Mixing
Find transformation that maps region between two circles to vertical strip
$begingroup$
I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.
I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.
All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(
complex-analysis
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
$endgroup$
add a comment |
$begingroup$
I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.
I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.
All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(
complex-analysis
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
$endgroup$
$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48
$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53
add a comment |
$begingroup$
I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.
I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.
All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(
complex-analysis
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
$endgroup$
I'm having lots of trouble with these conformal transformations. I have no idea how to transform conformally the region $A={zinmathbb{C}:|z-1|>1}cap{zinmathbb{C}:|z-2|<2}$ into the vertical strip $B={zinmathbb{C}:0< text{Re}(z)<1}$.
I think we need to map the point $0$ to $infty$, because we might want the centers of the circles inside the region to be in the same vertical line in $B$. So $1/z$ would probably be a good place to start, but I can't go any further. We should probably require that this transformation $T(z)$ maps the boundary of $A$ to the boundary of $B$ as well.
All the books I've read about this are not of much help, every transformation seems to appear out of nowhere with little explanation. If anyone has any suggestion that will enlighten my brain, please do help me :(
complex-analysis
complex-analysis
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
edited Jan 25 at 20:48
AstlyDichrar
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
asked Jan 25 at 20:33
AstlyDichrarAstlyDichrar
42248
42248
$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48
$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53
add a comment |
$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48
$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53
$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48
$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48
$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53
$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
$endgroup$
$begingroup$
I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
$endgroup$
– AstlyDichrar
Jan 25 at 21:06
$begingroup$
Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
$endgroup$
– Ted Shifrin
Jan 25 at 21:09
add a comment |
$begingroup$
Hint:
Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.
answered Jan 25 at 20:54
BernardBernard
123k741117
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087594%2ffind-transformation-that-maps-region-between-two-circles-to-vertical-strip%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
$endgroup$
$begingroup$
I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
$endgroup$
– AstlyDichrar
Jan 25 at 21:06
$begingroup$
Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
$endgroup$
– Ted Shifrin
Jan 25 at 21:09
add a comment |
$begingroup$
HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
$endgroup$
$begingroup$
I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
$endgroup$
– AstlyDichrar
Jan 25 at 21:06
$begingroup$
Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
$endgroup$
– Ted Shifrin
Jan 25 at 21:09
add a comment |
$begingroup$
HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
$endgroup$
HINT: Since you're looking for a linear fractional transformation (which is a good bet if you want to map circles to lines), you need to specify where three particular points map. We already said that you want $0rightsquigarrowinfty$. What about choosing one other point on each circle?
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
answered Jan 25 at 20:54
Ted ShifrinTed Shifrin
64.4k44692
64.4k44692
$begingroup$
I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
$endgroup$
– AstlyDichrar
Jan 25 at 21:06
$begingroup$
Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
$endgroup$
– Ted Shifrin
Jan 25 at 21:09
add a comment |
$begingroup$
I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
$endgroup$
– AstlyDichrar
Jan 25 at 21:06
$begingroup$
Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
$endgroup$
– Ted Shifrin
Jan 25 at 21:09
$begingroup$
I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
$endgroup$
– AstlyDichrar
Jan 25 at 21:06
$begingroup$
I tried using a lemma in Bak & Newman's Complex Analysis which says that the unique bilinear mapping that sends $z_1, z_2, z_3$ to $infty, 0, 1$ respectively is $T(z)=frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$. In this case, we would want $z_1 = 0$, $z_2=2$, $z_3=4$, I think. This means $T(z)=frac{2z-4}{z}$ if I made the calculations correctly, does that seem good? Unfortunately I don't have any solutions, and I can't find a straightforward way to check if this is correct.
$endgroup$
– AstlyDichrar
Jan 25 at 21:06
$begingroup$
Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
$endgroup$
– Ted Shifrin
Jan 25 at 21:09
$begingroup$
Yes, that looks right. You should be able to check that the mapping is correct. For example, the inverse mapping $z=frac 4{2-w}$ does indeed send the imaginary axis to the circle $|z-1|=1$.
$endgroup$
– Ted Shifrin
Jan 25 at 21:09
add a comment |
$begingroup$
Hint:
Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.
answered Jan 25 at 20:54
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.
answered Jan 25 at 20:54
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.
answered Jan 25 at 20:54
BernardBernard
123k741117
$endgroup$
Hint:
Inversion with ratio $k>0$ and centre $O$ maps a circle passing through $O$ to the line through the intersection points of the circle with the circle of radius $sqrt k$ centred at $O$.
answered Jan 25 at 20:54
BernardBernard
123k741117
answered Jan 25 at 20:54
BernardBernard
123k741117
answered Jan 25 at 20:54
BernardBernard
123k741117
answered Jan 25 at 20:54
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3087594%2ffind-transformation-that-maps-region-between-two-circles-to-vertical-strip%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Yes, I'll correct.
$endgroup$
– AstlyDichrar
Jan 25 at 20:48
$begingroup$
Since $0$ goes to $infty$, $1/z$ is a good start. Note that $l_1:|z-1|=1$ and $l_2:|z-2|=2$ are mapped to parallel lines via $zmapsto 1/z$ because they should intersect at $infty$. Now, through rotation and translation we can find $a/z+b$ with the required property.
$endgroup$
– Song
Jan 25 at 20:53