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When is composition of meromorphic functions meromorphic

Explicit Representations of Meromorphic Functions as Quotients of Entire FunctionsMeromorphic functions on the unit diskCan all meromorphic functions be described by analytic parametrizations?Tilings and meromorphic functionsMeromorphic functions of several variablesfind all meromorphic functions s.t. |f|=1 where |z|=1Can a meromorphic function have removable singularities?Meromorphic functions on $hatmathbbC$ are rational functionsTerminology for a set of meromorphic functions with controlled poles and ordersWhy is the derevative of meromorphic function is meromorphic?

6

$begingroup$

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

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asked Mar 19 at 9:13

user2316602user2316602

606

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add a comment | 

6

$begingroup$

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

share|cite|improve this question

asked Mar 19 at 9:13

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

share|cite|improve this question

asked Mar 19 at 9:13

user2316602user2316602

606

$endgroup$

share|cite|improve this question

asked Mar 19 at 9:13

user2316602user2316602

606

share|cite|improve this question

asked Mar 19 at 9:13

user2316602user2316602

606

asked Mar 19 at 9:13

user2316602user2316602

606

asked Mar 19 at 9:13

user2316602user2316602

606

2 Answers
2

active

oldest

votes

9

$begingroup$

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 $Bbb C$.

share|cite|improve this answer

answered Mar 19 at 9:45

Martin RMartin R

30.2k33558

$endgroup$

add a comment | 

4

$begingroup$

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 $f = e^z$, $g = frac1z$, ...

share|cite|improve this answer

edited Mar 19 at 9:32

answered Mar 19 at 9:23

rabotarabota

14.3k32782

$endgroup$

add a comment | 

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9

$begingroup$

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 $Bbb C$.

share|cite|improve this answer

answered Mar 19 at 9:45

Martin RMartin R

30.2k33558

$endgroup$

add a comment | 

9

$begingroup$

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 $Bbb C$.

share|cite|improve this answer

answered Mar 19 at 9:45

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 $Bbb C$.

share|cite|improve this answer

answered Mar 19 at 9:45

Martin RMartin R

30.2k33558

$endgroup$

share|cite|improve this answer

answered Mar 19 at 9:45

Martin RMartin R

30.2k33558

answered Mar 19 at 9:45

Martin RMartin R

30.2k33558

answered Mar 19 at 9:45

Martin RMartin R

30.2k33558

4

$begingroup$

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 $f = e^z$, $g = frac1z$, ...

share|cite|improve this answer

edited Mar 19 at 9:32

answered Mar 19 at 9:23

rabotarabota

14.3k32782

$endgroup$

add a comment | 

4

$begingroup$

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 $f = e^z$, $g = frac1z$, ...

share|cite|improve this answer

edited Mar 19 at 9:32

answered Mar 19 at 9:23

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 $f = e^z$, $g = frac1z$, ...

share|cite|improve this answer

edited Mar 19 at 9:32

answered Mar 19 at 9:23

rabotarabota

14.3k32782

$endgroup$

share|cite|improve this answer

edited Mar 19 at 9:32

answered Mar 19 at 9:23

rabotarabota

14.3k32782

answered Mar 19 at 9:23

rabotarabota

14.3k32782

answered Mar 19 at 9:23

rabotarabota

14.3k32782