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Are branch cuts always ‘cancellable’?
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Many functions can be analytically continued to $mathbb C$ except the branch cut.
However, it appears to me that for every function $f(z)$ that has a branch cut, there always exists a non-constant meromorphic/entire function $g(z)$ such that $f(g(z))$ can be analytically continued to the whole $mathbb C$.
For example,
$$sqrt x^2=x$$
$$ln e^x=x$$
$$arccoscos x =x$$
$$lnln e^e^x=x$$
$$operatornameW(xe^x)=x$$
More complicated examples:
$$f(x)=sqrt(x+1)(x+3)=sqrt(x+2)^2-1$$
$$g(x)=-2+cosh x$$
$$f(x)=x^alphaqquadalphainmathbb C$$
$$g(x)=e^x$$
Is this true?
complex-analysis
asked Aug 1 at 4:07
Szeto
3,8431421
add a comment |Â
up vote
9
down vote
favorite
Many functions can be analytically continued to $mathbb C$ except the branch cut.
However, it appears to me that for every function $f(z)$ that has a branch cut, there always exists a non-constant meromorphic/entire function $g(z)$ such that $f(g(z))$ can be analytically continued to the whole $mathbb C$.
For example,
$$sqrt x^2=x$$
$$ln e^x=x$$
$$arccoscos x =x$$
$$lnln e^e^x=x$$
$$operatornameW(xe^x)=x$$
More complicated examples:
$$f(x)=sqrt(x+1)(x+3)=sqrt(x+2)^2-1$$
$$g(x)=-2+cosh x$$
$$f(x)=x^alphaqquadalphainmathbb C$$
$$g(x)=e^x$$
Is this true?
complex-analysis
asked Aug 1 at 4:07
Szeto
3,8431421
How about $f(z) = z^2/3$?
– Sangchul Lee
Aug 1 at 5:09
$z=x^3 dots dots$
– Fabian
Aug 1 at 5:10
1
Typically, functions with branch cuts appear as inverse of holomorphic function on all of $mathbbC$ (this covers all your examples). The question is interesting, as it asks the reverse.
– Fabian
Aug 1 at 5:14
@Fabian Please see my newly added example.
– Szeto
Aug 1 at 5:37
@Fabian, Ah, I thought that $g$ must qualify as a local inverse of $f$, i.e. $f(g(z)) = z$. My bad for not carefully reading the question :s
– Sangchul Lee
Aug 1 at 5:44
add a comment |Â
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Many functions can be analytically continued to $mathbb C$ except the branch cut.
However, it appears to me that for every function $f(z)$ that has a branch cut, there always exists a non-constant meromorphic/entire function $g(z)$ such that $f(g(z))$ can be analytically continued to the whole $mathbb C$.
For example,
$$sqrt x^2=x$$
$$ln e^x=x$$
$$arccoscos x =x$$
$$lnln e^e^x=x$$
$$operatornameW(xe^x)=x$$
More complicated examples:
$$f(x)=sqrt(x+1)(x+3)=sqrt(x+2)^2-1$$
$$g(x)=-2+cosh x$$
$$f(x)=x^alphaqquadalphainmathbb C$$
$$g(x)=e^x$$
Is this true?
complex-analysis
asked Aug 1 at 4:07
Szeto
3,8431421
Many functions can be analytically continued to $mathbb C$ except the branch cut.
However, it appears to me that for every function $f(z)$ that has a branch cut, there always exists a non-constant meromorphic/entire function $g(z)$ such that $f(g(z))$ can be analytically continued to the whole $mathbb C$.
For example,
$$sqrt x^2=x$$
$$ln e^x=x$$
$$arccoscos x =x$$
$$lnln e^e^x=x$$
$$operatornameW(xe^x)=x$$
More complicated examples:
$$f(x)=sqrt(x+1)(x+3)=sqrt(x+2)^2-1$$
$$g(x)=-2+cosh x$$
$$f(x)=x^alphaqquadalphainmathbb C$$
$$g(x)=e^x$$
Is this true?
complex-analysis
asked Aug 1 at 4:07
Szeto
3,8431421
edited Aug 1 at 5:44
asked Aug 1 at 4:07
Szeto
3,8431421
asked Aug 1 at 4:07
Szeto
3,8431421
asked Aug 1 at 4:07
Szeto
3,8431421
3,8431421
How about $f(z) = z^2/3$?
– Sangchul Lee
Aug 1 at 5:09
$z=x^3 dots dots$
– Fabian
Aug 1 at 5:10
1
Typically, functions with branch cuts appear as inverse of holomorphic function on all of $mathbbC$ (this covers all your examples). The question is interesting, as it asks the reverse.
– Fabian
Aug 1 at 5:14
@Fabian Please see my newly added example.
– Szeto
Aug 1 at 5:37
@Fabian, Ah, I thought that $g$ must qualify as a local inverse of $f$, i.e. $f(g(z)) = z$. My bad for not carefully reading the question :s
– Sangchul Lee
Aug 1 at 5:44
add a comment |Â
How about $f(z) = z^2/3$?
– Sangchul Lee
Aug 1 at 5:09
$z=x^3 dots dots$
– Fabian
Aug 1 at 5:10
1
Typically, functions with branch cuts appear as inverse of holomorphic function on all of $mathbbC$ (this covers all your examples). The question is interesting, as it asks the reverse.
– Fabian
Aug 1 at 5:14
@Fabian Please see my newly added example.
– Szeto
Aug 1 at 5:37
@Fabian, Ah, I thought that $g$ must qualify as a local inverse of $f$, i.e. $f(g(z)) = z$. My bad for not carefully reading the question :s
– Sangchul Lee
Aug 1 at 5:44
How about $f(z) = z^2/3$?
– Sangchul Lee
Aug 1 at 5:09
How about $f(z) = z^2/3$?
– Sangchul Lee
Aug 1 at 5:09
$z=x^3 dots dots$
– Fabian
Aug 1 at 5:10
$z=x^3 dots dots$
– Fabian
Aug 1 at 5:10
1
1
Typically, functions with branch cuts appear as inverse of holomorphic function on all of $mathbbC$ (this covers all your examples). The question is interesting, as it asks the reverse.
– Fabian
Aug 1 at 5:14
Typically, functions with branch cuts appear as inverse of holomorphic function on all of $mathbbC$ (this covers all your examples). The question is interesting, as it asks the reverse.
– Fabian
Aug 1 at 5:14
@Fabian Please see my newly added example.
– Szeto
Aug 1 at 5:37
@Fabian Please see my newly added example.
– Szeto
Aug 1 at 5:37
@Fabian, Ah, I thought that $g$ must qualify as a local inverse of $f$, i.e. $f(g(z)) = z$. My bad for not carefully reading the question :s
– Sangchul Lee
Aug 1 at 5:44
@Fabian, Ah, I thought that $g$ must qualify as a local inverse of $f$, i.e. $f(g(z)) = z$. My bad for not carefully reading the question :s
– Sangchul Lee
Aug 1 at 5:44
add a comment |Â
1 Answer
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1
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There is indeed always a function such as you describe.
First, pick an open disk in $mathbbC$ where $f$ is analytic. Since $f$ is non-constant, there is a point $a$ in this region where $f'(a)neq 0$. Thus, by the Lagrange Inversion Theorem, there is a local inverse $g(z)$ which is analytic in a neighborhood of $f(a)$. In this neighborhood, $f(g(z))=z$, and the identity function $z$ can of course be extended to the complex plane.
Now, as you likely noticed, this restriction of the domain to an open disk before analytic continuation is necessary. Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$. By Picard's theorem there is no entire function on $mathbbC$ which can avoid more than one point, and no meromorphic function on $mathbbC$ which can avoid more than two. Thus for any entire or meromorphic $g$, either $g$ is constant or the preimage of the branch cut under $g$ must be a branch cut for $g$. The second case contradicts the meromorphicity of $g$. Thus there is no non-constant meromorphic $g$ such that $f(g(z))$ is meromorphic on all of $mathbbC$.
answered Aug 1 at 7:28
Grant B.
653414
Thank you. I will think about it for a while to see if I have any doubts.
– Szeto
Aug 1 at 8:15
‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $sqrt x$.
– Szeto
Aug 1 at 9:50
Using the standard branch cut for $sqrt$ along the negative real axis, the identity $sqrtx^2=x$ is only valid for $textRe(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $textRe(x)=0$.
– Grant B.
Aug 1 at 10:22
To clarify, by "avoid the cut", I mean there must not exist a path in $mathbbC$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $sqrt$, so $sqrtx^2$ is discontinuous along the imaginary axis.
– Grant B.
Aug 1 at 20:12
By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane?
– Szeto
Aug 1 at 23:32
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There is indeed always a function such as you describe.
First, pick an open disk in $mathbbC$ where $f$ is analytic. Since $f$ is non-constant, there is a point $a$ in this region where $f'(a)neq 0$. Thus, by the Lagrange Inversion Theorem, there is a local inverse $g(z)$ which is analytic in a neighborhood of $f(a)$. In this neighborhood, $f(g(z))=z$, and the identity function $z$ can of course be extended to the complex plane.
Now, as you likely noticed, this restriction of the domain to an open disk before analytic continuation is necessary. Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$. By Picard's theorem there is no entire function on $mathbbC$ which can avoid more than one point, and no meromorphic function on $mathbbC$ which can avoid more than two. Thus for any entire or meromorphic $g$, either $g$ is constant or the preimage of the branch cut under $g$ must be a branch cut for $g$. The second case contradicts the meromorphicity of $g$. Thus there is no non-constant meromorphic $g$ such that $f(g(z))$ is meromorphic on all of $mathbbC$.
answered Aug 1 at 7:28
Grant B.
653414
Thank you. I will think about it for a while to see if I have any doubts.
– Szeto
Aug 1 at 8:15
‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $sqrt x$.
– Szeto
Aug 1 at 9:50
Using the standard branch cut for $sqrt$ along the negative real axis, the identity $sqrtx^2=x$ is only valid for $textRe(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $textRe(x)=0$.
– Grant B.
Aug 1 at 10:22
To clarify, by "avoid the cut", I mean there must not exist a path in $mathbbC$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $sqrt$, so $sqrtx^2$ is discontinuous along the imaginary axis.
– Grant B.
Aug 1 at 20:12
By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane?
– Szeto
Aug 1 at 23:32
 |Â
show 2 more comments
up vote
1
down vote
There is indeed always a function such as you describe.
First, pick an open disk in $mathbbC$ where $f$ is analytic. Since $f$ is non-constant, there is a point $a$ in this region where $f'(a)neq 0$. Thus, by the Lagrange Inversion Theorem, there is a local inverse $g(z)$ which is analytic in a neighborhood of $f(a)$. In this neighborhood, $f(g(z))=z$, and the identity function $z$ can of course be extended to the complex plane.
Now, as you likely noticed, this restriction of the domain to an open disk before analytic continuation is necessary. Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$. By Picard's theorem there is no entire function on $mathbbC$ which can avoid more than one point, and no meromorphic function on $mathbbC$ which can avoid more than two. Thus for any entire or meromorphic $g$, either $g$ is constant or the preimage of the branch cut under $g$ must be a branch cut for $g$. The second case contradicts the meromorphicity of $g$. Thus there is no non-constant meromorphic $g$ such that $f(g(z))$ is meromorphic on all of $mathbbC$.
answered Aug 1 at 7:28
Grant B.
653414
Thank you. I will think about it for a while to see if I have any doubts.
– Szeto
Aug 1 at 8:15
‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $sqrt x$.
– Szeto
Aug 1 at 9:50
Using the standard branch cut for $sqrt$ along the negative real axis, the identity $sqrtx^2=x$ is only valid for $textRe(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $textRe(x)=0$.
– Grant B.
Aug 1 at 10:22
To clarify, by "avoid the cut", I mean there must not exist a path in $mathbbC$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $sqrt$, so $sqrtx^2$ is discontinuous along the imaginary axis.
– Grant B.
Aug 1 at 20:12
By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane?
– Szeto
Aug 1 at 23:32
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
There is indeed always a function such as you describe.
First, pick an open disk in $mathbbC$ where $f$ is analytic. Since $f$ is non-constant, there is a point $a$ in this region where $f'(a)neq 0$. Thus, by the Lagrange Inversion Theorem, there is a local inverse $g(z)$ which is analytic in a neighborhood of $f(a)$. In this neighborhood, $f(g(z))=z$, and the identity function $z$ can of course be extended to the complex plane.
Now, as you likely noticed, this restriction of the domain to an open disk before analytic continuation is necessary. Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$. By Picard's theorem there is no entire function on $mathbbC$ which can avoid more than one point, and no meromorphic function on $mathbbC$ which can avoid more than two. Thus for any entire or meromorphic $g$, either $g$ is constant or the preimage of the branch cut under $g$ must be a branch cut for $g$. The second case contradicts the meromorphicity of $g$. Thus there is no non-constant meromorphic $g$ such that $f(g(z))$ is meromorphic on all of $mathbbC$.
answered Aug 1 at 7:28
Grant B.
653414
There is indeed always a function such as you describe.
First, pick an open disk in $mathbbC$ where $f$ is analytic. Since $f$ is non-constant, there is a point $a$ in this region where $f'(a)neq 0$. Thus, by the Lagrange Inversion Theorem, there is a local inverse $g(z)$ which is analytic in a neighborhood of $f(a)$. In this neighborhood, $f(g(z))=z$, and the identity function $z$ can of course be extended to the complex plane.
Now, as you likely noticed, this restriction of the domain to an open disk before analytic continuation is necessary. Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$. By Picard's theorem there is no entire function on $mathbbC$ which can avoid more than one point, and no meromorphic function on $mathbbC$ which can avoid more than two. Thus for any entire or meromorphic $g$, either $g$ is constant or the preimage of the branch cut under $g$ must be a branch cut for $g$. The second case contradicts the meromorphicity of $g$. Thus there is no non-constant meromorphic $g$ such that $f(g(z))$ is meromorphic on all of $mathbbC$.
answered Aug 1 at 7:28
Grant B.
653414
edited Aug 1 at 9:08
answered Aug 1 at 7:28
Grant B.
653414
answered Aug 1 at 7:28
Grant B.
653414
answered Aug 1 at 7:28
Grant B.
653414
653414
Thank you. I will think about it for a while to see if I have any doubts.
– Szeto
Aug 1 at 8:15
‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $sqrt x$.
– Szeto
Aug 1 at 9:50
Using the standard branch cut for $sqrt$ along the negative real axis, the identity $sqrtx^2=x$ is only valid for $textRe(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $textRe(x)=0$.
– Grant B.
Aug 1 at 10:22
To clarify, by "avoid the cut", I mean there must not exist a path in $mathbbC$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $sqrt$, so $sqrtx^2$ is discontinuous along the imaginary axis.
– Grant B.
Aug 1 at 20:12
By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane?
– Szeto
Aug 1 at 23:32
 |Â
show 2 more comments
Thank you. I will think about it for a while to see if I have any doubts.
– Szeto
Aug 1 at 8:15
‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $sqrt x$.
– Szeto
Aug 1 at 9:50
Using the standard branch cut for $sqrt$ along the negative real axis, the identity $sqrtx^2=x$ is only valid for $textRe(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $textRe(x)=0$.
– Grant B.
Aug 1 at 10:22
To clarify, by "avoid the cut", I mean there must not exist a path in $mathbbC$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $sqrt$, so $sqrtx^2$ is discontinuous along the imaginary axis.
– Grant B.
Aug 1 at 20:12
By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane?
– Szeto
Aug 1 at 23:32
Thank you. I will think about it for a while to see if I have any doubts.
– Szeto
Aug 1 at 8:15
Thank you. I will think about it for a while to see if I have any doubts.
– Szeto
Aug 1 at 8:15
‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $sqrt x$.
– Szeto
Aug 1 at 9:50
‘Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$.’ Umm...not necessarily. In my first example, $x^2$ does not avoid $0$, which is the branch point of $sqrt x$.
– Szeto
Aug 1 at 9:50
Using the standard branch cut for $sqrt$ along the negative real axis, the identity $sqrtx^2=x$ is only valid for $textRe(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $textRe(x)=0$.
– Grant B.
Aug 1 at 10:22
Using the standard branch cut for $sqrt$ along the negative real axis, the identity $sqrtx^2=x$ is only valid for $textRe(x)>0$. (And the point $x=0$ as you said, but no neighborhood around it.) The discontinuity in the composite function is along the line $textRe(x)=0$.
– Grant B.
Aug 1 at 10:22
To clarify, by "avoid the cut", I mean there must not exist a path in $mathbbC$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $sqrt$, so $sqrtx^2$ is discontinuous along the imaginary axis.
– Grant B.
Aug 1 at 20:12
To clarify, by "avoid the cut", I mean there must not exist a path in $mathbbC$ such that the image of the path under $g$ crosses the branch cut. For $x^2$ any path crossing the imaginary axis will cross the branch cut of $sqrt$, so $sqrtx^2$ is discontinuous along the imaginary axis.
– Grant B.
Aug 1 at 20:12
By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane?
– Szeto
Aug 1 at 23:32
By the way, is it guaranteed that $g’(z)$ can be analytically continued to the whole complex plane?
– Szeto
Aug 1 at 23:32
 |Â
show 2 more comments
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How about $f(z) = z^2/3$?
– Sangchul Lee
Aug 1 at 5:09
$z=x^3 dots dots$
– Fabian
Aug 1 at 5:10
1
Typically, functions with branch cuts appear as inverse of holomorphic function on all of $mathbbC$ (this covers all your examples). The question is interesting, as it asks the reverse.
– Fabian
Aug 1 at 5:14
@Fabian Please see my newly added example.
– Szeto
Aug 1 at 5:37
@Fabian, Ah, I thought that $g$ must qualify as a local inverse of $f$, i.e. $f(g(z)) = z$. My bad for not carefully reading the question :s
– Sangchul Lee
Aug 1 at 5:44