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Limit at infinity for complex functions

Whats an extended complex plane?continuity at infinityAsymmetry in the complex plane.No essential singularity at infinity implies polynomialmeromorphic functions on the extended complex planeOne side half twist mobius band are one to one transformation on extended complex planeIn complex variables, what is the behavior of e^z as x,y->positve or negative infinity?'Extended complex plane and infinityProving a generalization of Liouville's TheoremProblems to understand complex infinity limits.

3

1

$begingroup$

Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$. Show that $f(mathbb{C})=mathbb{C}$.

First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$, but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$?

complex-analysis

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asked 39 mins ago

Fluffy SkyeFluffy Skye

1459

$endgroup$

add a comment | 

3

1

$begingroup$

Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$. Show that $f(mathbb{C})=mathbb{C}$.

First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$, but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$?

complex-analysis

share|cite|improve this question

asked 39 mins ago

Fluffy SkyeFluffy Skye

1459

$endgroup$

add a comment | 

$begingroup$

Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$. Show that $f(mathbb{C})=mathbb{C}$.

First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$, but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$?

complex-analysis

share|cite|improve this question

asked 39 mins ago

Fluffy SkyeFluffy Skye

1459

$endgroup$

share|cite|improve this question

asked 39 mins ago

Fluffy SkyeFluffy Skye

1459

share|cite|improve this question

asked 39 mins ago

Fluffy SkyeFluffy Skye

1459

asked 39 mins ago

Fluffy SkyeFluffy Skye

1459

asked 39 mins ago

Fluffy SkyeFluffy Skye

1459

3 Answers
3

active

oldest

votes

5

$begingroup$

The niceties of the one-point compactification of $Bbb C$ aside, consider:

$f(z) ne w, ; forall z in Bbb C; tag 1$

$f(z) - w ne 0, ; forall z in Bbb C; tag 2$

$(f(z) - w)^{-1}$ is then entire; since

$f(z) to infty ; text{as} ; z to infty tag 3$

$(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.

share|cite|improve this answer

answered 20 mins ago

Robert LewisRobert Lewis

47.1k23067

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice answer! +1 vote.
  $endgroup$
  – Kavi Rama Murthy
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: quite a complement, considering the source! Cheers!
  $endgroup$
  – Robert Lewis
  18 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
$
g(z)=f(frac1{z})
$ for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
$$
g(z)=sum_{k=-n}^infty a_kz^k.
$$ (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$

share|cite|improve this answer

edited 2 mins ago

answered 27 mins ago

SongSong

14.9k1635

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
  $endgroup$
  – nicomezi
  13 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
  $endgroup$
  – nicomezi
  8 mins ago
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.

Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by

$g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.

Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.

share|cite|improve this answer

answered 11 mins ago

FredFred

46.9k1848

$endgroup$

add a comment | 

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5

$begingroup$

The niceties of the one-point compactification of $Bbb C$ aside, consider:

$f(z) ne w, ; forall z in Bbb C; tag 1$

$f(z) - w ne 0, ; forall z in Bbb C; tag 2$

$(f(z) - w)^{-1}$ is then entire; since

$f(z) to infty ; text{as} ; z to infty tag 3$

$(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.

share|cite|improve this answer

answered 20 mins ago

Robert LewisRobert Lewis

47.1k23067

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice answer! +1 vote.
  $endgroup$
  – Kavi Rama Murthy
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: quite a complement, considering the source! Cheers!
  $endgroup$
  – Robert Lewis
  18 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

The niceties of the one-point compactification of $Bbb C$ aside, consider:

$f(z) ne w, ; forall z in Bbb C; tag 1$

$f(z) - w ne 0, ; forall z in Bbb C; tag 2$

$(f(z) - w)^{-1}$ is then entire; since

$f(z) to infty ; text{as} ; z to infty tag 3$

$(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.

share|cite|improve this answer

answered 20 mins ago

Robert LewisRobert Lewis

47.1k23067

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice answer! +1 vote.
  $endgroup$
  – Kavi Rama Murthy
  19 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviRamaMurthy: quite a complement, considering the source! Cheers!
  $endgroup$
  – Robert Lewis
  18 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

The niceties of the one-point compactification of $Bbb C$ aside, consider:

$f(z) ne w, ; forall z in Bbb C; tag 1$

$f(z) - w ne 0, ; forall z in Bbb C; tag 2$

$(f(z) - w)^{-1}$ is then entire; since

$f(z) to infty ; text{as} ; z to infty tag 3$

$(f(z) - w)^{-1}$ is bounded; hence by Liouville's theorem, constant; but then we cannot have (3); thus (1) fails.

share|cite|improve this answer

answered 20 mins ago

Robert LewisRobert Lewis

47.1k23067

$endgroup$

share|cite|improve this answer

answered 20 mins ago

Robert LewisRobert Lewis

47.1k23067

answered 20 mins ago

Robert LewisRobert Lewis

47.1k23067

answered 20 mins ago

Robert LewisRobert Lewis

47.1k23067

1

$begingroup$

The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
$
g(z)=f(frac1{z})
$ for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
$$
g(z)=sum_{k=-n}^infty a_kz^k.
$$ (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$

share|cite|improve this answer

edited 2 mins ago

answered 27 mins ago

SongSong

14.9k1635

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
  $endgroup$
  – nicomezi
  13 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
  $endgroup$
  – nicomezi
  8 mins ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
$
g(z)=f(frac1{z})
$ for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
$$
g(z)=sum_{k=-n}^infty a_kz^k.
$$ (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$

share|cite|improve this answer

edited 2 mins ago

answered 27 mins ago

SongSong

14.9k1635

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The statement right after $lim_{z to 0} |g(z)| = infty$ is not that clear to me. It would require more arguments.
  $endgroup$
  – nicomezi
  13 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you but I was rather talking about what is after that limit (the fact that $g$ has an $n$-th pole). I am not saying it is false. Since we have the limit on $f$ at $infty$ it cannot be an essential singularity. But it seems quite hard (compared to others answers) to write this rigorously.
  $endgroup$
  – nicomezi
  8 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

The statement $lim_{ztoinfty}f(z)=infty$ means $|f(z)|to infty$ as $|z|toinfty$. It can be seen as convergence to $infty inBbb C_infty$ where $Bbb C_infty$ is a one-point compactification of $Bbb C$. In fact, this property implies that $f(z)=p(z)$ for some non-constant polynomial $p$. By fundamental theorem of algebra, it follows $f(Bbb C)=Bbb C$. To see this, let
$
g(z)=f(frac1{z})
$ for $zne 0$. Since $lim_{zto 0}|g(z)|=infty$ (which is true because as $zto 0$, $1/z to infty$ and $g(z)=f(1/z)toinfty$ by the assumption), it follows that $g$ has an $n$-th pole at $0$, i.e. for some $nge 1$, $a_{-n}ne 0$ and for all $zne 0$,
$$
g(z)=sum_{k=-n}^infty a_kz^k.
$$ (Here, $g$ having a pole is a classical result. Since $1/g$ tends to $0$ as $zto 0$, it has a removable singularity at $0$, which is also an $n$-th zero. It follows $1/g(z)=z^nh(z)$ for some $hne 0$ on a neighborhood of $0$. So $g(z) = z^{-n}frac1{h(z)}$ has an $n$-th pole.) Thus $f(z)=g(frac1{z})=sum_{k=0}^n a_{-k}z^k +sum_{i>0}a_{i}z^{-i}$, and since $f$ is entire, it follows $$f(z)=sum_{k=0}^n a_{-k}z^k=p(z).$$

share|cite|improve this answer

edited 2 mins ago

answered 27 mins ago

SongSong

14.9k1635

$endgroup$

share|cite|improve this answer

edited 2 mins ago

answered 27 mins ago

SongSong

14.9k1635

answered 27 mins ago

SongSong

14.9k1635

answered 27 mins ago

SongSong

14.9k1635

0

$begingroup$

Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.

Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by

$g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.

Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.

share|cite|improve this answer

answered 11 mins ago

FredFred

46.9k1848

$endgroup$

add a comment | 

0

$begingroup$

Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.

Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by

$g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.

Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.

share|cite|improve this answer

answered 11 mins ago

FredFred

46.9k1848

$endgroup$

add a comment | 

$begingroup$

Let $g(z):=f(1/z)$ for $z ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) to infty$ as $z to 0$, $g$ has a pole at $0$.

Let $f(z)=sum_{n=0}^{infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is given by

$g(z)=sum_{n=0}^{infty}a_n frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.

Hence $f$ is a non- constant poynomial. Now use the fundamental theorem of algebra.

share|cite|improve this answer

answered 11 mins ago

FredFred

46.9k1848

$endgroup$

share|cite|improve this answer

answered 11 mins ago

FredFred

46.9k1848

answered 11 mins ago

FredFred

46.9k1848

answered 11 mins ago

FredFred

46.9k1848