $mathbb Ccup{infty}$ is usually called the extended complex plane or the Riemann sphere.
– Did
Oct 5 '11 at 9:12
1
As an explicit counterexample to the title question, consider $frac{1}{2 - e^z}$.
– Qiaochu Yuan
Oct 5 '11 at 15:22
10
10
The title of the post asks a different question than the body and the question in the title is inaccurate: a meromorphic function in the complex plane MAY have infinitely many poles. The space $mathbb Ccup{infty}$ is usually called the extended complex plane or the Riemann sphere.
– Did
Oct 5 '11 at 9:12
The title of the post asks a different question than the body and the question in the title is inaccurate: a meromorphic function in the complex plane MAY have infinitely many poles. The space $mathbb Ccup{infty}$ is usually called the extended complex plane or the Riemann sphere.
– Did
Oct 5 '11 at 9:12
1
1
As an explicit counterexample to the title question, consider $frac{1}{2 - e^z}$.
– Qiaochu Yuan
Oct 5 '11 at 15:22
As an explicit counterexample to the title question, consider $frac{1}{2 - e^z}$.
– Qiaochu Yuan
Oct 5 '11 at 15:22
add a comment |
2 Answers
2
active
oldest
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By definition a pole is an isolated singularity, and so each pole has a neighborhood containing no other poles besides itself. Thus the set of all poles is discrete.
The set of poles is also closed, since its complement, the set of points at which $f$ is holomorphic, is open. (Any point where $f$ is holomorphic has a neighborhood restricted to which $f$ is holomorphic.)
Since $mathbb C cup {infty}$ is compact, any discrete and closed subset is discrete and compact, hence finite.
(Note that the reasoning of the first two paragraphs applies to a meromorphic function on any open subset of the Riemann sphere. E.g. a meromorphic function on $mathbb C$ can have infinitely many poles, but they must form a closed and discrete subset of $mathbb C$, and hence there can only be finitely many in any given bounded subset, i.e. they must accumulate at $infty$. A typical example is given by $f(z) = 1/sin z$.)
answered Oct 5 '11 at 3:55
Matt E
104k8217385
As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets).
– AnonymousCoward
Oct 5 '11 at 4:27
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere.
– AnonymousCoward
Oct 5 '11 at 4:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf
– wqr
Mar 29 '13 at 3:30
add a comment |
Poles are isolated singularities, therefore they are at most countable. (There exists an injective map between the isolated singularities and $mathbb{Q}^2$)
A meromorphic function $f$ in the extended complex plane either has pole at infinity or is holomorphic at infinity. In either case there is a radius $R$ large enough so that in the region $|z|>R$ there is at most one pole, which is at infinity.
We now consider the other poles which can only be in set $K={|z|leq R}$. Suppose there are countably infinite of them, then they form a bounded sequence in $mathbb{R}^2$. By Bolzano-Weierstrass Theorem, there must be a convergent subsequence, which contradicts the fact that all poles are isolated singularities. So there can only be finite of them.
answered Nov 21 '18 at 3:14
Tian He
277
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
By definition a pole is an isolated singularity, and so each pole has a neighborhood containing no other poles besides itself. Thus the set of all poles is discrete.
The set of poles is also closed, since its complement, the set of points at which $f$ is holomorphic, is open. (Any point where $f$ is holomorphic has a neighborhood restricted to which $f$ is holomorphic.)
Since $mathbb C cup {infty}$ is compact, any discrete and closed subset is discrete and compact, hence finite.
(Note that the reasoning of the first two paragraphs applies to a meromorphic function on any open subset of the Riemann sphere. E.g. a meromorphic function on $mathbb C$ can have infinitely many poles, but they must form a closed and discrete subset of $mathbb C$, and hence there can only be finitely many in any given bounded subset, i.e. they must accumulate at $infty$. A typical example is given by $f(z) = 1/sin z$.)
answered Oct 5 '11 at 3:55
Matt E
104k8217385
As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets).
– AnonymousCoward
Oct 5 '11 at 4:27
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere.
– AnonymousCoward
Oct 5 '11 at 4:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf
– wqr
Mar 29 '13 at 3:30
add a comment |
By definition a pole is an isolated singularity, and so each pole has a neighborhood containing no other poles besides itself. Thus the set of all poles is discrete.
The set of poles is also closed, since its complement, the set of points at which $f$ is holomorphic, is open. (Any point where $f$ is holomorphic has a neighborhood restricted to which $f$ is holomorphic.)
Since $mathbb C cup {infty}$ is compact, any discrete and closed subset is discrete and compact, hence finite.
(Note that the reasoning of the first two paragraphs applies to a meromorphic function on any open subset of the Riemann sphere. E.g. a meromorphic function on $mathbb C$ can have infinitely many poles, but they must form a closed and discrete subset of $mathbb C$, and hence there can only be finitely many in any given bounded subset, i.e. they must accumulate at $infty$. A typical example is given by $f(z) = 1/sin z$.)
answered Oct 5 '11 at 3:55
Matt E
104k8217385
As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets).
– AnonymousCoward
Oct 5 '11 at 4:27
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere.
– AnonymousCoward
Oct 5 '11 at 4:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf
– wqr
Mar 29 '13 at 3:30
add a comment |
By definition a pole is an isolated singularity, and so each pole has a neighborhood containing no other poles besides itself. Thus the set of all poles is discrete.
The set of poles is also closed, since its complement, the set of points at which $f$ is holomorphic, is open. (Any point where $f$ is holomorphic has a neighborhood restricted to which $f$ is holomorphic.)
Since $mathbb C cup {infty}$ is compact, any discrete and closed subset is discrete and compact, hence finite.
(Note that the reasoning of the first two paragraphs applies to a meromorphic function on any open subset of the Riemann sphere. E.g. a meromorphic function on $mathbb C$ can have infinitely many poles, but they must form a closed and discrete subset of $mathbb C$, and hence there can only be finitely many in any given bounded subset, i.e. they must accumulate at $infty$. A typical example is given by $f(z) = 1/sin z$.)
answered Oct 5 '11 at 3:55
Matt E
104k8217385
By definition a pole is an isolated singularity, and so each pole has a neighborhood containing no other poles besides itself. Thus the set of all poles is discrete.
The set of poles is also closed, since its complement, the set of points at which $f$ is holomorphic, is open. (Any point where $f$ is holomorphic has a neighborhood restricted to which $f$ is holomorphic.)
Since $mathbb C cup {infty}$ is compact, any discrete and closed subset is discrete and compact, hence finite.
(Note that the reasoning of the first two paragraphs applies to a meromorphic function on any open subset of the Riemann sphere. E.g. a meromorphic function on $mathbb C$ can have infinitely many poles, but they must form a closed and discrete subset of $mathbb C$, and hence there can only be finitely many in any given bounded subset, i.e. they must accumulate at $infty$. A typical example is given by $f(z) = 1/sin z$.)
answered Oct 5 '11 at 3:55
Matt E
104k8217385
edited Oct 5 '11 at 14:58
answered Oct 5 '11 at 3:55
Matt E
104k8217385
answered Oct 5 '11 at 3:55
Matt E
104k8217385
answered Oct 5 '11 at 3:55
Matt E
104k8217385
104k8217385
As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets).
– AnonymousCoward
Oct 5 '11 at 4:27
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere.
– AnonymousCoward
Oct 5 '11 at 4:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf
– wqr
Mar 29 '13 at 3:30
add a comment |
As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets).
– AnonymousCoward
Oct 5 '11 at 4:27
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere.
– AnonymousCoward
Oct 5 '11 at 4:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf
– wqr
Mar 29 '13 at 3:30
As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets).
– AnonymousCoward
Oct 5 '11 at 4:27
As an exercise, extend this example and reasoning to show all nonconstant holomorphic maps between compact complex curves are finite (ie all fibers are finite sets).
– AnonymousCoward
Oct 5 '11 at 4:27
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere.
– AnonymousCoward
Oct 5 '11 at 4:30
I guess this is if you are thinking of meromorphic functions as holomorphic maps to the Riemann sphere.
– AnonymousCoward
Oct 5 '11 at 4:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf
– wqr
Mar 29 '13 at 3:30
here there is a proof in the first paragraph: math.wisc.edu/~dummit/sets/110b-1s.pdf
– wqr
Mar 29 '13 at 3:30
add a comment |
Poles are isolated singularities, therefore they are at most countable. (There exists an injective map between the isolated singularities and $mathbb{Q}^2$)
A meromorphic function $f$ in the extended complex plane either has pole at infinity or is holomorphic at infinity. In either case there is a radius $R$ large enough so that in the region $|z|>R$ there is at most one pole, which is at infinity.
We now consider the other poles which can only be in set $K={|z|leq R}$. Suppose there are countably infinite of them, then they form a bounded sequence in $mathbb{R}^2$. By Bolzano-Weierstrass Theorem, there must be a convergent subsequence, which contradicts the fact that all poles are isolated singularities. So there can only be finite of them.
answered Nov 21 '18 at 3:14
Tian He
277
add a comment |
Poles are isolated singularities, therefore they are at most countable. (There exists an injective map between the isolated singularities and $mathbb{Q}^2$)
A meromorphic function $f$ in the extended complex plane either has pole at infinity or is holomorphic at infinity. In either case there is a radius $R$ large enough so that in the region $|z|>R$ there is at most one pole, which is at infinity.
We now consider the other poles which can only be in set $K={|z|leq R}$. Suppose there are countably infinite of them, then they form a bounded sequence in $mathbb{R}^2$. By Bolzano-Weierstrass Theorem, there must be a convergent subsequence, which contradicts the fact that all poles are isolated singularities. So there can only be finite of them.
answered Nov 21 '18 at 3:14
Tian He
277
add a comment |
Poles are isolated singularities, therefore they are at most countable. (There exists an injective map between the isolated singularities and $mathbb{Q}^2$)
A meromorphic function $f$ in the extended complex plane either has pole at infinity or is holomorphic at infinity. In either case there is a radius $R$ large enough so that in the region $|z|>R$ there is at most one pole, which is at infinity.
We now consider the other poles which can only be in set $K={|z|leq R}$. Suppose there are countably infinite of them, then they form a bounded sequence in $mathbb{R}^2$. By Bolzano-Weierstrass Theorem, there must be a convergent subsequence, which contradicts the fact that all poles are isolated singularities. So there can only be finite of them.
answered Nov 21 '18 at 3:14
Tian He
277
Poles are isolated singularities, therefore they are at most countable. (There exists an injective map between the isolated singularities and $mathbb{Q}^2$)
A meromorphic function $f$ in the extended complex plane either has pole at infinity or is holomorphic at infinity. In either case there is a radius $R$ large enough so that in the region $|z|>R$ there is at most one pole, which is at infinity.
We now consider the other poles which can only be in set $K={|z|leq R}$. Suppose there are countably infinite of them, then they form a bounded sequence in $mathbb{R}^2$. By Bolzano-Weierstrass Theorem, there must be a convergent subsequence, which contradicts the fact that all poles are isolated singularities. So there can only be finite of them.
answered Nov 21 '18 at 3:14
Tian He
277
edited Nov 21 '18 at 3:21
answered Nov 21 '18 at 3:14
Tian He
277
answered Nov 21 '18 at 3:14
Tian He
277
answered Nov 21 '18 at 3:14
Tian He
277
277
add a comment |
add a comment |
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10
The title of the post asks a different question than the body and the question in the title is inaccurate: a meromorphic function in the complex plane MAY have infinitely many poles. The space $mathbb Ccup{infty}$ is usually called the extended complex plane or the Riemann sphere.
– Did
Oct 5 '11 at 9:12
1
As an explicit counterexample to the title question, consider $frac{1}{2 - e^z}$.
– Qiaochu Yuan
Oct 5 '11 at 15:22