Mixing
Is there a holomorphic function such that $(f(z))^3=z-z^2$
$begingroup$
Is there a holomorphic function $f:C-[0,1]$ such that $(f(z))^3=z-z^2$ for all $zin C-[0,1]$
My intuition tells me that not really, for instance $$(z-z^2)^{frac{1}{3}}$$
does not have a unique branch on this set, but I do not know how to formally prove it.
complex-analysis complex-numbers
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
$endgroup$
add a comment |
$begingroup$
Is there a holomorphic function $f:C-[0,1]$ such that $(f(z))^3=z-z^2$ for all $zin C-[0,1]$
My intuition tells me that not really, for instance $$(z-z^2)^{frac{1}{3}}$$
does not have a unique branch on this set, but I do not know how to formally prove it.
complex-analysis complex-numbers
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
$endgroup$
add a comment |
$begingroup$
Is there a holomorphic function $f:C-[0,1]$ such that $(f(z))^3=z-z^2$ for all $zin C-[0,1]$
My intuition tells me that not really, for instance $$(z-z^2)^{frac{1}{3}}$$
does not have a unique branch on this set, but I do not know how to formally prove it.
complex-analysis complex-numbers
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
$endgroup$
Is there a holomorphic function $f:C-[0,1]$ such that $(f(z))^3=z-z^2$ for all $zin C-[0,1]$
My intuition tells me that not really, for instance $$(z-z^2)^{frac{1}{3}}$$
does not have a unique branch on this set, but I do not know how to formally prove it.
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
asked Feb 1 at 17:29
ryszard egginkryszard eggink
420110
420110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The function $zto z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
which means that we can write $f=z^{frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $frac{f}{g}$ would be holomorphic but it is impossible, since $z^{frac{2}{3}}$ is not holomorphic in infinity?
$endgroup$
– ryszard eggink
Feb 1 at 17:42
$begingroup$
@ryszardeggink or consider $hcolon zmapsto frac1{(1/z)+(1/z)^2}$ on $Bbb C-({0}cup [1,infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+ldots$. But what should the power sereis of $zmapsto frac1{f(1/z)}$ look like?
$endgroup$
– Hagen von Eitzen
Feb 1 at 19:26
add a comment |
$begingroup$
The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.
What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function $zto z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
which means that we can write $f=z^{frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $frac{f}{g}$ would be holomorphic but it is impossible, since $z^{frac{2}{3}}$ is not holomorphic in infinity?
$endgroup$
– ryszard eggink
Feb 1 at 17:42
$begingroup$
@ryszardeggink or consider $hcolon zmapsto frac1{(1/z)+(1/z)^2}$ on $Bbb C-({0}cup [1,infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+ldots$. But what should the power sereis of $zmapsto frac1{f(1/z)}$ look like?
$endgroup$
– Hagen von Eitzen
Feb 1 at 19:26
add a comment |
$begingroup$
The function $zto z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
which means that we can write $f=z^{frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $frac{f}{g}$ would be holomorphic but it is impossible, since $z^{frac{2}{3}}$ is not holomorphic in infinity?
$endgroup$
– ryszard eggink
Feb 1 at 17:42
$begingroup$
@ryszardeggink or consider $hcolon zmapsto frac1{(1/z)+(1/z)^2}$ on $Bbb C-({0}cup [1,infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+ldots$. But what should the power sereis of $zmapsto frac1{f(1/z)}$ look like?
$endgroup$
– Hagen von Eitzen
Feb 1 at 19:26
add a comment |
$begingroup$
The function $zto z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
The function $zto z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Feb 1 at 17:37
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
$begingroup$
which means that we can write $f=z^{frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $frac{f}{g}$ would be holomorphic but it is impossible, since $z^{frac{2}{3}}$ is not holomorphic in infinity?
$endgroup$
– ryszard eggink
Feb 1 at 17:42
$begingroup$
@ryszardeggink or consider $hcolon zmapsto frac1{(1/z)+(1/z)^2}$ on $Bbb C-({0}cup [1,infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+ldots$. But what should the power sereis of $zmapsto frac1{f(1/z)}$ look like?
$endgroup$
– Hagen von Eitzen
Feb 1 at 19:26
add a comment |
$begingroup$
which means that we can write $f=z^{frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $frac{f}{g}$ would be holomorphic but it is impossible, since $z^{frac{2}{3}}$ is not holomorphic in infinity?
$endgroup$
– ryszard eggink
Feb 1 at 17:42
$begingroup$
@ryszardeggink or consider $hcolon zmapsto frac1{(1/z)+(1/z)^2}$ on $Bbb C-({0}cup [1,infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+ldots$. But what should the power sereis of $zmapsto frac1{f(1/z)}$ look like?
$endgroup$
– Hagen von Eitzen
Feb 1 at 19:26
$begingroup$
which means that we can write $f=z^{frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $frac{f}{g}$ would be holomorphic but it is impossible, since $z^{frac{2}{3}}$ is not holomorphic in infinity?
$endgroup$
– ryszard eggink
Feb 1 at 17:42
$begingroup$
which means that we can write $f=z^{frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $frac{f}{g}$ would be holomorphic but it is impossible, since $z^{frac{2}{3}}$ is not holomorphic in infinity?
$endgroup$
– ryszard eggink
Feb 1 at 17:42
$begingroup$
@ryszardeggink or consider $hcolon zmapsto frac1{(1/z)+(1/z)^2}$ on $Bbb C-({0}cup [1,infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+ldots$. But what should the power sereis of $zmapsto frac1{f(1/z)}$ look like?
$endgroup$
– Hagen von Eitzen
Feb 1 at 19:26
$begingroup$
@ryszardeggink or consider $hcolon zmapsto frac1{(1/z)+(1/z)^2}$ on $Bbb C-({0}cup [1,infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+ldots$. But what should the power sereis of $zmapsto frac1{f(1/z)}$ look like?
$endgroup$
– Hagen von Eitzen
Feb 1 at 19:26
add a comment |
$begingroup$
The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.
What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
$endgroup$
add a comment |
$begingroup$
The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.
What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
$endgroup$
add a comment |
$begingroup$
The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.
What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
$endgroup$
The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.
What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
answered Feb 1 at 17:41
Henning MakholmHenning Makholm
243k17311555
243k17311555
add a comment |
add a comment |
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