When is composition of meromorphic functions meromorphic
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When I compose a meromorphic and a holomorphic function, I get a meromorphic function. Are there other cases when a composition of two meromorphic functions is meromorphic? For example, if I compose a holomorphic and a meromorphic function? Or does it hold that the composition is meromorphic in general?
complex-analysis function-and-relation-composition holomorphic-functions meromorphic-functions
asked 2 days ago
user2316602user2316602
606
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When I compose a meromorphic and a holomorphic function, I get a meromorphic function. Are there other cases when a composition of two meromorphic functions is meromorphic? For example, if I compose a holomorphic and a meromorphic function? Or does it hold that the composition is meromorphic in general?
complex-analysis function-and-relation-composition holomorphic-functions meromorphic-functions
asked 2 days ago
user2316602user2316602
606
$endgroup$
add a comment |
$begingroup$
When I compose a meromorphic and a holomorphic function, I get a meromorphic function. Are there other cases when a composition of two meromorphic functions is meromorphic? For example, if I compose a holomorphic and a meromorphic function? Or does it hold that the composition is meromorphic in general?
complex-analysis function-and-relation-composition holomorphic-functions meromorphic-functions
asked 2 days ago
user2316602user2316602
606
$endgroup$
When I compose a meromorphic and a holomorphic function, I get a meromorphic function. Are there other cases when a composition of two meromorphic functions is meromorphic? For example, if I compose a holomorphic and a meromorphic function? Or does it hold that the composition is meromorphic in general?
complex-analysis function-and-relation-composition holomorphic-functions meromorphic-functions
complex-analysis function-and-relation-composition holomorphic-functions meromorphic-functions
asked 2 days ago
user2316602user2316602
606
asked 2 days ago
user2316602user2316602
606
asked 2 days ago
user2316602user2316602
606
asked 2 days ago
user2316602user2316602
606
asked 2 days ago
user2316602user2316602
606
606
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2 Answers
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Assume that $f$ and $g$ are meromorphic in $Bbb C$, and that the composition $h = f circ g$ is also meromorphic in $Bbb C$.
If $f$ is not a rational function then it has an essential singularity at $z= infty$. If in addition, $g$ has a pole at $z_0$, then the Casorati-Weierstraß theorem shows that $lim_{zto z_0} f(g(z))$ does not exist in the extended complex plane. So this can not happen.
Therefore $f circ g$ is meromorphic in $Bbb C$ if and only if
$f$ is a rational function, or
$g$ is holomorphic in $Bbb C$.
answered 2 days ago
Martin RMartin R
30.2k33558
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add a comment |
$begingroup$
The composition of holomorphic functions is holomorphic. As a special case, a holomorphic function followed by a meromorphic function is meromorphic.
I say 'special case' because we can think of a meromorphic function as a holomorphic function $mathbb C to P^1(mathbb C)$.
The converse is often false. In fact, when:
$f$ is entire and there is a value which it reaches infinitely often
$g$ is meromorphic with at least one pole
then $f circ g$ is not meromorphic unless $f$ is constant. Indeed, $f circ g$ reaches a value infinitely often in any neighborhood of the pole of $g$, which can only be if $f circ g$ is constant.
Examples: $f = sin z$, $g = frac1z$, or $f = e^z$, $g = frac1z$, ...
answered 2 days ago
rabotarabota
14.3k32782
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2 Answers
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2 Answers
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$begingroup$
Assume that $f$ and $g$ are meromorphic in $Bbb C$, and that the composition $h = f circ g$ is also meromorphic in $Bbb C$.
If $f$ is not a rational function then it has an essential singularity at $z= infty$. If in addition, $g$ has a pole at $z_0$, then the Casorati-Weierstraß theorem shows that $lim_{zto z_0} f(g(z))$ does not exist in the extended complex plane. So this can not happen.
Therefore $f circ g$ is meromorphic in $Bbb C$ if and only if
$f$ is a rational function, or
$g$ is holomorphic in $Bbb C$.
answered 2 days ago
Martin RMartin R
30.2k33558
$endgroup$
add a comment |
$begingroup$
Assume that $f$ and $g$ are meromorphic in $Bbb C$, and that the composition $h = f circ g$ is also meromorphic in $Bbb C$.
If $f$ is not a rational function then it has an essential singularity at $z= infty$. If in addition, $g$ has a pole at $z_0$, then the Casorati-Weierstraß theorem shows that $lim_{zto z_0} f(g(z))$ does not exist in the extended complex plane. So this can not happen.
Therefore $f circ g$ is meromorphic in $Bbb C$ if and only if
$f$ is a rational function, or
$g$ is holomorphic in $Bbb C$.
answered 2 days ago
Martin RMartin R
30.2k33558
$endgroup$
add a comment |
$begingroup$
Assume that $f$ and $g$ are meromorphic in $Bbb C$, and that the composition $h = f circ g$ is also meromorphic in $Bbb C$.
If $f$ is not a rational function then it has an essential singularity at $z= infty$. If in addition, $g$ has a pole at $z_0$, then the Casorati-Weierstraß theorem shows that $lim_{zto z_0} f(g(z))$ does not exist in the extended complex plane. So this can not happen.
Therefore $f circ g$ is meromorphic in $Bbb C$ if and only if
$f$ is a rational function, or
$g$ is holomorphic in $Bbb C$.
answered 2 days ago
Martin RMartin R
30.2k33558
$endgroup$
Assume that $f$ and $g$ are meromorphic in $Bbb C$, and that the composition $h = f circ g$ is also meromorphic in $Bbb C$.
If $f$ is not a rational function then it has an essential singularity at $z= infty$. If in addition, $g$ has a pole at $z_0$, then the Casorati-Weierstraß theorem shows that $lim_{zto z_0} f(g(z))$ does not exist in the extended complex plane. So this can not happen.
Therefore $f circ g$ is meromorphic in $Bbb C$ if and only if
$f$ is a rational function, or
$g$ is holomorphic in $Bbb C$.
answered 2 days ago
Martin RMartin R
30.2k33558
answered 2 days ago
Martin RMartin R
30.2k33558
answered 2 days ago
Martin RMartin R
30.2k33558
answered 2 days ago
Martin RMartin R
30.2k33558
30.2k33558
add a comment |
add a comment |
$begingroup$
The composition of holomorphic functions is holomorphic. As a special case, a holomorphic function followed by a meromorphic function is meromorphic.
I say 'special case' because we can think of a meromorphic function as a holomorphic function $mathbb C to P^1(mathbb C)$.
The converse is often false. In fact, when:
$f$ is entire and there is a value which it reaches infinitely often
$g$ is meromorphic with at least one pole
then $f circ g$ is not meromorphic unless $f$ is constant. Indeed, $f circ g$ reaches a value infinitely often in any neighborhood of the pole of $g$, which can only be if $f circ g$ is constant.
Examples: $f = sin z$, $g = frac1z$, or $f = e^z$, $g = frac1z$, ...
answered 2 days ago
rabotarabota
14.3k32782
$endgroup$
add a comment |
$begingroup$
The composition of holomorphic functions is holomorphic. As a special case, a holomorphic function followed by a meromorphic function is meromorphic.
I say 'special case' because we can think of a meromorphic function as a holomorphic function $mathbb C to P^1(mathbb C)$.
The converse is often false. In fact, when:
$f$ is entire and there is a value which it reaches infinitely often
$g$ is meromorphic with at least one pole
then $f circ g$ is not meromorphic unless $f$ is constant. Indeed, $f circ g$ reaches a value infinitely often in any neighborhood of the pole of $g$, which can only be if $f circ g$ is constant.
Examples: $f = sin z$, $g = frac1z$, or $f = e^z$, $g = frac1z$, ...
answered 2 days ago
rabotarabota
14.3k32782
$endgroup$
add a comment |
$begingroup$
The composition of holomorphic functions is holomorphic. As a special case, a holomorphic function followed by a meromorphic function is meromorphic.
I say 'special case' because we can think of a meromorphic function as a holomorphic function $mathbb C to P^1(mathbb C)$.
The converse is often false. In fact, when:
$f$ is entire and there is a value which it reaches infinitely often
$g$ is meromorphic with at least one pole
then $f circ g$ is not meromorphic unless $f$ is constant. Indeed, $f circ g$ reaches a value infinitely often in any neighborhood of the pole of $g$, which can only be if $f circ g$ is constant.
Examples: $f = sin z$, $g = frac1z$, or $f = e^z$, $g = frac1z$, ...
answered 2 days ago
rabotarabota
14.3k32782
$endgroup$
The composition of holomorphic functions is holomorphic. As a special case, a holomorphic function followed by a meromorphic function is meromorphic.
I say 'special case' because we can think of a meromorphic function as a holomorphic function $mathbb C to P^1(mathbb C)$.
The converse is often false. In fact, when:
$f$ is entire and there is a value which it reaches infinitely often
$g$ is meromorphic with at least one pole
then $f circ g$ is not meromorphic unless $f$ is constant. Indeed, $f circ g$ reaches a value infinitely often in any neighborhood of the pole of $g$, which can only be if $f circ g$ is constant.
Examples: $f = sin z$, $g = frac1z$, or $f = e^z$, $g = frac1z$, ...
answered 2 days ago
rabotarabota
14.3k32782
edited 2 days ago
answered 2 days ago
rabotarabota
14.3k32782
answered 2 days ago
rabotarabota
14.3k32782
answered 2 days ago
rabotarabota
14.3k32782
14.3k32782
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add a comment |
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