Finding a conformal map from this domain into the unit disc
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Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.
Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.
I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?
Help is appreciated.
complex-analysis
asked Aug 19 at 9:12
Kamil
1,90421237
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up vote
1
down vote
favorite
Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.
Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.
I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?
Help is appreciated.
complex-analysis
asked Aug 19 at 9:12
Kamil
1,90421237
Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23
In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26
What book are you using?
– quasi
Aug 19 at 9:27
'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29
It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.
Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.
I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?
Help is appreciated.
complex-analysis
asked Aug 19 at 9:12
Kamil
1,90421237
Problem: Find a conformal map $f$ from $ A = left z in mathbbC mid textIm(z) > 0, $ into the unit disk.
Attempt: I started off with a map $F_1: z mapsto z + frac1z$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$.
I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| > 1$ condition?
Help is appreciated.
complex-analysis
asked Aug 19 at 9:12
Kamil
1,90421237
asked Aug 19 at 9:12
Kamil
1,90421237
asked Aug 19 at 9:12
Kamil
1,90421237
asked Aug 19 at 9:12
Kamil
1,90421237
1,90421237
Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23
In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26
What book are you using?
– quasi
Aug 19 at 9:27
'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29
It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43
add a comment |Â
Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23
In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26
What book are you using?
– quasi
Aug 19 at 9:27
'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29
It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43
Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23
Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23
In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26
In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26
What book are you using?
– quasi
Aug 19 at 9:27
What book are you using?
– quasi
Aug 19 at 9:27
'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29
'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29
It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43
It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Hint: With conformal mapping
$$w=left(dfracz+1z-1right)^2$$
you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.
answered Aug 19 at 9:44
Nosrati
20.7k41644
Great. Could you explain to me how you arrived at this map? What was your logic?
– Kamil
Aug 19 at 11:02
what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
– Nosrati
Aug 19 at 11:06
The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
– Kamil
Aug 19 at 11:39
yes, it is.....
– Nosrati
Aug 19 at 11:45
add a comment |Â
up vote
-1
down vote
It seems that your region $A$ is the limit as $R to + infty$ of
$$
A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form
answered Aug 19 at 9:26
Laurent
127
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint: With conformal mapping
$$w=left(dfracz+1z-1right)^2$$
you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.
answered Aug 19 at 9:44
Nosrati
20.7k41644
Great. Could you explain to me how you arrived at this map? What was your logic?
– Kamil
Aug 19 at 11:02
what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
– Nosrati
Aug 19 at 11:06
The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
– Kamil
Aug 19 at 11:39
yes, it is.....
– Nosrati
Aug 19 at 11:45
add a comment |Â
up vote
1
down vote
accepted
Hint: With conformal mapping
$$w=left(dfracz+1z-1right)^2$$
you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.
answered Aug 19 at 9:44
Nosrati
20.7k41644
Great. Could you explain to me how you arrived at this map? What was your logic?
– Kamil
Aug 19 at 11:02
what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
– Nosrati
Aug 19 at 11:06
The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
– Kamil
Aug 19 at 11:39
yes, it is.....
– Nosrati
Aug 19 at 11:45
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint: With conformal mapping
$$w=left(dfracz+1z-1right)^2$$
you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.
answered Aug 19 at 9:44
Nosrati
20.7k41644
Hint: With conformal mapping
$$w=left(dfracz+1z-1right)^2$$
you map region $A=>1$ to lower half plane $zinmathbbC:bf Im z<0$.
answered Aug 19 at 9:44
Nosrati
20.7k41644
answered Aug 19 at 9:44
Nosrati
20.7k41644
answered Aug 19 at 9:44
Nosrati
20.7k41644
answered Aug 19 at 9:44
Nosrati
20.7k41644
20.7k41644
Great. Could you explain to me how you arrived at this map? What was your logic?
– Kamil
Aug 19 at 11:02
what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
– Nosrati
Aug 19 at 11:06
The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
– Kamil
Aug 19 at 11:39
yes, it is.....
– Nosrati
Aug 19 at 11:45
add a comment |Â
Great. Could you explain to me how you arrived at this map? What was your logic?
– Kamil
Aug 19 at 11:02
what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
– Nosrati
Aug 19 at 11:06
The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
– Kamil
Aug 19 at 11:39
yes, it is.....
– Nosrati
Aug 19 at 11:45
Great. Could you explain to me how you arrived at this map? What was your logic?
– Kamil
Aug 19 at 11:02
Great. Could you explain to me how you arrived at this map? What was your logic?
– Kamil
Aug 19 at 11:02
what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
– Nosrati
Aug 19 at 11:06
what does mapping $frac1+z1-z$ do with upper unit circle. then square it and let $ztofrac1z$ to bring inner points to outer of semicircle.
– Nosrati
Aug 19 at 11:06
The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
– Kamil
Aug 19 at 11:39
The function $(1+z)/(1-z)$ maps the upper unit circle to the positive imaginary axis. Then I think under this mapping, the whole domain $A$ gets mapped to the quadrant $leftw = u + iv: u < 0, v > 0 right$. Is this correct?
– Kamil
Aug 19 at 11:39
yes, it is.....
– Nosrati
Aug 19 at 11:45
yes, it is.....
– Nosrati
Aug 19 at 11:45
add a comment |Â
up vote
-1
down vote
It seems that your region $A$ is the limit as $R to + infty$ of
$$
A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form
answered Aug 19 at 9:26
Laurent
127
add a comment |Â
up vote
-1
down vote
It seems that your region $A$ is the limit as $R to + infty$ of
$$
A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form
answered Aug 19 at 9:26
Laurent
127
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
It seems that your region $A$ is the limit as $R to + infty$ of
$$
A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form
answered Aug 19 at 9:26
Laurent
127
It seems that your region $A$ is the limit as $R to + infty$ of
$$
A_R=z=rexp(itheta), theta in (0,pi), r in (1,R) .
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form
answered Aug 19 at 9:26
Laurent
127
answered Aug 19 at 9:26
Laurent
127
answered Aug 19 at 9:26
Laurent
127
answered Aug 19 at 9:26
Laurent
127
127
add a comment |Â
add a comment |Â
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Just use the map $zmapsto 1/z$. It maps $A$ conformally into the standard unit disk. As stated, the map need not be surjective.
– quasi
Aug 19 at 9:23
In our book we defined a conformal map as 'holomorphic bijection'.
– Kamil
Aug 19 at 9:26
What book are you using?
– quasi
Aug 19 at 9:27
'Complex Analysis', by Stein and Shakarchi. Princeton lectures in Analysis
– Kamil
Aug 19 at 9:29
It's not always defined that way. For example: en.wikipedia.org/wiki/Conformal_map Perhaps edit the question to require $f$ to be bijective.
– quasi
Aug 19 at 9:43