consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$.choose the correct...
$begingroup$
consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to
choose the correct options
$1.$Circle
$2.$ a line
$3.$ a line passing through the origin
$4.$ a line not passing through the origin
My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$
that $f(z) = costheta - isintheta$
After that im not able to proceed further pliz help me
any hints/solution will be appreciated thanks u
complex-analysis complex-numbers
$endgroup$
add a comment |
$begingroup$
consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to
choose the correct options
$1.$Circle
$2.$ a line
$3.$ a line passing through the origin
$4.$ a line not passing through the origin
My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$
that $f(z) = costheta - isintheta$
After that im not able to proceed further pliz help me
any hints/solution will be appreciated thanks u
complex-analysis complex-numbers
$endgroup$
add a comment |
$begingroup$
consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to
choose the correct options
$1.$Circle
$2.$ a line
$3.$ a line passing through the origin
$4.$ a line not passing through the origin
My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$
that $f(z) = costheta - isintheta$
After that im not able to proceed further pliz help me
any hints/solution will be appreciated thanks u
complex-analysis complex-numbers
$endgroup$
consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to
choose the correct options
$1.$Circle
$2.$ a line
$3.$ a line passing through the origin
$4.$ a line not passing through the origin
My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$
that $f(z) = costheta - isintheta$
After that im not able to proceed further pliz help me
any hints/solution will be appreciated thanks u
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Dec 13 '18 at 12:30
José Carlos Santos
164k22131235
164k22131235
asked Dec 13 '18 at 12:23
jasminejasmine
1,794418
1,794418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.
$endgroup$
$begingroup$
im not getting @Jose carlos sir ....can u elaborate more,,,,,
$endgroup$
– jasmine
Dec 13 '18 at 12:35
1
$begingroup$
Don't you get that your circle doesn't pass through the origin?
$endgroup$
– José Carlos Santos
Dec 13 '18 at 12:38
$begingroup$
my confusion is that why its is not map circle to circle ?
$endgroup$
– jasmine
Dec 13 '18 at 12:47
1
$begingroup$
It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 13:00
1
$begingroup$
It matches my eyes, doesn't it?! ;-)
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:24
|
show 2 more comments
$begingroup$
You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.
Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.
$endgroup$
$begingroup$
im not getting @Jose carlos sir ....can u elaborate more,,,,,
$endgroup$
– jasmine
Dec 13 '18 at 12:35
1
$begingroup$
Don't you get that your circle doesn't pass through the origin?
$endgroup$
– José Carlos Santos
Dec 13 '18 at 12:38
$begingroup$
my confusion is that why its is not map circle to circle ?
$endgroup$
– jasmine
Dec 13 '18 at 12:47
1
$begingroup$
It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 13:00
1
$begingroup$
It matches my eyes, doesn't it?! ;-)
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:24
|
show 2 more comments
$begingroup$
What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.
$endgroup$
$begingroup$
im not getting @Jose carlos sir ....can u elaborate more,,,,,
$endgroup$
– jasmine
Dec 13 '18 at 12:35
1
$begingroup$
Don't you get that your circle doesn't pass through the origin?
$endgroup$
– José Carlos Santos
Dec 13 '18 at 12:38
$begingroup$
my confusion is that why its is not map circle to circle ?
$endgroup$
– jasmine
Dec 13 '18 at 12:47
1
$begingroup$
It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 13:00
1
$begingroup$
It matches my eyes, doesn't it?! ;-)
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:24
|
show 2 more comments
$begingroup$
What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.
$endgroup$
What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.
answered Dec 13 '18 at 12:29
José Carlos SantosJosé Carlos Santos
164k22131235
164k22131235
$begingroup$
im not getting @Jose carlos sir ....can u elaborate more,,,,,
$endgroup$
– jasmine
Dec 13 '18 at 12:35
1
$begingroup$
Don't you get that your circle doesn't pass through the origin?
$endgroup$
– José Carlos Santos
Dec 13 '18 at 12:38
$begingroup$
my confusion is that why its is not map circle to circle ?
$endgroup$
– jasmine
Dec 13 '18 at 12:47
1
$begingroup$
It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 13:00
1
$begingroup$
It matches my eyes, doesn't it?! ;-)
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:24
|
show 2 more comments
$begingroup$
im not getting @Jose carlos sir ....can u elaborate more,,,,,
$endgroup$
– jasmine
Dec 13 '18 at 12:35
1
$begingroup$
Don't you get that your circle doesn't pass through the origin?
$endgroup$
– José Carlos Santos
Dec 13 '18 at 12:38
$begingroup$
my confusion is that why its is not map circle to circle ?
$endgroup$
– jasmine
Dec 13 '18 at 12:47
1
$begingroup$
It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 13:00
1
$begingroup$
It matches my eyes, doesn't it?! ;-)
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:24
$begingroup$
im not getting @Jose carlos sir ....can u elaborate more,,,,,
$endgroup$
– jasmine
Dec 13 '18 at 12:35
$begingroup$
im not getting @Jose carlos sir ....can u elaborate more,,,,,
$endgroup$
– jasmine
Dec 13 '18 at 12:35
1
1
$begingroup$
Don't you get that your circle doesn't pass through the origin?
$endgroup$
– José Carlos Santos
Dec 13 '18 at 12:38
$begingroup$
Don't you get that your circle doesn't pass through the origin?
$endgroup$
– José Carlos Santos
Dec 13 '18 at 12:38
$begingroup$
my confusion is that why its is not map circle to circle ?
$endgroup$
– jasmine
Dec 13 '18 at 12:47
$begingroup$
my confusion is that why its is not map circle to circle ?
$endgroup$
– jasmine
Dec 13 '18 at 12:47
1
1
$begingroup$
It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 13:00
$begingroup$
It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
$endgroup$
– José Carlos Santos
Dec 13 '18 at 13:00
1
1
$begingroup$
It matches my eyes, doesn't it?! ;-)
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:24
$begingroup$
It matches my eyes, doesn't it?! ;-)
$endgroup$
– José Carlos Santos
Dec 13 '18 at 14:24
|
show 2 more comments
$begingroup$
You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.
Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.
$endgroup$
add a comment |
$begingroup$
You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.
Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.
$endgroup$
add a comment |
$begingroup$
You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.
Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.
$endgroup$
You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.
Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.
answered Dec 13 '18 at 12:45
FredFred
47.1k1848
47.1k1848
add a comment |
add a comment |
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