Application of Casorati WeierstrassLooser Conditions on Casorati-Weierstrasstype of singularityHow to find...
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Application of Casorati Weierstrass
Looser Conditions on Casorati-Weierstrasstype of singularityHow to find the Casorati-Weierstrass' Theorem ? Can we find the phenomenon from a classical function: $expleft(frac{1}{z}right)$?A problem related to the Casorati–Weierstrass theoremWhen is the converse of Casorati–Weierstrass false?Proof of Casorati-Weierstrass?Meremorphic function on $mathbb{D}setminus{0}$ has dense image?Are the following theorems' converses also true?Proving Riemann's Theorem through Casorati- WeierstrassProof of Casorati-Weierstrass
$begingroup$
I have the following Theorem in my lecture notes:
If $f:D to mathbb{C}$ is holomorphic on $D-{z_0}$ where $D$ is open and connected and $z_0$ is an isolated singularity of f, then: $f$ has a pole at $z_0$ iff $lim_{z to z_0} |f(z)| = infty$.
One implication is easy but I have issues proving the converse. I believe it has to do with Casorati-Weierstrass Theorem. That is, if $z_0$ was an essential singularity then for all $varepsilon >0$, $f(B(z_0,varepsilon)-{z_0})$ would be dense in $mathbb{C}$, but I don't see how to get a contradiction from there. On top of that, I don't even see how $e^{1/z}$ isn't a contradiction to the theorem in the first place. Am I missing something really obvious, or do I miss some assumptions? Thank you for your help.
complex-analysis holomorphic-functions singularity
asked 15 hours ago
ChloChlo
704
$endgroup$
add a comment |
$begingroup$
I have the following Theorem in my lecture notes:
If $f:D to mathbb{C}$ is holomorphic on $D-{z_0}$ where $D$ is open and connected and $z_0$ is an isolated singularity of f, then: $f$ has a pole at $z_0$ iff $lim_{z to z_0} |f(z)| = infty$.
One implication is easy but I have issues proving the converse. I believe it has to do with Casorati-Weierstrass Theorem. That is, if $z_0$ was an essential singularity then for all $varepsilon >0$, $f(B(z_0,varepsilon)-{z_0})$ would be dense in $mathbb{C}$, but I don't see how to get a contradiction from there. On top of that, I don't even see how $e^{1/z}$ isn't a contradiction to the theorem in the first place. Am I missing something really obvious, or do I miss some assumptions? Thank you for your help.
complex-analysis holomorphic-functions singularity
asked 15 hours ago
ChloChlo
704
$endgroup$
$begingroup$
??? Where I come from $lim_{zto z_0}|f(z)|=infty$ is the definition of "$f$ has a pole"! What definition are you using?
$endgroup$
– David C. Ullrich
14 hours ago
add a comment |
$begingroup$
I have the following Theorem in my lecture notes:
If $f:D to mathbb{C}$ is holomorphic on $D-{z_0}$ where $D$ is open and connected and $z_0$ is an isolated singularity of f, then: $f$ has a pole at $z_0$ iff $lim_{z to z_0} |f(z)| = infty$.
One implication is easy but I have issues proving the converse. I believe it has to do with Casorati-Weierstrass Theorem. That is, if $z_0$ was an essential singularity then for all $varepsilon >0$, $f(B(z_0,varepsilon)-{z_0})$ would be dense in $mathbb{C}$, but I don't see how to get a contradiction from there. On top of that, I don't even see how $e^{1/z}$ isn't a contradiction to the theorem in the first place. Am I missing something really obvious, or do I miss some assumptions? Thank you for your help.
complex-analysis holomorphic-functions singularity
asked 15 hours ago
ChloChlo
704
$endgroup$
I have the following Theorem in my lecture notes:
If $f:D to mathbb{C}$ is holomorphic on $D-{z_0}$ where $D$ is open and connected and $z_0$ is an isolated singularity of f, then: $f$ has a pole at $z_0$ iff $lim_{z to z_0} |f(z)| = infty$.
One implication is easy but I have issues proving the converse. I believe it has to do with Casorati-Weierstrass Theorem. That is, if $z_0$ was an essential singularity then for all $varepsilon >0$, $f(B(z_0,varepsilon)-{z_0})$ would be dense in $mathbb{C}$, but I don't see how to get a contradiction from there. On top of that, I don't even see how $e^{1/z}$ isn't a contradiction to the theorem in the first place. Am I missing something really obvious, or do I miss some assumptions? Thank you for your help.
complex-analysis holomorphic-functions singularity
complex-analysis holomorphic-functions singularity
asked 15 hours ago
ChloChlo
704
asked 15 hours ago
ChloChlo
704
asked 15 hours ago
ChloChlo
704
asked 15 hours ago
ChloChlo
704
asked 15 hours ago
ChloChlo
704
704
$begingroup$
??? Where I come from $lim_{zto z_0}|f(z)|=infty$ is the definition of "$f$ has a pole"! What definition are you using?
$endgroup$
– David C. Ullrich
14 hours ago
add a comment |
$begingroup$
??? Where I come from $lim_{zto z_0}|f(z)|=infty$ is the definition of "$f$ has a pole"! What definition are you using?
$endgroup$
– David C. Ullrich
14 hours ago
$begingroup$
??? Where I come from $lim_{zto z_0}|f(z)|=infty$ is the definition of "$f$ has a pole"! What definition are you using?
$endgroup$
– David C. Ullrich
14 hours ago
$begingroup$
??? Where I come from $lim_{zto z_0}|f(z)|=infty$ is the definition of "$f$ has a pole"! What definition are you using?
$endgroup$
– David C. Ullrich
14 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $|f(z)| to infty$ as $z to z_0$ then $f(B(z_0,epsilon)setminus {z_0})$ does not intersect the open unit disk if $epsilon$ is small enough. Since any dense set has to intersect this ball it follows that $z_0$ is not an essential singularity.
The functio $e^{1/z}$ does not have the property $|f(z)| to infty$ as $z to 0$. For example this function has modulus $1$ when $z =frac 1 {2npi i}$.
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
I see thank you very much!
$endgroup$
– Chlo
15 hours ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
If $|f(z)| to infty$ as $z to z_0$ then $f(B(z_0,epsilon)setminus {z_0})$ does not intersect the open unit disk if $epsilon$ is small enough. Since any dense set has to intersect this ball it follows that $z_0$ is not an essential singularity.
The functio $e^{1/z}$ does not have the property $|f(z)| to infty$ as $z to 0$. For example this function has modulus $1$ when $z =frac 1 {2npi i}$.
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
I see thank you very much!
$endgroup$
– Chlo
15 hours ago
add a comment |
$begingroup$
If $|f(z)| to infty$ as $z to z_0$ then $f(B(z_0,epsilon)setminus {z_0})$ does not intersect the open unit disk if $epsilon$ is small enough. Since any dense set has to intersect this ball it follows that $z_0$ is not an essential singularity.
The functio $e^{1/z}$ does not have the property $|f(z)| to infty$ as $z to 0$. For example this function has modulus $1$ when $z =frac 1 {2npi i}$.
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
$begingroup$
I see thank you very much!
$endgroup$
– Chlo
15 hours ago
add a comment |
$begingroup$
If $|f(z)| to infty$ as $z to z_0$ then $f(B(z_0,epsilon)setminus {z_0})$ does not intersect the open unit disk if $epsilon$ is small enough. Since any dense set has to intersect this ball it follows that $z_0$ is not an essential singularity.
The functio $e^{1/z}$ does not have the property $|f(z)| to infty$ as $z to 0$. For example this function has modulus $1$ when $z =frac 1 {2npi i}$.
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
$endgroup$
If $|f(z)| to infty$ as $z to z_0$ then $f(B(z_0,epsilon)setminus {z_0})$ does not intersect the open unit disk if $epsilon$ is small enough. Since any dense set has to intersect this ball it follows that $z_0$ is not an essential singularity.
The functio $e^{1/z}$ does not have the property $|f(z)| to infty$ as $z to 0$. For example this function has modulus $1$ when $z =frac 1 {2npi i}$.
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
answered 15 hours ago
Kavi Rama MurthyKavi Rama Murthy
65.4k42766
65.4k42766
$begingroup$
I see thank you very much!
$endgroup$
– Chlo
15 hours ago
add a comment |
$begingroup$
I see thank you very much!
$endgroup$
– Chlo
15 hours ago
$begingroup$
I see thank you very much!
$endgroup$
– Chlo
15 hours ago
$begingroup$
I see thank you very much!
$endgroup$
– Chlo
15 hours ago
add a comment |
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$begingroup$
??? Where I come from $lim_{zto z_0}|f(z)|=infty$ is the definition of "$f$ has a pole"! What definition are you using?
$endgroup$
– David C. Ullrich
14 hours ago