Find Conformal Mapping Between Circles
$begingroup$
How would I go about finding the conformal mapping bewtween $|z| = 2$ and $|z-1| = 1$? I have done the following so far:
Step 1: map regions between circles onto a strip:
$$frac{az + b}{cz +d} = frac{iz}{z-2}$$
Step 2: I assume this is the step you rotate it into a horizontal strip, although I am not sure how.
Step 3: Transform from strip to upper half plane using $e^z$ (should I use $e^z$ or $e^{iz}$).
Step 4: Transform from upper half plane onto unit disc using Möbius transformation:
$$frac{z-1}{z+1}$$
Then do $f_4 circ f_3 circ f_2 circ f_1$. Is this correct? Am I missing anything? Please let me know! Thank you!!
complex-analysis analysis conformal-geometry
asked Dec 11 '18 at 9:56
moloculemolocule
776
$endgroup$
add a comment |
$begingroup$
How would I go about finding the conformal mapping bewtween $|z| = 2$ and $|z-1| = 1$? I have done the following so far:
Step 1: map regions between circles onto a strip:
$$frac{az + b}{cz +d} = frac{iz}{z-2}$$
Step 2: I assume this is the step you rotate it into a horizontal strip, although I am not sure how.
Step 3: Transform from strip to upper half plane using $e^z$ (should I use $e^z$ or $e^{iz}$).
Step 4: Transform from upper half plane onto unit disc using Möbius transformation:
$$frac{z-1}{z+1}$$
Then do $f_4 circ f_3 circ f_2 circ f_1$. Is this correct? Am I missing anything? Please let me know! Thank you!!
complex-analysis analysis conformal-geometry
asked Dec 11 '18 at 9:56
moloculemolocule
776
$endgroup$
add a comment |
$begingroup$
How would I go about finding the conformal mapping bewtween $|z| = 2$ and $|z-1| = 1$? I have done the following so far:
Step 1: map regions between circles onto a strip:
$$frac{az + b}{cz +d} = frac{iz}{z-2}$$
Step 2: I assume this is the step you rotate it into a horizontal strip, although I am not sure how.
Step 3: Transform from strip to upper half plane using $e^z$ (should I use $e^z$ or $e^{iz}$).
Step 4: Transform from upper half plane onto unit disc using Möbius transformation:
$$frac{z-1}{z+1}$$
Then do $f_4 circ f_3 circ f_2 circ f_1$. Is this correct? Am I missing anything? Please let me know! Thank you!!
complex-analysis analysis conformal-geometry
asked Dec 11 '18 at 9:56
moloculemolocule
776
$endgroup$
How would I go about finding the conformal mapping bewtween $|z| = 2$ and $|z-1| = 1$? I have done the following so far:
Step 1: map regions between circles onto a strip:
$$frac{az + b}{cz +d} = frac{iz}{z-2}$$
Step 2: I assume this is the step you rotate it into a horizontal strip, although I am not sure how.
Step 3: Transform from strip to upper half plane using $e^z$ (should I use $e^z$ or $e^{iz}$).
Step 4: Transform from upper half plane onto unit disc using Möbius transformation:
$$frac{z-1}{z+1}$$
Then do $f_4 circ f_3 circ f_2 circ f_1$. Is this correct? Am I missing anything? Please let me know! Thank you!!
complex-analysis analysis conformal-geometry
complex-analysis analysis conformal-geometry
asked Dec 11 '18 at 9:56
moloculemolocule
776
asked Dec 11 '18 at 9:56
moloculemolocule
776
asked Dec 11 '18 at 9:56
moloculemolocule
776
asked Dec 11 '18 at 9:56
moloculemolocule
776
asked Dec 11 '18 at 9:56
moloculemolocule
776
776
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general, when dealing with circles/planes use Mobius Transforms. You simply need to carry three points on the starting circle to three points on the end circle, and you are guaranteed a unique transformation. Any good complex analysis text will outline the details and provide basic formulas to use.
In this case, we can save some time and simply consult a list of known transforms, e.g. here, and find that:
$f_1(z)=frac{z-1}{z+1}$ carries the right half plane to the unit disk
$f_2(z) = z+1/2,$ carries the right half plane to ${z : operatorname{Re}(z) > 1/2}$
$f_3(z) = 1/z,$ carries ${z : operatorname{Re}(z) > 1/2}$ to the disk $|z-1| = 1$
In other words,
$$text{unit disk} stackrel{f_1^{-1}}{longrightarrow} text{right half plane} stackrel{f_2}{longrightarrow} {z : operatorname{Re}(z) > 1/2} stackrel{f_3}{longrightarrow} {z : |z-1| = 1}$$
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
$begingroup$
Thank you for this answer!! I was wondering how I could get the region between the two circles $|z| = 2$ and $|z-1| = 1$. Would I just follow the same logic then? I think I am just confused how to get from between circles to upper half plane or right half plane.
$endgroup$
– molocule
Dec 11 '18 at 10:31
$begingroup$
@molocule I'm not sure what you mean by "the region between the circles"... are you referring to $mathbb C setminus {z : |z| = 1} setminus {z : |z-1| = 1}$ ? To map circles to half planes you need to use Mobius Transforms. In my post I list a transformation taking the right half plane to the unit disk. If you want to take the upper half plane to the unit disk, use the Cayley Transform. If you want to work with rotated half planes, you need to build your own transform...
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:37
$begingroup$
... either see the Wikipedia page on Mobius Transforms for an outline of methods, or better yet consult your favorite complex analysis textbook. Mobius Transforms are uniquely determined by where three points map to, and the formulas are well known.
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:38
$begingroup$
Ah, I see! Thank you!
$endgroup$
– molocule
Dec 11 '18 at 16:20
add a comment |
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1 Answer
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1 Answer
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$begingroup$
In general, when dealing with circles/planes use Mobius Transforms. You simply need to carry three points on the starting circle to three points on the end circle, and you are guaranteed a unique transformation. Any good complex analysis text will outline the details and provide basic formulas to use.
In this case, we can save some time and simply consult a list of known transforms, e.g. here, and find that:
$f_1(z)=frac{z-1}{z+1}$ carries the right half plane to the unit disk
$f_2(z) = z+1/2,$ carries the right half plane to ${z : operatorname{Re}(z) > 1/2}$
$f_3(z) = 1/z,$ carries ${z : operatorname{Re}(z) > 1/2}$ to the disk $|z-1| = 1$
In other words,
$$text{unit disk} stackrel{f_1^{-1}}{longrightarrow} text{right half plane} stackrel{f_2}{longrightarrow} {z : operatorname{Re}(z) > 1/2} stackrel{f_3}{longrightarrow} {z : |z-1| = 1}$$
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
$begingroup$
Thank you for this answer!! I was wondering how I could get the region between the two circles $|z| = 2$ and $|z-1| = 1$. Would I just follow the same logic then? I think I am just confused how to get from between circles to upper half plane or right half plane.
$endgroup$
– molocule
Dec 11 '18 at 10:31
$begingroup$
@molocule I'm not sure what you mean by "the region between the circles"... are you referring to $mathbb C setminus {z : |z| = 1} setminus {z : |z-1| = 1}$ ? To map circles to half planes you need to use Mobius Transforms. In my post I list a transformation taking the right half plane to the unit disk. If you want to take the upper half plane to the unit disk, use the Cayley Transform. If you want to work with rotated half planes, you need to build your own transform...
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:37
$begingroup$
... either see the Wikipedia page on Mobius Transforms for an outline of methods, or better yet consult your favorite complex analysis textbook. Mobius Transforms are uniquely determined by where three points map to, and the formulas are well known.
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:38
$begingroup$
Ah, I see! Thank you!
$endgroup$
– molocule
Dec 11 '18 at 16:20
add a comment |
$begingroup$
In general, when dealing with circles/planes use Mobius Transforms. You simply need to carry three points on the starting circle to three points on the end circle, and you are guaranteed a unique transformation. Any good complex analysis text will outline the details and provide basic formulas to use.
In this case, we can save some time and simply consult a list of known transforms, e.g. here, and find that:
$f_1(z)=frac{z-1}{z+1}$ carries the right half plane to the unit disk
$f_2(z) = z+1/2,$ carries the right half plane to ${z : operatorname{Re}(z) > 1/2}$
$f_3(z) = 1/z,$ carries ${z : operatorname{Re}(z) > 1/2}$ to the disk $|z-1| = 1$
In other words,
$$text{unit disk} stackrel{f_1^{-1}}{longrightarrow} text{right half plane} stackrel{f_2}{longrightarrow} {z : operatorname{Re}(z) > 1/2} stackrel{f_3}{longrightarrow} {z : |z-1| = 1}$$
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
$begingroup$
Thank you for this answer!! I was wondering how I could get the region between the two circles $|z| = 2$ and $|z-1| = 1$. Would I just follow the same logic then? I think I am just confused how to get from between circles to upper half plane or right half plane.
$endgroup$
– molocule
Dec 11 '18 at 10:31
$begingroup$
@molocule I'm not sure what you mean by "the region between the circles"... are you referring to $mathbb C setminus {z : |z| = 1} setminus {z : |z-1| = 1}$ ? To map circles to half planes you need to use Mobius Transforms. In my post I list a transformation taking the right half plane to the unit disk. If you want to take the upper half plane to the unit disk, use the Cayley Transform. If you want to work with rotated half planes, you need to build your own transform...
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:37
$begingroup$
... either see the Wikipedia page on Mobius Transforms for an outline of methods, or better yet consult your favorite complex analysis textbook. Mobius Transforms are uniquely determined by where three points map to, and the formulas are well known.
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:38
$begingroup$
Ah, I see! Thank you!
$endgroup$
– molocule
Dec 11 '18 at 16:20
add a comment |
$begingroup$
In general, when dealing with circles/planes use Mobius Transforms. You simply need to carry three points on the starting circle to three points on the end circle, and you are guaranteed a unique transformation. Any good complex analysis text will outline the details and provide basic formulas to use.
In this case, we can save some time and simply consult a list of known transforms, e.g. here, and find that:
$f_1(z)=frac{z-1}{z+1}$ carries the right half plane to the unit disk
$f_2(z) = z+1/2,$ carries the right half plane to ${z : operatorname{Re}(z) > 1/2}$
$f_3(z) = 1/z,$ carries ${z : operatorname{Re}(z) > 1/2}$ to the disk $|z-1| = 1$
In other words,
$$text{unit disk} stackrel{f_1^{-1}}{longrightarrow} text{right half plane} stackrel{f_2}{longrightarrow} {z : operatorname{Re}(z) > 1/2} stackrel{f_3}{longrightarrow} {z : |z-1| = 1}$$
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
$endgroup$
In general, when dealing with circles/planes use Mobius Transforms. You simply need to carry three points on the starting circle to three points on the end circle, and you are guaranteed a unique transformation. Any good complex analysis text will outline the details and provide basic formulas to use.
In this case, we can save some time and simply consult a list of known transforms, e.g. here, and find that:
$f_1(z)=frac{z-1}{z+1}$ carries the right half plane to the unit disk
$f_2(z) = z+1/2,$ carries the right half plane to ${z : operatorname{Re}(z) > 1/2}$
$f_3(z) = 1/z,$ carries ${z : operatorname{Re}(z) > 1/2}$ to the disk $|z-1| = 1$
In other words,
$$text{unit disk} stackrel{f_1^{-1}}{longrightarrow} text{right half plane} stackrel{f_2}{longrightarrow} {z : operatorname{Re}(z) > 1/2} stackrel{f_3}{longrightarrow} {z : |z-1| = 1}$$
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
answered Dec 11 '18 at 10:25
Brevan EllefsenBrevan Ellefsen
11.7k31649
11.7k31649
$begingroup$
Thank you for this answer!! I was wondering how I could get the region between the two circles $|z| = 2$ and $|z-1| = 1$. Would I just follow the same logic then? I think I am just confused how to get from between circles to upper half plane or right half plane.
$endgroup$
– molocule
Dec 11 '18 at 10:31
$begingroup$
@molocule I'm not sure what you mean by "the region between the circles"... are you referring to $mathbb C setminus {z : |z| = 1} setminus {z : |z-1| = 1}$ ? To map circles to half planes you need to use Mobius Transforms. In my post I list a transformation taking the right half plane to the unit disk. If you want to take the upper half plane to the unit disk, use the Cayley Transform. If you want to work with rotated half planes, you need to build your own transform...
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:37
$begingroup$
... either see the Wikipedia page on Mobius Transforms for an outline of methods, or better yet consult your favorite complex analysis textbook. Mobius Transforms are uniquely determined by where three points map to, and the formulas are well known.
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:38
$begingroup$
Ah, I see! Thank you!
$endgroup$
– molocule
Dec 11 '18 at 16:20
add a comment |
$begingroup$
Thank you for this answer!! I was wondering how I could get the region between the two circles $|z| = 2$ and $|z-1| = 1$. Would I just follow the same logic then? I think I am just confused how to get from between circles to upper half plane or right half plane.
$endgroup$
– molocule
Dec 11 '18 at 10:31
$begingroup$
@molocule I'm not sure what you mean by "the region between the circles"... are you referring to $mathbb C setminus {z : |z| = 1} setminus {z : |z-1| = 1}$ ? To map circles to half planes you need to use Mobius Transforms. In my post I list a transformation taking the right half plane to the unit disk. If you want to take the upper half plane to the unit disk, use the Cayley Transform. If you want to work with rotated half planes, you need to build your own transform...
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:37
$begingroup$
... either see the Wikipedia page on Mobius Transforms for an outline of methods, or better yet consult your favorite complex analysis textbook. Mobius Transforms are uniquely determined by where three points map to, and the formulas are well known.
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:38
$begingroup$
Ah, I see! Thank you!
$endgroup$
– molocule
Dec 11 '18 at 16:20
$begingroup$
Thank you for this answer!! I was wondering how I could get the region between the two circles $|z| = 2$ and $|z-1| = 1$. Would I just follow the same logic then? I think I am just confused how to get from between circles to upper half plane or right half plane.
$endgroup$
– molocule
Dec 11 '18 at 10:31
$begingroup$
Thank you for this answer!! I was wondering how I could get the region between the two circles $|z| = 2$ and $|z-1| = 1$. Would I just follow the same logic then? I think I am just confused how to get from between circles to upper half plane or right half plane.
$endgroup$
– molocule
Dec 11 '18 at 10:31
$begingroup$
@molocule I'm not sure what you mean by "the region between the circles"... are you referring to $mathbb C setminus {z : |z| = 1} setminus {z : |z-1| = 1}$ ? To map circles to half planes you need to use Mobius Transforms. In my post I list a transformation taking the right half plane to the unit disk. If you want to take the upper half plane to the unit disk, use the Cayley Transform. If you want to work with rotated half planes, you need to build your own transform...
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:37
$begingroup$
@molocule I'm not sure what you mean by "the region between the circles"... are you referring to $mathbb C setminus {z : |z| = 1} setminus {z : |z-1| = 1}$ ? To map circles to half planes you need to use Mobius Transforms. In my post I list a transformation taking the right half plane to the unit disk. If you want to take the upper half plane to the unit disk, use the Cayley Transform. If you want to work with rotated half planes, you need to build your own transform...
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:37
$begingroup$
... either see the Wikipedia page on Mobius Transforms for an outline of methods, or better yet consult your favorite complex analysis textbook. Mobius Transforms are uniquely determined by where three points map to, and the formulas are well known.
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:38
$begingroup$
... either see the Wikipedia page on Mobius Transforms for an outline of methods, or better yet consult your favorite complex analysis textbook. Mobius Transforms are uniquely determined by where three points map to, and the formulas are well known.
$endgroup$
– Brevan Ellefsen
Dec 11 '18 at 10:38
$begingroup$
Ah, I see! Thank you!
$endgroup$
– molocule
Dec 11 '18 at 16:20
$begingroup$
Ah, I see! Thank you!
$endgroup$
– molocule
Dec 11 '18 at 16:20
add a comment |
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