Casorati-Weierstraß theorem - dense
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I try to understand the Casorati-Weierstraß theorem. But I don't understand when a picture is dense in C. $e^{1/z}$ is dense, $1/z$ isn't. But why?
Thanks.
density-function
edited Dec 3 '18 at 22:28
Andrews
5331318
asked Dec 3 '18 at 21:42
malilinimalilini
44
$endgroup$
add a comment |
$begingroup$
I try to understand the Casorati-Weierstraß theorem. But I don't understand when a picture is dense in C. $e^{1/z}$ is dense, $1/z$ isn't. But why?
Thanks.
density-function
edited Dec 3 '18 at 22:28
Andrews
5331318
asked Dec 3 '18 at 21:42
malilinimalilini
44
$endgroup$
add a comment |
$begingroup$
I try to understand the Casorati-Weierstraß theorem. But I don't understand when a picture is dense in C. $e^{1/z}$ is dense, $1/z$ isn't. But why?
Thanks.
density-function
edited Dec 3 '18 at 22:28
Andrews
5331318
asked Dec 3 '18 at 21:42
malilinimalilini
44
$endgroup$
I try to understand the Casorati-Weierstraß theorem. But I don't understand when a picture is dense in C. $e^{1/z}$ is dense, $1/z$ isn't. But why?
Thanks.
density-function
density-function
edited Dec 3 '18 at 22:28
Andrews
5331318
asked Dec 3 '18 at 21:42
malilinimalilini
44
edited Dec 3 '18 at 22:28
Andrews
5331318
asked Dec 3 '18 at 21:42
malilinimalilini
44
edited Dec 3 '18 at 22:28
Andrews
5331318
edited Dec 3 '18 at 22:28
Andrews
5331318
edited Dec 3 '18 at 22:28
Andrews
5331318
5331318
asked Dec 3 '18 at 21:42
malilinimalilini
44
asked Dec 3 '18 at 21:42
malilinimalilini
44
asked Dec 3 '18 at 21:42
malilinimalilini
44
44
add a comment |
add a comment |
1 Answer
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No function is (or isn't) dense. That theorem says that if $V$ is a neighborhood of an essential singularity at $a$ of a function $f$, then $f(Vsetminus{a})$ is dense. Well, $0$ is not an essential singularity of $frac1z$ (it's a simple pole), but it is an essential singularity of $expleft(frac1zright)$.
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
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add a comment |
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1 Answer
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$begingroup$
No function is (or isn't) dense. That theorem says that if $V$ is a neighborhood of an essential singularity at $a$ of a function $f$, then $f(Vsetminus{a})$ is dense. Well, $0$ is not an essential singularity of $frac1z$ (it's a simple pole), but it is an essential singularity of $expleft(frac1zright)$.
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
No function is (or isn't) dense. That theorem says that if $V$ is a neighborhood of an essential singularity at $a$ of a function $f$, then $f(Vsetminus{a})$ is dense. Well, $0$ is not an essential singularity of $frac1z$ (it's a simple pole), but it is an essential singularity of $expleft(frac1zright)$.
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
No function is (or isn't) dense. That theorem says that if $V$ is a neighborhood of an essential singularity at $a$ of a function $f$, then $f(Vsetminus{a})$ is dense. Well, $0$ is not an essential singularity of $frac1z$ (it's a simple pole), but it is an essential singularity of $expleft(frac1zright)$.
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
No function is (or isn't) dense. That theorem says that if $V$ is a neighborhood of an essential singularity at $a$ of a function $f$, then $f(Vsetminus{a})$ is dense. Well, $0$ is not an essential singularity of $frac1z$ (it's a simple pole), but it is an essential singularity of $expleft(frac1zright)$.
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 3 '18 at 21:48
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
add a comment |
add a comment |
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