Then choose the correct option regarding Picard theorem
$begingroup$
let $f$ be a non constant entire function and let $ E$ be the image of $ f$.
Then
choose the correct option
$1$. $E$ is an open set
$2$. $E cap {z : |z| < 1} $ is empty
$3.$$E cap mathbb{R}$ is non empty
$4.$ $E$ is a bounded set
I know that by open mapping theorem, only option $1)$ will be correct
im confused at other option
pliz help me....
complex-analysis
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
$endgroup$
add a comment |
$begingroup$
let $f$ be a non constant entire function and let $ E$ be the image of $ f$.
Then
choose the correct option
$1$. $E$ is an open set
$2$. $E cap {z : |z| < 1} $ is empty
$3.$$E cap mathbb{R}$ is non empty
$4.$ $E$ is a bounded set
I know that by open mapping theorem, only option $1)$ will be correct
im confused at other option
pliz help me....
complex-analysis
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
$endgroup$
$begingroup$
Are we considering "points at infinity" to be a part of the complex plane? Either way, $E$, if bounded, is bounded only by directed infinities. I don't know about 2) and 3). Wouldn't you need to know more about the function to say if these were true or false?
$endgroup$
– R. Burton
Dec 13 '18 at 22:38
add a comment |
$begingroup$
let $f$ be a non constant entire function and let $ E$ be the image of $ f$.
Then
choose the correct option
$1$. $E$ is an open set
$2$. $E cap {z : |z| < 1} $ is empty
$3.$$E cap mathbb{R}$ is non empty
$4.$ $E$ is a bounded set
I know that by open mapping theorem, only option $1)$ will be correct
im confused at other option
pliz help me....
complex-analysis
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
$endgroup$
let $f$ be a non constant entire function and let $ E$ be the image of $ f$.
Then
choose the correct option
$1$. $E$ is an open set
$2$. $E cap {z : |z| < 1} $ is empty
$3.$$E cap mathbb{R}$ is non empty
$4.$ $E$ is a bounded set
I know that by open mapping theorem, only option $1)$ will be correct
im confused at other option
pliz help me....
complex-analysis
complex-analysis
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
edited Dec 14 '18 at 6:22
jasmine
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
asked Dec 13 '18 at 22:10
jasminejasmine
1,766417
1,766417
$begingroup$
Are we considering "points at infinity" to be a part of the complex plane? Either way, $E$, if bounded, is bounded only by directed infinities. I don't know about 2) and 3). Wouldn't you need to know more about the function to say if these were true or false?
$endgroup$
– R. Burton
Dec 13 '18 at 22:38
add a comment |
$begingroup$
Are we considering "points at infinity" to be a part of the complex plane? Either way, $E$, if bounded, is bounded only by directed infinities. I don't know about 2) and 3). Wouldn't you need to know more about the function to say if these were true or false?
$endgroup$
– R. Burton
Dec 13 '18 at 22:38
$begingroup$
Are we considering "points at infinity" to be a part of the complex plane? Either way, $E$, if bounded, is bounded only by directed infinities. I don't know about 2) and 3). Wouldn't you need to know more about the function to say if these were true or false?
$endgroup$
– R. Burton
Dec 13 '18 at 22:38
$begingroup$
Are we considering "points at infinity" to be a part of the complex plane? Either way, $E$, if bounded, is bounded only by directed infinities. I don't know about 2) and 3). Wouldn't you need to know more about the function to say if these were true or false?
$endgroup$
– R. Burton
Dec 13 '18 at 22:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
3) is also true. Picard's Theorem says if $f$ is a non-constant entire function then $E=mathbb C$ or $E=mathbb C setminus {c}$ for some complex number $c$. So $E$ must contain all real numbers except possibly one. 2) and 4) are false: take $f(z)=z$ for 2) and use Louivile's Theorem for 4). Alternative argument for 3): If $E cap mathbb R =emptyset$ then $E=E_1 cup E_2$ where $E_1={z: Im, z>0}$ and $E_1={z: Im, z<0}$. This gives a contradiction to the fact that $E$ is connected.
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
add a comment |
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$begingroup$
3) is also true. Picard's Theorem says if $f$ is a non-constant entire function then $E=mathbb C$ or $E=mathbb C setminus {c}$ for some complex number $c$. So $E$ must contain all real numbers except possibly one. 2) and 4) are false: take $f(z)=z$ for 2) and use Louivile's Theorem for 4). Alternative argument for 3): If $E cap mathbb R =emptyset$ then $E=E_1 cup E_2$ where $E_1={z: Im, z>0}$ and $E_1={z: Im, z<0}$. This gives a contradiction to the fact that $E$ is connected.
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
add a comment |
$begingroup$
3) is also true. Picard's Theorem says if $f$ is a non-constant entire function then $E=mathbb C$ or $E=mathbb C setminus {c}$ for some complex number $c$. So $E$ must contain all real numbers except possibly one. 2) and 4) are false: take $f(z)=z$ for 2) and use Louivile's Theorem for 4). Alternative argument for 3): If $E cap mathbb R =emptyset$ then $E=E_1 cup E_2$ where $E_1={z: Im, z>0}$ and $E_1={z: Im, z<0}$. This gives a contradiction to the fact that $E$ is connected.
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
add a comment |
$begingroup$
3) is also true. Picard's Theorem says if $f$ is a non-constant entire function then $E=mathbb C$ or $E=mathbb C setminus {c}$ for some complex number $c$. So $E$ must contain all real numbers except possibly one. 2) and 4) are false: take $f(z)=z$ for 2) and use Louivile's Theorem for 4). Alternative argument for 3): If $E cap mathbb R =emptyset$ then $E=E_1 cup E_2$ where $E_1={z: Im, z>0}$ and $E_1={z: Im, z<0}$. This gives a contradiction to the fact that $E$ is connected.
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
$endgroup$
3) is also true. Picard's Theorem says if $f$ is a non-constant entire function then $E=mathbb C$ or $E=mathbb C setminus {c}$ for some complex number $c$. So $E$ must contain all real numbers except possibly one. 2) and 4) are false: take $f(z)=z$ for 2) and use Louivile's Theorem for 4). Alternative argument for 3): If $E cap mathbb R =emptyset$ then $E=E_1 cup E_2$ where $E_1={z: Im, z>0}$ and $E_1={z: Im, z<0}$. This gives a contradiction to the fact that $E$ is connected.
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
answered Dec 13 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
60.5k42161
60.5k42161
add a comment |
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$begingroup$
Are we considering "points at infinity" to be a part of the complex plane? Either way, $E$, if bounded, is bounded only by directed infinities. I don't know about 2) and 3). Wouldn't you need to know more about the function to say if these were true or false?
$endgroup$
– R. Burton
Dec 13 '18 at 22:38