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Definitions of analytic, regular, holomorphic, differentiable, conformal: what implies what and do any imply that a function is a bijection?
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I'm looking back at some complex analysis and have gotten myself a little muddled in all of the definitions analytic/ regular/ holomorphic/ differentiable/ conformal...
In particular, at the moment I'm thinking about conformal functions $f(z)$ on open sets in $mathbbC$. Many of the conformal maps I'm using are bijections on $hatmathbbC$ e.g. Mobius Transformations. But other conformal maps are only bijections when restricted to a certain set e.g. $f(z)=z^2$ from $z:vert mathrmarg(z) vert < fracpi4 $ to $z: vert mathrmarg(z) vert < fracpi2 $ is conformal and a bijection but $f(z)=z^2$ from $z:vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is not a bijection.
Does bijection-ness feature of any of the definitions of analytic/ regular/ holomorphic/ differentiable/ conformal? Is it a result of any of the definitions? (e.g. might "analytic" imply "bijection"?) Is there a specific name for conformal functions like $f(z)=z^2$ that can be restricted so that they become bijections? Thanks for any help!
complex-analysis conformal-geometry holomorphic-functions analytic-functions
asked Aug 7 at 14:19
Churchill
625
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up vote
7
down vote
favorite
I'm looking back at some complex analysis and have gotten myself a little muddled in all of the definitions analytic/ regular/ holomorphic/ differentiable/ conformal...
In particular, at the moment I'm thinking about conformal functions $f(z)$ on open sets in $mathbbC$. Many of the conformal maps I'm using are bijections on $hatmathbbC$ e.g. Mobius Transformations. But other conformal maps are only bijections when restricted to a certain set e.g. $f(z)=z^2$ from $z:vert mathrmarg(z) vert < fracpi4 $ to $z: vert mathrmarg(z) vert < fracpi2 $ is conformal and a bijection but $f(z)=z^2$ from $z:vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is not a bijection.
Does bijection-ness feature of any of the definitions of analytic/ regular/ holomorphic/ differentiable/ conformal? Is it a result of any of the definitions? (e.g. might "analytic" imply "bijection"?) Is there a specific name for conformal functions like $f(z)=z^2$ that can be restricted so that they become bijections? Thanks for any help!
complex-analysis conformal-geometry holomorphic-functions analytic-functions
asked Aug 7 at 14:19
Churchill
625
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I'm looking back at some complex analysis and have gotten myself a little muddled in all of the definitions analytic/ regular/ holomorphic/ differentiable/ conformal...
In particular, at the moment I'm thinking about conformal functions $f(z)$ on open sets in $mathbbC$. Many of the conformal maps I'm using are bijections on $hatmathbbC$ e.g. Mobius Transformations. But other conformal maps are only bijections when restricted to a certain set e.g. $f(z)=z^2$ from $z:vert mathrmarg(z) vert < fracpi4 $ to $z: vert mathrmarg(z) vert < fracpi2 $ is conformal and a bijection but $f(z)=z^2$ from $z:vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is not a bijection.
Does bijection-ness feature of any of the definitions of analytic/ regular/ holomorphic/ differentiable/ conformal? Is it a result of any of the definitions? (e.g. might "analytic" imply "bijection"?) Is there a specific name for conformal functions like $f(z)=z^2$ that can be restricted so that they become bijections? Thanks for any help!
complex-analysis conformal-geometry holomorphic-functions analytic-functions
asked Aug 7 at 14:19
Churchill
625
I'm looking back at some complex analysis and have gotten myself a little muddled in all of the definitions analytic/ regular/ holomorphic/ differentiable/ conformal...
In particular, at the moment I'm thinking about conformal functions $f(z)$ on open sets in $mathbbC$. Many of the conformal maps I'm using are bijections on $hatmathbbC$ e.g. Mobius Transformations. But other conformal maps are only bijections when restricted to a certain set e.g. $f(z)=z^2$ from $z:vert mathrmarg(z) vert < fracpi4 $ to $z: vert mathrmarg(z) vert < fracpi2 $ is conformal and a bijection but $f(z)=z^2$ from $z:vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is not a bijection.
Does bijection-ness feature of any of the definitions of analytic/ regular/ holomorphic/ differentiable/ conformal? Is it a result of any of the definitions? (e.g. might "analytic" imply "bijection"?) Is there a specific name for conformal functions like $f(z)=z^2$ that can be restricted so that they become bijections? Thanks for any help!
complex-analysis conformal-geometry holomorphic-functions analytic-functions
asked Aug 7 at 14:19
Churchill
625
asked Aug 7 at 14:19
Churchill
625
asked Aug 7 at 14:19
Churchill
625
asked Aug 7 at 14:19
Churchill
625
625
add a comment |Â
add a comment |Â
2 Answers
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Firstly, let me clear up the definitions for you. Let $U$ be an open set. Then:
An analytic function on $U$ (real or complex) is infinitely differentiable and equal to its Taylor series in a neighbourhood of every point in $U$.
A holomorphic function on $U$ (necessarily complex) is complex differentiable in a neighbourhood of every point in $U$.
A conformal function is a holomorphic function whose derivative is non-zero on $U$.
Avoid using the term "differentiable function" when working with complex functions. Use one of the terms above instead. I would also avoid "regular": it means the same as "analytic" but isn't well used.
For all functions (real or complex), analytic implies holomorphic. For complex functions, Cauchy proved that holomorphic implies analytic (which I still find astounding)! Hence conformal also implies holomorphic and analytic.
None of the definitions involve a function being a bijection, nor do any of them imply that a function is a bijection. You offer an example yourself: $f(z)=z^2$ from $z: vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is conformal, hence also holomorphic and analytic, but is not a bijection.
However, in Latrace's answer to this question they prove that:
If $f$ is conformal in an open set $G$, then for each $a in G$ there exists an $r > 0$ such that the restriction of $f$ to $D(a;r)$ is one-to-one (where $D(a;r)$ represents the open disk centered at $a$ of radius $r$).
... so in fact every conformal function can be restricted so that they become a bijection.
answered Aug 7 at 14:31
Malkin
1,514623
add a comment |Â
up vote
2
down vote
None of these types of functions have to be bijective. However, for every point $z_0$ of the domain of any conformal map $f$, there is a neighborhood $U$ of $z_0$ such that the restriction of $f$ to $U$ is injective.
On the other hand, note that constant functions are analytic, holomorphic and differentiable.
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
Firstly, let me clear up the definitions for you. Let $U$ be an open set. Then:
An analytic function on $U$ (real or complex) is infinitely differentiable and equal to its Taylor series in a neighbourhood of every point in $U$.
A holomorphic function on $U$ (necessarily complex) is complex differentiable in a neighbourhood of every point in $U$.
A conformal function is a holomorphic function whose derivative is non-zero on $U$.
Avoid using the term "differentiable function" when working with complex functions. Use one of the terms above instead. I would also avoid "regular": it means the same as "analytic" but isn't well used.
For all functions (real or complex), analytic implies holomorphic. For complex functions, Cauchy proved that holomorphic implies analytic (which I still find astounding)! Hence conformal also implies holomorphic and analytic.
None of the definitions involve a function being a bijection, nor do any of them imply that a function is a bijection. You offer an example yourself: $f(z)=z^2$ from $z: vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is conformal, hence also holomorphic and analytic, but is not a bijection.
However, in Latrace's answer to this question they prove that:
If $f$ is conformal in an open set $G$, then for each $a in G$ there exists an $r > 0$ such that the restriction of $f$ to $D(a;r)$ is one-to-one (where $D(a;r)$ represents the open disk centered at $a$ of radius $r$).
... so in fact every conformal function can be restricted so that they become a bijection.
answered Aug 7 at 14:31
Malkin
1,514623
add a comment |Â
up vote
12
down vote
accepted
Firstly, let me clear up the definitions for you. Let $U$ be an open set. Then:
An analytic function on $U$ (real or complex) is infinitely differentiable and equal to its Taylor series in a neighbourhood of every point in $U$.
A holomorphic function on $U$ (necessarily complex) is complex differentiable in a neighbourhood of every point in $U$.
A conformal function is a holomorphic function whose derivative is non-zero on $U$.
Avoid using the term "differentiable function" when working with complex functions. Use one of the terms above instead. I would also avoid "regular": it means the same as "analytic" but isn't well used.
For all functions (real or complex), analytic implies holomorphic. For complex functions, Cauchy proved that holomorphic implies analytic (which I still find astounding)! Hence conformal also implies holomorphic and analytic.
None of the definitions involve a function being a bijection, nor do any of them imply that a function is a bijection. You offer an example yourself: $f(z)=z^2$ from $z: vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is conformal, hence also holomorphic and analytic, but is not a bijection.
However, in Latrace's answer to this question they prove that:
If $f$ is conformal in an open set $G$, then for each $a in G$ there exists an $r > 0$ such that the restriction of $f$ to $D(a;r)$ is one-to-one (where $D(a;r)$ represents the open disk centered at $a$ of radius $r$).
... so in fact every conformal function can be restricted so that they become a bijection.
answered Aug 7 at 14:31
Malkin
1,514623
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
Firstly, let me clear up the definitions for you. Let $U$ be an open set. Then:
An analytic function on $U$ (real or complex) is infinitely differentiable and equal to its Taylor series in a neighbourhood of every point in $U$.
A holomorphic function on $U$ (necessarily complex) is complex differentiable in a neighbourhood of every point in $U$.
A conformal function is a holomorphic function whose derivative is non-zero on $U$.
Avoid using the term "differentiable function" when working with complex functions. Use one of the terms above instead. I would also avoid "regular": it means the same as "analytic" but isn't well used.
For all functions (real or complex), analytic implies holomorphic. For complex functions, Cauchy proved that holomorphic implies analytic (which I still find astounding)! Hence conformal also implies holomorphic and analytic.
None of the definitions involve a function being a bijection, nor do any of them imply that a function is a bijection. You offer an example yourself: $f(z)=z^2$ from $z: vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is conformal, hence also holomorphic and analytic, but is not a bijection.
However, in Latrace's answer to this question they prove that:
If $f$ is conformal in an open set $G$, then for each $a in G$ there exists an $r > 0$ such that the restriction of $f$ to $D(a;r)$ is one-to-one (where $D(a;r)$ represents the open disk centered at $a$ of radius $r$).
... so in fact every conformal function can be restricted so that they become a bijection.
answered Aug 7 at 14:31
Malkin
1,514623
Firstly, let me clear up the definitions for you. Let $U$ be an open set. Then:
An analytic function on $U$ (real or complex) is infinitely differentiable and equal to its Taylor series in a neighbourhood of every point in $U$.
A holomorphic function on $U$ (necessarily complex) is complex differentiable in a neighbourhood of every point in $U$.
A conformal function is a holomorphic function whose derivative is non-zero on $U$.
Avoid using the term "differentiable function" when working with complex functions. Use one of the terms above instead. I would also avoid "regular": it means the same as "analytic" but isn't well used.
For all functions (real or complex), analytic implies holomorphic. For complex functions, Cauchy proved that holomorphic implies analytic (which I still find astounding)! Hence conformal also implies holomorphic and analytic.
None of the definitions involve a function being a bijection, nor do any of them imply that a function is a bijection. You offer an example yourself: $f(z)=z^2$ from $z: vert mathrmarg(z) vert < frac2pi3 $ to $mathbbC-0$ is conformal, hence also holomorphic and analytic, but is not a bijection.
However, in Latrace's answer to this question they prove that:
If $f$ is conformal in an open set $G$, then for each $a in G$ there exists an $r > 0$ such that the restriction of $f$ to $D(a;r)$ is one-to-one (where $D(a;r)$ represents the open disk centered at $a$ of radius $r$).
... so in fact every conformal function can be restricted so that they become a bijection.
answered Aug 7 at 14:31
Malkin
1,514623
edited Aug 16 at 17:45
answered Aug 7 at 14:31
Malkin
1,514623
answered Aug 7 at 14:31
Malkin
1,514623
answered Aug 7 at 14:31
Malkin
1,514623
1,514623
add a comment |Â
add a comment |Â
up vote
2
down vote
None of these types of functions have to be bijective. However, for every point $z_0$ of the domain of any conformal map $f$, there is a neighborhood $U$ of $z_0$ such that the restriction of $f$ to $U$ is injective.
On the other hand, note that constant functions are analytic, holomorphic and differentiable.
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
add a comment |Â
up vote
2
down vote
None of these types of functions have to be bijective. However, for every point $z_0$ of the domain of any conformal map $f$, there is a neighborhood $U$ of $z_0$ such that the restriction of $f$ to $U$ is injective.
On the other hand, note that constant functions are analytic, holomorphic and differentiable.
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
add a comment |Â
up vote
2
down vote
up vote
2
down vote
None of these types of functions have to be bijective. However, for every point $z_0$ of the domain of any conformal map $f$, there is a neighborhood $U$ of $z_0$ such that the restriction of $f$ to $U$ is injective.
On the other hand, note that constant functions are analytic, holomorphic and differentiable.
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
None of these types of functions have to be bijective. However, for every point $z_0$ of the domain of any conformal map $f$, there is a neighborhood $U$ of $z_0$ such that the restriction of $f$ to $U$ is injective.
On the other hand, note that constant functions are analytic, holomorphic and differentiable.
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
answered Aug 7 at 14:26
José Carlos Santos
118k16101182
118k16101182
add a comment |Â
add a comment |Â
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