What is the difference between a holomorphic function and a meromorphic function?
As far as I can tell, if a function is holomorphic on its domain, then it's also meromorphic and vice versa. Can someone tell me what the difference between these two properties are (if any)? A counter-example and an explanation of why it's a counter-example would be nice.
complex-analysis
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
add a comment |
As far as I can tell, if a function is holomorphic on its domain, then it's also meromorphic and vice versa. Can someone tell me what the difference between these two properties are (if any)? A counter-example and an explanation of why it's a counter-example would be nice.
complex-analysis
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
1
It might be helpful to explain why you think they are the same thing. What definitions are you using?
– Antonio Vargas
Dec 30 '13 at 21:20
1
You cannot tell the difference because you do not add a letter for the domain: holomorphic on $D$ implies meromorphic on $D$, but meromorphic on $D$ implies holomorphic only on some subset of $D$.
– Phira
Dec 30 '13 at 21:31
add a comment |
As far as I can tell, if a function is holomorphic on its domain, then it's also meromorphic and vice versa. Can someone tell me what the difference between these two properties are (if any)? A counter-example and an explanation of why it's a counter-example would be nice.
complex-analysis
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
As far as I can tell, if a function is holomorphic on its domain, then it's also meromorphic and vice versa. Can someone tell me what the difference between these two properties are (if any)? A counter-example and an explanation of why it's a counter-example would be nice.
complex-analysis
complex-analysis
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
asked Dec 30 '13 at 21:16
Tac-Tics
1,181822
1,181822
1
It might be helpful to explain why you think they are the same thing. What definitions are you using?
– Antonio Vargas
Dec 30 '13 at 21:20
1
You cannot tell the difference because you do not add a letter for the domain: holomorphic on $D$ implies meromorphic on $D$, but meromorphic on $D$ implies holomorphic only on some subset of $D$.
– Phira
Dec 30 '13 at 21:31
add a comment |
1
It might be helpful to explain why you think they are the same thing. What definitions are you using?
– Antonio Vargas
Dec 30 '13 at 21:20
1
You cannot tell the difference because you do not add a letter for the domain: holomorphic on $D$ implies meromorphic on $D$, but meromorphic on $D$ implies holomorphic only on some subset of $D$.
– Phira
Dec 30 '13 at 21:31
1
1
It might be helpful to explain why you think they are the same thing. What definitions are you using?
– Antonio Vargas
Dec 30 '13 at 21:20
It might be helpful to explain why you think they are the same thing. What definitions are you using?
– Antonio Vargas
Dec 30 '13 at 21:20
1
1
You cannot tell the difference because you do not add a letter for the domain: holomorphic on $D$ implies meromorphic on $D$, but meromorphic on $D$ implies holomorphic only on some subset of $D$.
– Phira
Dec 30 '13 at 21:31
You cannot tell the difference because you do not add a letter for the domain: holomorphic on $D$ implies meromorphic on $D$, but meromorphic on $D$ implies holomorphic only on some subset of $D$.
– Phira
Dec 30 '13 at 21:31
add a comment |
2 Answers
2
active
oldest
votes
Every holomorphic function is meromorphic, but not vice versa. A meromorphic function that is not holomorphic has poles in its domain, e.g. $dfrac1z$ is holomorphic on $mathbb{C}setminus {0}$, but it is meromorphic on all of $mathbb{C}$.
If $f$ is meromorphic on a region $U$, there is a closed discrete subset $P subset U$ such that $flvert_{Usetminus P}$ is holomorphic on $Usetminus P$, and $f$ has a pole in every $pin P$. If $P = varnothing$, then $f$ is holomorphic on $U$, otherwise not.
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
add a comment |
A meromorphic function is allowed to take the value $infty$ (this is an unsigned complex infinity), while a holomorphic function is not. Since infinite values are allowed but not required, every holomorphic function is meromorphic, but not the other way around.
(It is standard to say that the function is undefined rather than that its value is infinite, but it's important that the limit be infinite, that is, the limit of the reciprocal must be zero. In the 19th century, mathematicians were less shy about simply saying that the value is infinite.)
For example, $f(z) = 1/z$ is not a holomorphism, because $f(0)$ is undefined (as a finite complex number), but it is a meromorphism, because $lim_{zto0} (1/f(z)) = 0$, so we can say that $f(0) = infty$. (Of course, you also have to check that the function is complex-differentiable away from zero.) However, $g(z) = exp(1/z)$ is not meromorphic, because neither $lim_{zto0} g(z)$ nor $lim_{zto0} (1/g(z))$ exists, so there is no value (finite or infinite) that we can assign to $g(0)$.
A meromorphic function is holomorphic on a smaller domain, but there is more to being meromorphic than that. For example, both functions $f$ and $g$ above are holomorphic on the domain $mathbb{C} setminus {0}$, but only $f$ is meromorphic on all of $mathbb{C}$.
answered Feb 9 at 13:34
Toby Bartels
630513
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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votes
Every holomorphic function is meromorphic, but not vice versa. A meromorphic function that is not holomorphic has poles in its domain, e.g. $dfrac1z$ is holomorphic on $mathbb{C}setminus {0}$, but it is meromorphic on all of $mathbb{C}$.
If $f$ is meromorphic on a region $U$, there is a closed discrete subset $P subset U$ such that $flvert_{Usetminus P}$ is holomorphic on $Usetminus P$, and $f$ has a pole in every $pin P$. If $P = varnothing$, then $f$ is holomorphic on $U$, otherwise not.
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
add a comment |
Every holomorphic function is meromorphic, but not vice versa. A meromorphic function that is not holomorphic has poles in its domain, e.g. $dfrac1z$ is holomorphic on $mathbb{C}setminus {0}$, but it is meromorphic on all of $mathbb{C}$.
If $f$ is meromorphic on a region $U$, there is a closed discrete subset $P subset U$ such that $flvert_{Usetminus P}$ is holomorphic on $Usetminus P$, and $f$ has a pole in every $pin P$. If $P = varnothing$, then $f$ is holomorphic on $U$, otherwise not.
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
add a comment |
Every holomorphic function is meromorphic, but not vice versa. A meromorphic function that is not holomorphic has poles in its domain, e.g. $dfrac1z$ is holomorphic on $mathbb{C}setminus {0}$, but it is meromorphic on all of $mathbb{C}$.
If $f$ is meromorphic on a region $U$, there is a closed discrete subset $P subset U$ such that $flvert_{Usetminus P}$ is holomorphic on $Usetminus P$, and $f$ has a pole in every $pin P$. If $P = varnothing$, then $f$ is holomorphic on $U$, otherwise not.
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
Every holomorphic function is meromorphic, but not vice versa. A meromorphic function that is not holomorphic has poles in its domain, e.g. $dfrac1z$ is holomorphic on $mathbb{C}setminus {0}$, but it is meromorphic on all of $mathbb{C}$.
If $f$ is meromorphic on a region $U$, there is a closed discrete subset $P subset U$ such that $flvert_{Usetminus P}$ is holomorphic on $Usetminus P$, and $f$ has a pole in every $pin P$. If $P = varnothing$, then $f$ is holomorphic on $U$, otherwise not.
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
answered Dec 30 '13 at 21:22
Daniel Fischer♦
173k16159281
173k16159281
add a comment |
add a comment |
A meromorphic function is allowed to take the value $infty$ (this is an unsigned complex infinity), while a holomorphic function is not. Since infinite values are allowed but not required, every holomorphic function is meromorphic, but not the other way around.
(It is standard to say that the function is undefined rather than that its value is infinite, but it's important that the limit be infinite, that is, the limit of the reciprocal must be zero. In the 19th century, mathematicians were less shy about simply saying that the value is infinite.)
For example, $f(z) = 1/z$ is not a holomorphism, because $f(0)$ is undefined (as a finite complex number), but it is a meromorphism, because $lim_{zto0} (1/f(z)) = 0$, so we can say that $f(0) = infty$. (Of course, you also have to check that the function is complex-differentiable away from zero.) However, $g(z) = exp(1/z)$ is not meromorphic, because neither $lim_{zto0} g(z)$ nor $lim_{zto0} (1/g(z))$ exists, so there is no value (finite or infinite) that we can assign to $g(0)$.
A meromorphic function is holomorphic on a smaller domain, but there is more to being meromorphic than that. For example, both functions $f$ and $g$ above are holomorphic on the domain $mathbb{C} setminus {0}$, but only $f$ is meromorphic on all of $mathbb{C}$.
answered Feb 9 at 13:34
Toby Bartels
630513
add a comment |
A meromorphic function is allowed to take the value $infty$ (this is an unsigned complex infinity), while a holomorphic function is not. Since infinite values are allowed but not required, every holomorphic function is meromorphic, but not the other way around.
(It is standard to say that the function is undefined rather than that its value is infinite, but it's important that the limit be infinite, that is, the limit of the reciprocal must be zero. In the 19th century, mathematicians were less shy about simply saying that the value is infinite.)
For example, $f(z) = 1/z$ is not a holomorphism, because $f(0)$ is undefined (as a finite complex number), but it is a meromorphism, because $lim_{zto0} (1/f(z)) = 0$, so we can say that $f(0) = infty$. (Of course, you also have to check that the function is complex-differentiable away from zero.) However, $g(z) = exp(1/z)$ is not meromorphic, because neither $lim_{zto0} g(z)$ nor $lim_{zto0} (1/g(z))$ exists, so there is no value (finite or infinite) that we can assign to $g(0)$.
A meromorphic function is holomorphic on a smaller domain, but there is more to being meromorphic than that. For example, both functions $f$ and $g$ above are holomorphic on the domain $mathbb{C} setminus {0}$, but only $f$ is meromorphic on all of $mathbb{C}$.
answered Feb 9 at 13:34
Toby Bartels
630513
add a comment |
A meromorphic function is allowed to take the value $infty$ (this is an unsigned complex infinity), while a holomorphic function is not. Since infinite values are allowed but not required, every holomorphic function is meromorphic, but not the other way around.
(It is standard to say that the function is undefined rather than that its value is infinite, but it's important that the limit be infinite, that is, the limit of the reciprocal must be zero. In the 19th century, mathematicians were less shy about simply saying that the value is infinite.)
For example, $f(z) = 1/z$ is not a holomorphism, because $f(0)$ is undefined (as a finite complex number), but it is a meromorphism, because $lim_{zto0} (1/f(z)) = 0$, so we can say that $f(0) = infty$. (Of course, you also have to check that the function is complex-differentiable away from zero.) However, $g(z) = exp(1/z)$ is not meromorphic, because neither $lim_{zto0} g(z)$ nor $lim_{zto0} (1/g(z))$ exists, so there is no value (finite or infinite) that we can assign to $g(0)$.
A meromorphic function is holomorphic on a smaller domain, but there is more to being meromorphic than that. For example, both functions $f$ and $g$ above are holomorphic on the domain $mathbb{C} setminus {0}$, but only $f$ is meromorphic on all of $mathbb{C}$.
answered Feb 9 at 13:34
Toby Bartels
630513
A meromorphic function is allowed to take the value $infty$ (this is an unsigned complex infinity), while a holomorphic function is not. Since infinite values are allowed but not required, every holomorphic function is meromorphic, but not the other way around.
(It is standard to say that the function is undefined rather than that its value is infinite, but it's important that the limit be infinite, that is, the limit of the reciprocal must be zero. In the 19th century, mathematicians were less shy about simply saying that the value is infinite.)
For example, $f(z) = 1/z$ is not a holomorphism, because $f(0)$ is undefined (as a finite complex number), but it is a meromorphism, because $lim_{zto0} (1/f(z)) = 0$, so we can say that $f(0) = infty$. (Of course, you also have to check that the function is complex-differentiable away from zero.) However, $g(z) = exp(1/z)$ is not meromorphic, because neither $lim_{zto0} g(z)$ nor $lim_{zto0} (1/g(z))$ exists, so there is no value (finite or infinite) that we can assign to $g(0)$.
A meromorphic function is holomorphic on a smaller domain, but there is more to being meromorphic than that. For example, both functions $f$ and $g$ above are holomorphic on the domain $mathbb{C} setminus {0}$, but only $f$ is meromorphic on all of $mathbb{C}$.
answered Feb 9 at 13:34
Toby Bartels
630513
answered Feb 9 at 13:34
Toby Bartels
630513
answered Feb 9 at 13:34
Toby Bartels
630513
answered Feb 9 at 13:34
Toby Bartels
630513
630513
add a comment |
add a comment |
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1
It might be helpful to explain why you think they are the same thing. What definitions are you using?
– Antonio Vargas
Dec 30 '13 at 21:20
1
You cannot tell the difference because you do not add a letter for the domain: holomorphic on $D$ implies meromorphic on $D$, but meromorphic on $D$ implies holomorphic only on some subset of $D$.
– Phira
Dec 30 '13 at 21:31