Image of unit disk under inversion - Mobius transformations
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I am trying to find the image of ${z in mathbb{C} : |z| < 1}$ under $f(z) = frac{1}{z}$
Let $$w = frac{1}{z} Rightarrow z = frac{1}{w} = frac{1}{u+iv} = frac{u-iv}{u^2 + v^2}$$
if $w = u +iv$.
Then considering the boundary first, $$|z| =1 Rightarrow Big|frac{u}{u^2+v^2} -ifrac{v}{u^2+v^2}Big| =1 $$ Using Pythagoras gives $$left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1$$ but this is not the equation of a unit circle because I have this factor of $(u^2 + v^2)^2$ lying about, which is what the answer should be...
What's happening?
complex-analysis mobius-transformation
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
$endgroup$
add a comment |
$begingroup$
I am trying to find the image of ${z in mathbb{C} : |z| < 1}$ under $f(z) = frac{1}{z}$
Let $$w = frac{1}{z} Rightarrow z = frac{1}{w} = frac{1}{u+iv} = frac{u-iv}{u^2 + v^2}$$
if $w = u +iv$.
Then considering the boundary first, $$|z| =1 Rightarrow Big|frac{u}{u^2+v^2} -ifrac{v}{u^2+v^2}Big| =1 $$ Using Pythagoras gives $$left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1$$ but this is not the equation of a unit circle because I have this factor of $(u^2 + v^2)^2$ lying about, which is what the answer should be...
What's happening?
complex-analysis mobius-transformation
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
$endgroup$
add a comment |
$begingroup$
I am trying to find the image of ${z in mathbb{C} : |z| < 1}$ under $f(z) = frac{1}{z}$
Let $$w = frac{1}{z} Rightarrow z = frac{1}{w} = frac{1}{u+iv} = frac{u-iv}{u^2 + v^2}$$
if $w = u +iv$.
Then considering the boundary first, $$|z| =1 Rightarrow Big|frac{u}{u^2+v^2} -ifrac{v}{u^2+v^2}Big| =1 $$ Using Pythagoras gives $$left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1$$ but this is not the equation of a unit circle because I have this factor of $(u^2 + v^2)^2$ lying about, which is what the answer should be...
What's happening?
complex-analysis mobius-transformation
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
$endgroup$
I am trying to find the image of ${z in mathbb{C} : |z| < 1}$ under $f(z) = frac{1}{z}$
Let $$w = frac{1}{z} Rightarrow z = frac{1}{w} = frac{1}{u+iv} = frac{u-iv}{u^2 + v^2}$$
if $w = u +iv$.
Then considering the boundary first, $$|z| =1 Rightarrow Big|frac{u}{u^2+v^2} -ifrac{v}{u^2+v^2}Big| =1 $$ Using Pythagoras gives $$left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1$$ but this is not the equation of a unit circle because I have this factor of $(u^2 + v^2)^2$ lying about, which is what the answer should be...
What's happening?
complex-analysis mobius-transformation
complex-analysis mobius-transformation
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
asked Jan 22 at 14:06
PhysicsMathsLovePhysicsMathsLove
1,208414
1,208414
add a comment |
add a comment |
1 Answer
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Of course, it makes no sense to talk about the image of $0$ under $frac1z$, but the image of ${zinmathbb{C},|,lvert zrvert<1}setminus{0}$ is ${zinmathbb{C},|,lvert zrvert>1}$.
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Why does my method not yield the unit circle as the boundary?
$endgroup$
– PhysicsMathsLove
Jan 22 at 14:11
1
$begingroup$
begin{align}left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1&ifffrac{u^2+v^2}{(u^2+v^2)^2}=1\&ifffrac1{u^2+v^2}=1\&iff u^2+v^2=1.end{align}
$endgroup$
– José Carlos Santos
Jan 22 at 14:17
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
Of course, it makes no sense to talk about the image of $0$ under $frac1z$, but the image of ${zinmathbb{C},|,lvert zrvert<1}setminus{0}$ is ${zinmathbb{C},|,lvert zrvert>1}$.
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Why does my method not yield the unit circle as the boundary?
$endgroup$
– PhysicsMathsLove
Jan 22 at 14:11
1
$begingroup$
begin{align}left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1&ifffrac{u^2+v^2}{(u^2+v^2)^2}=1\&ifffrac1{u^2+v^2}=1\&iff u^2+v^2=1.end{align}
$endgroup$
– José Carlos Santos
Jan 22 at 14:17
add a comment |
$begingroup$
Of course, it makes no sense to talk about the image of $0$ under $frac1z$, but the image of ${zinmathbb{C},|,lvert zrvert<1}setminus{0}$ is ${zinmathbb{C},|,lvert zrvert>1}$.
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
Why does my method not yield the unit circle as the boundary?
$endgroup$
– PhysicsMathsLove
Jan 22 at 14:11
1
$begingroup$
begin{align}left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1&ifffrac{u^2+v^2}{(u^2+v^2)^2}=1\&ifffrac1{u^2+v^2}=1\&iff u^2+v^2=1.end{align}
$endgroup$
– José Carlos Santos
Jan 22 at 14:17
add a comment |
$begingroup$
Of course, it makes no sense to talk about the image of $0$ under $frac1z$, but the image of ${zinmathbb{C},|,lvert zrvert<1}setminus{0}$ is ${zinmathbb{C},|,lvert zrvert>1}$.
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
Of course, it makes no sense to talk about the image of $0$ under $frac1z$, but the image of ${zinmathbb{C},|,lvert zrvert<1}setminus{0}$ is ${zinmathbb{C},|,lvert zrvert>1}$.
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 14:08
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
Why does my method not yield the unit circle as the boundary?
$endgroup$
– PhysicsMathsLove
Jan 22 at 14:11
1
$begingroup$
begin{align}left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1&ifffrac{u^2+v^2}{(u^2+v^2)^2}=1\&ifffrac1{u^2+v^2}=1\&iff u^2+v^2=1.end{align}
$endgroup$
– José Carlos Santos
Jan 22 at 14:17
add a comment |
$begingroup$
Why does my method not yield the unit circle as the boundary?
$endgroup$
– PhysicsMathsLove
Jan 22 at 14:11
1
$begingroup$
begin{align}left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1&ifffrac{u^2+v^2}{(u^2+v^2)^2}=1\&ifffrac1{u^2+v^2}=1\&iff u^2+v^2=1.end{align}
$endgroup$
– José Carlos Santos
Jan 22 at 14:17
$begingroup$
Why does my method not yield the unit circle as the boundary?
$endgroup$
– PhysicsMathsLove
Jan 22 at 14:11
$begingroup$
Why does my method not yield the unit circle as the boundary?
$endgroup$
– PhysicsMathsLove
Jan 22 at 14:11
1
1
$begingroup$
begin{align}left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1&ifffrac{u^2+v^2}{(u^2+v^2)^2}=1\&ifffrac1{u^2+v^2}=1\&iff u^2+v^2=1.end{align}
$endgroup$
– José Carlos Santos
Jan 22 at 14:17
$begingroup$
begin{align}left(frac{u}{u^2 +v^2}right)^2 + left(frac{v}{u^2 +v^2}right)^2 = 1&ifffrac{u^2+v^2}{(u^2+v^2)^2}=1\&ifffrac1{u^2+v^2}=1\&iff u^2+v^2=1.end{align}
$endgroup$
– José Carlos Santos
Jan 22 at 14:17
add a comment |
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