Example of holomorphic function with no natural root
$begingroup$
If $f$ is holomorphic over a simply connected set $OmegasubsetBbb C$ and $f(z)ne0 forall zinOmega$ then it is known that for any $ninBbb N^*$ there exists a holomorphic function $g$ such that $g^n=f$.
But this isn't necessarily true if one of the properties [$fne0 forall zinOmega$] or [$Omega$ simply connected] isn't true.
- Why is $zmapsto z$ over $Bbb C$ a counter example? Namely, why is $pm z^{1/2}=pm e^{log zover 2 }$ not holomorphic?
- Is there an example where $fne 0forall zin Omega$ but the fact that $Omega$ isn't simply connected messes things up?
complex-analysis examples-counterexamples holomorphic-functions
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
$endgroup$
add a comment |
$begingroup$
If $f$ is holomorphic over a simply connected set $OmegasubsetBbb C$ and $f(z)ne0 forall zinOmega$ then it is known that for any $ninBbb N^*$ there exists a holomorphic function $g$ such that $g^n=f$.
But this isn't necessarily true if one of the properties [$fne0 forall zinOmega$] or [$Omega$ simply connected] isn't true.
- Why is $zmapsto z$ over $Bbb C$ a counter example? Namely, why is $pm z^{1/2}=pm e^{log zover 2 }$ not holomorphic?
- Is there an example where $fne 0forall zin Omega$ but the fact that $Omega$ isn't simply connected messes things up?
complex-analysis examples-counterexamples holomorphic-functions
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
$endgroup$
$begingroup$
Half of the Arg mod $2pi$ is ambiguous. This ambiguity leads to a singularly.
$endgroup$
– Charlie Frohman
Dec 1 '18 at 16:14
$begingroup$
@CharlieFrohman Could you expand a little bit on this? Why should there be a singularity? Isn't it just a discontinuity?
$endgroup$
– John Cataldo
Dec 1 '18 at 16:17
add a comment |
$begingroup$
If $f$ is holomorphic over a simply connected set $OmegasubsetBbb C$ and $f(z)ne0 forall zinOmega$ then it is known that for any $ninBbb N^*$ there exists a holomorphic function $g$ such that $g^n=f$.
But this isn't necessarily true if one of the properties [$fne0 forall zinOmega$] or [$Omega$ simply connected] isn't true.
- Why is $zmapsto z$ over $Bbb C$ a counter example? Namely, why is $pm z^{1/2}=pm e^{log zover 2 }$ not holomorphic?
- Is there an example where $fne 0forall zin Omega$ but the fact that $Omega$ isn't simply connected messes things up?
complex-analysis examples-counterexamples holomorphic-functions
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
$endgroup$
If $f$ is holomorphic over a simply connected set $OmegasubsetBbb C$ and $f(z)ne0 forall zinOmega$ then it is known that for any $ninBbb N^*$ there exists a holomorphic function $g$ such that $g^n=f$.
But this isn't necessarily true if one of the properties [$fne0 forall zinOmega$] or [$Omega$ simply connected] isn't true.
- Why is $zmapsto z$ over $Bbb C$ a counter example? Namely, why is $pm z^{1/2}=pm e^{log zover 2 }$ not holomorphic?
- Is there an example where $fne 0forall zin Omega$ but the fact that $Omega$ isn't simply connected messes things up?
complex-analysis examples-counterexamples holomorphic-functions
complex-analysis examples-counterexamples holomorphic-functions
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
asked Dec 1 '18 at 16:08
John CataldoJohn Cataldo
1,1151316
1,1151316
$begingroup$
Half of the Arg mod $2pi$ is ambiguous. This ambiguity leads to a singularly.
$endgroup$
– Charlie Frohman
Dec 1 '18 at 16:14
$begingroup$
@CharlieFrohman Could you expand a little bit on this? Why should there be a singularity? Isn't it just a discontinuity?
$endgroup$
– John Cataldo
Dec 1 '18 at 16:17
add a comment |
$begingroup$
Half of the Arg mod $2pi$ is ambiguous. This ambiguity leads to a singularly.
$endgroup$
– Charlie Frohman
Dec 1 '18 at 16:14
$begingroup$
@CharlieFrohman Could you expand a little bit on this? Why should there be a singularity? Isn't it just a discontinuity?
$endgroup$
– John Cataldo
Dec 1 '18 at 16:17
$begingroup$
Half of the Arg mod $2pi$ is ambiguous. This ambiguity leads to a singularly.
$endgroup$
– Charlie Frohman
Dec 1 '18 at 16:14
$begingroup$
Half of the Arg mod $2pi$ is ambiguous. This ambiguity leads to a singularly.
$endgroup$
– Charlie Frohman
Dec 1 '18 at 16:14
$begingroup$
@CharlieFrohman Could you expand a little bit on this? Why should there be a singularity? Isn't it just a discontinuity?
$endgroup$
– John Cataldo
Dec 1 '18 at 16:17
$begingroup$
@CharlieFrohman Could you expand a little bit on this? Why should there be a singularity? Isn't it just a discontinuity?
$endgroup$
– John Cataldo
Dec 1 '18 at 16:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $V=Bbb Csetminus {0}$. Of course there is no $fin H(V)$ with $f(z)^2=z$. People show this by considering a function $z^{1/2}$ holomorphic near $1$, say, and then showing that if you try to continue it on a loop around the origin what you get is discontinuous.
An argument that seems cleaner and simpler to me: Recall first that if $fin H(V)$ and $gamma$ is any closed curve in $V$ then $$frac1{2pi i}int_gammafrac{f'(z)}{f(z)},dzinBbb Z,$$by the argument principle. But if $f(z)^2=z$ then $2f(z)f'(z)=1$, so $f'(z)/f(z)=1/(2 f(z)^2)=1/(2z)$and hence $$frac1{2pi i}int_{|z|=1}frac{f'(z)}{f(z)},dz=frac12.$$
(Considering the argument principle, this really is the same proof: If $gamma$ is some parametrization of the unit circle then this shows that $fcirc gamma$ cannot be a closed curve, since the winding number about the origin would be $1/2$. In fact $1/2$ is exactly right - as $z$ traverses the unit circle, $z^{1/2}$ only goes halfway around the circle, ending at the negative of the initial point.)
About square roots in general:
Exercise. Suppose $Omega$ is an open set in the plane and $gin H(Omega)$, not vanishing identically on any component of $Omega$. There exists $fin H(Omega)$ with $f^2=g$ if and only if $int_gamma g'/g$ is an even integer for every closed curve $gamma$ in $Omega$.
Similarly for the non-existence of a logarithm: If $fin H(V)$ and $gamma$ is a closed curve in $V$ then $$int_gamma f'(z),dz=0,$$but $int_{|z|=1}frac1z,dzne0$, hence there is no $fin H(V)$ with $f'(z)=1/z$.
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
add a comment |
$begingroup$
There are no such functions as $pm z^{1/2}$ and $pm e^{frac{log z}2}$, because
$z^{1/2}$ can be any sequare root of $z$. So there it doesnt represent an individual complex number.- each complex number $z$ has infinitely many logarithms. Which one do ou have in mind.
Concerning your final question,$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb C\z&mapsto&zend{array}$$has no holomorphic square root.
edited Dec 1 '18 at 16:24
mercio
44.6k256110
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
$begingroup$
I mean the principal definition $log(z)=log(|z|)+itext{arg}(z)$ where arg(z) is the argument ranging in $(-pi,pi]$
$endgroup$
– John Cataldo
Dec 1 '18 at 16:15
1
$begingroup$
That function isn't even continuous. Therefore, it is not holomorphic.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 16:17
$begingroup$
That explains why just saying "let $f(z)=z^{1/2}$" does not suffice to define a square root, but I don't see how it's a proof that there is no $f$ with $f^2=z$...
$endgroup$
– David C. Ullrich
Dec 1 '18 at 16:42
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $V=Bbb Csetminus {0}$. Of course there is no $fin H(V)$ with $f(z)^2=z$. People show this by considering a function $z^{1/2}$ holomorphic near $1$, say, and then showing that if you try to continue it on a loop around the origin what you get is discontinuous.
An argument that seems cleaner and simpler to me: Recall first that if $fin H(V)$ and $gamma$ is any closed curve in $V$ then $$frac1{2pi i}int_gammafrac{f'(z)}{f(z)},dzinBbb Z,$$by the argument principle. But if $f(z)^2=z$ then $2f(z)f'(z)=1$, so $f'(z)/f(z)=1/(2 f(z)^2)=1/(2z)$and hence $$frac1{2pi i}int_{|z|=1}frac{f'(z)}{f(z)},dz=frac12.$$
(Considering the argument principle, this really is the same proof: If $gamma$ is some parametrization of the unit circle then this shows that $fcirc gamma$ cannot be a closed curve, since the winding number about the origin would be $1/2$. In fact $1/2$ is exactly right - as $z$ traverses the unit circle, $z^{1/2}$ only goes halfway around the circle, ending at the negative of the initial point.)
About square roots in general:
Exercise. Suppose $Omega$ is an open set in the plane and $gin H(Omega)$, not vanishing identically on any component of $Omega$. There exists $fin H(Omega)$ with $f^2=g$ if and only if $int_gamma g'/g$ is an even integer for every closed curve $gamma$ in $Omega$.
Similarly for the non-existence of a logarithm: If $fin H(V)$ and $gamma$ is a closed curve in $V$ then $$int_gamma f'(z),dz=0,$$but $int_{|z|=1}frac1z,dzne0$, hence there is no $fin H(V)$ with $f'(z)=1/z$.
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
add a comment |
$begingroup$
Let $V=Bbb Csetminus {0}$. Of course there is no $fin H(V)$ with $f(z)^2=z$. People show this by considering a function $z^{1/2}$ holomorphic near $1$, say, and then showing that if you try to continue it on a loop around the origin what you get is discontinuous.
An argument that seems cleaner and simpler to me: Recall first that if $fin H(V)$ and $gamma$ is any closed curve in $V$ then $$frac1{2pi i}int_gammafrac{f'(z)}{f(z)},dzinBbb Z,$$by the argument principle. But if $f(z)^2=z$ then $2f(z)f'(z)=1$, so $f'(z)/f(z)=1/(2 f(z)^2)=1/(2z)$and hence $$frac1{2pi i}int_{|z|=1}frac{f'(z)}{f(z)},dz=frac12.$$
(Considering the argument principle, this really is the same proof: If $gamma$ is some parametrization of the unit circle then this shows that $fcirc gamma$ cannot be a closed curve, since the winding number about the origin would be $1/2$. In fact $1/2$ is exactly right - as $z$ traverses the unit circle, $z^{1/2}$ only goes halfway around the circle, ending at the negative of the initial point.)
About square roots in general:
Exercise. Suppose $Omega$ is an open set in the plane and $gin H(Omega)$, not vanishing identically on any component of $Omega$. There exists $fin H(Omega)$ with $f^2=g$ if and only if $int_gamma g'/g$ is an even integer for every closed curve $gamma$ in $Omega$.
Similarly for the non-existence of a logarithm: If $fin H(V)$ and $gamma$ is a closed curve in $V$ then $$int_gamma f'(z),dz=0,$$but $int_{|z|=1}frac1z,dzne0$, hence there is no $fin H(V)$ with $f'(z)=1/z$.
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
add a comment |
$begingroup$
Let $V=Bbb Csetminus {0}$. Of course there is no $fin H(V)$ with $f(z)^2=z$. People show this by considering a function $z^{1/2}$ holomorphic near $1$, say, and then showing that if you try to continue it on a loop around the origin what you get is discontinuous.
An argument that seems cleaner and simpler to me: Recall first that if $fin H(V)$ and $gamma$ is any closed curve in $V$ then $$frac1{2pi i}int_gammafrac{f'(z)}{f(z)},dzinBbb Z,$$by the argument principle. But if $f(z)^2=z$ then $2f(z)f'(z)=1$, so $f'(z)/f(z)=1/(2 f(z)^2)=1/(2z)$and hence $$frac1{2pi i}int_{|z|=1}frac{f'(z)}{f(z)},dz=frac12.$$
(Considering the argument principle, this really is the same proof: If $gamma$ is some parametrization of the unit circle then this shows that $fcirc gamma$ cannot be a closed curve, since the winding number about the origin would be $1/2$. In fact $1/2$ is exactly right - as $z$ traverses the unit circle, $z^{1/2}$ only goes halfway around the circle, ending at the negative of the initial point.)
About square roots in general:
Exercise. Suppose $Omega$ is an open set in the plane and $gin H(Omega)$, not vanishing identically on any component of $Omega$. There exists $fin H(Omega)$ with $f^2=g$ if and only if $int_gamma g'/g$ is an even integer for every closed curve $gamma$ in $Omega$.
Similarly for the non-existence of a logarithm: If $fin H(V)$ and $gamma$ is a closed curve in $V$ then $$int_gamma f'(z),dz=0,$$but $int_{|z|=1}frac1z,dzne0$, hence there is no $fin H(V)$ with $f'(z)=1/z$.
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
$endgroup$
Let $V=Bbb Csetminus {0}$. Of course there is no $fin H(V)$ with $f(z)^2=z$. People show this by considering a function $z^{1/2}$ holomorphic near $1$, say, and then showing that if you try to continue it on a loop around the origin what you get is discontinuous.
An argument that seems cleaner and simpler to me: Recall first that if $fin H(V)$ and $gamma$ is any closed curve in $V$ then $$frac1{2pi i}int_gammafrac{f'(z)}{f(z)},dzinBbb Z,$$by the argument principle. But if $f(z)^2=z$ then $2f(z)f'(z)=1$, so $f'(z)/f(z)=1/(2 f(z)^2)=1/(2z)$and hence $$frac1{2pi i}int_{|z|=1}frac{f'(z)}{f(z)},dz=frac12.$$
(Considering the argument principle, this really is the same proof: If $gamma$ is some parametrization of the unit circle then this shows that $fcirc gamma$ cannot be a closed curve, since the winding number about the origin would be $1/2$. In fact $1/2$ is exactly right - as $z$ traverses the unit circle, $z^{1/2}$ only goes halfway around the circle, ending at the negative of the initial point.)
About square roots in general:
Exercise. Suppose $Omega$ is an open set in the plane and $gin H(Omega)$, not vanishing identically on any component of $Omega$. There exists $fin H(Omega)$ with $f^2=g$ if and only if $int_gamma g'/g$ is an even integer for every closed curve $gamma$ in $Omega$.
Similarly for the non-existence of a logarithm: If $fin H(V)$ and $gamma$ is a closed curve in $V$ then $$int_gamma f'(z),dz=0,$$but $int_{|z|=1}frac1z,dzne0$, hence there is no $fin H(V)$ with $f'(z)=1/z$.
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
edited Dec 1 '18 at 16:54
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
answered Dec 1 '18 at 16:34
David C. UllrichDavid C. Ullrich
59.5k43893
59.5k43893
add a comment |
add a comment |
$begingroup$
There are no such functions as $pm z^{1/2}$ and $pm e^{frac{log z}2}$, because
$z^{1/2}$ can be any sequare root of $z$. So there it doesnt represent an individual complex number.- each complex number $z$ has infinitely many logarithms. Which one do ou have in mind.
Concerning your final question,$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb C\z&mapsto&zend{array}$$has no holomorphic square root.
edited Dec 1 '18 at 16:24
mercio
44.6k256110
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
$begingroup$
I mean the principal definition $log(z)=log(|z|)+itext{arg}(z)$ where arg(z) is the argument ranging in $(-pi,pi]$
$endgroup$
– John Cataldo
Dec 1 '18 at 16:15
1
$begingroup$
That function isn't even continuous. Therefore, it is not holomorphic.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 16:17
$begingroup$
That explains why just saying "let $f(z)=z^{1/2}$" does not suffice to define a square root, but I don't see how it's a proof that there is no $f$ with $f^2=z$...
$endgroup$
– David C. Ullrich
Dec 1 '18 at 16:42
add a comment |
$begingroup$
There are no such functions as $pm z^{1/2}$ and $pm e^{frac{log z}2}$, because
$z^{1/2}$ can be any sequare root of $z$. So there it doesnt represent an individual complex number.- each complex number $z$ has infinitely many logarithms. Which one do ou have in mind.
Concerning your final question,$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb C\z&mapsto&zend{array}$$has no holomorphic square root.
edited Dec 1 '18 at 16:24
mercio
44.6k256110
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
$begingroup$
I mean the principal definition $log(z)=log(|z|)+itext{arg}(z)$ where arg(z) is the argument ranging in $(-pi,pi]$
$endgroup$
– John Cataldo
Dec 1 '18 at 16:15
1
$begingroup$
That function isn't even continuous. Therefore, it is not holomorphic.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 16:17
$begingroup$
That explains why just saying "let $f(z)=z^{1/2}$" does not suffice to define a square root, but I don't see how it's a proof that there is no $f$ with $f^2=z$...
$endgroup$
– David C. Ullrich
Dec 1 '18 at 16:42
add a comment |
$begingroup$
There are no such functions as $pm z^{1/2}$ and $pm e^{frac{log z}2}$, because
$z^{1/2}$ can be any sequare root of $z$. So there it doesnt represent an individual complex number.- each complex number $z$ has infinitely many logarithms. Which one do ou have in mind.
Concerning your final question,$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb C\z&mapsto&zend{array}$$has no holomorphic square root.
edited Dec 1 '18 at 16:24
mercio
44.6k256110
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
$endgroup$
There are no such functions as $pm z^{1/2}$ and $pm e^{frac{log z}2}$, because
$z^{1/2}$ can be any sequare root of $z$. So there it doesnt represent an individual complex number.- each complex number $z$ has infinitely many logarithms. Which one do ou have in mind.
Concerning your final question,$$begin{array}{ccc}mathbb{C}setminus{0}&longrightarrow&mathbb C\z&mapsto&zend{array}$$has no holomorphic square root.
edited Dec 1 '18 at 16:24
mercio
44.6k256110
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
edited Dec 1 '18 at 16:24
mercio
44.6k256110
edited Dec 1 '18 at 16:24
mercio
44.6k256110
edited Dec 1 '18 at 16:24
mercio
44.6k256110
44.6k256110
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
answered Dec 1 '18 at 16:13
José Carlos SantosJosé Carlos Santos
154k22123226
154k22123226
$begingroup$
I mean the principal definition $log(z)=log(|z|)+itext{arg}(z)$ where arg(z) is the argument ranging in $(-pi,pi]$
$endgroup$
– John Cataldo
Dec 1 '18 at 16:15
1
$begingroup$
That function isn't even continuous. Therefore, it is not holomorphic.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 16:17
$begingroup$
That explains why just saying "let $f(z)=z^{1/2}$" does not suffice to define a square root, but I don't see how it's a proof that there is no $f$ with $f^2=z$...
$endgroup$
– David C. Ullrich
Dec 1 '18 at 16:42
add a comment |
$begingroup$
I mean the principal definition $log(z)=log(|z|)+itext{arg}(z)$ where arg(z) is the argument ranging in $(-pi,pi]$
$endgroup$
– John Cataldo
Dec 1 '18 at 16:15
1
$begingroup$
That function isn't even continuous. Therefore, it is not holomorphic.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 16:17
$begingroup$
That explains why just saying "let $f(z)=z^{1/2}$" does not suffice to define a square root, but I don't see how it's a proof that there is no $f$ with $f^2=z$...
$endgroup$
– David C. Ullrich
Dec 1 '18 at 16:42
$begingroup$
I mean the principal definition $log(z)=log(|z|)+itext{arg}(z)$ where arg(z) is the argument ranging in $(-pi,pi]$
$endgroup$
– John Cataldo
Dec 1 '18 at 16:15
$begingroup$
I mean the principal definition $log(z)=log(|z|)+itext{arg}(z)$ where arg(z) is the argument ranging in $(-pi,pi]$
$endgroup$
– John Cataldo
Dec 1 '18 at 16:15
1
1
$begingroup$
That function isn't even continuous. Therefore, it is not holomorphic.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 16:17
$begingroup$
That function isn't even continuous. Therefore, it is not holomorphic.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 16:17
$begingroup$
That explains why just saying "let $f(z)=z^{1/2}$" does not suffice to define a square root, but I don't see how it's a proof that there is no $f$ with $f^2=z$...
$endgroup$
– David C. Ullrich
Dec 1 '18 at 16:42
$begingroup$
That explains why just saying "let $f(z)=z^{1/2}$" does not suffice to define a square root, but I don't see how it's a proof that there is no $f$ with $f^2=z$...
$endgroup$
– David C. Ullrich
Dec 1 '18 at 16:42
add a comment |
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$begingroup$
Half of the Arg mod $2pi$ is ambiguous. This ambiguity leads to a singularly.
$endgroup$
– Charlie Frohman
Dec 1 '18 at 16:14
$begingroup$
@CharlieFrohman Could you expand a little bit on this? Why should there be a singularity? Isn't it just a discontinuity?
$endgroup$
– John Cataldo
Dec 1 '18 at 16:17