Mixing
Finding image of a set under bilinear transformation
$begingroup$
Let $T(z)=ifrac{z+1}{z-1}$, and $A={zinmathbb{C}:|z|>1}cup{infty}$.
How does one find $T(A)$? I tried finding $T^{-1}(z)$, because if $w$ is a point in $T(A)$, then $T^{-1}(w)in A$. I cannot get anything decent from this, but honestly I don't know if I'm proceeding correctly.
$T^{-1}(z)=frac{z+i}{z-i}$, so if $T^{-1}(w)in A$, we have $left|frac{w+i}{w-i}right|>1$... but I can't get anywhere decent from this.
Here are some calculations: let $w=x+iy$,
$$left|frac{w+i}{w-i}right|>1 Leftrightarrow left|frac{(w+i)(bar{w}+i)}{(w-i)(bar{w}+i)}right|>1 Leftrightarrow left|frac{x^2+y^2-1+2ix}{x^2+y^2-2y+1}right|>1$$
which eventually leads me to the inequality $y(x^2+(y-1)^2)>0$, which doesn't seem to be a reasonable answer for this kind of problem.
complex-analysis
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
$endgroup$
add a comment |
$begingroup$
Let $T(z)=ifrac{z+1}{z-1}$, and $A={zinmathbb{C}:|z|>1}cup{infty}$.
How does one find $T(A)$? I tried finding $T^{-1}(z)$, because if $w$ is a point in $T(A)$, then $T^{-1}(w)in A$. I cannot get anything decent from this, but honestly I don't know if I'm proceeding correctly.
$T^{-1}(z)=frac{z+i}{z-i}$, so if $T^{-1}(w)in A$, we have $left|frac{w+i}{w-i}right|>1$... but I can't get anywhere decent from this.
Here are some calculations: let $w=x+iy$,
$$left|frac{w+i}{w-i}right|>1 Leftrightarrow left|frac{(w+i)(bar{w}+i)}{(w-i)(bar{w}+i)}right|>1 Leftrightarrow left|frac{x^2+y^2-1+2ix}{x^2+y^2-2y+1}right|>1$$
which eventually leads me to the inequality $y(x^2+(y-1)^2)>0$, which doesn't seem to be a reasonable answer for this kind of problem.
complex-analysis
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
$endgroup$
$begingroup$
Wait... does this mean $T(A)={zinmathbb{C}:text{Im}(z)>0, zneq i}$?
$endgroup$
– AstlyDichrar
Jan 25 at 1:52
add a comment |
$begingroup$
Let $T(z)=ifrac{z+1}{z-1}$, and $A={zinmathbb{C}:|z|>1}cup{infty}$.
How does one find $T(A)$? I tried finding $T^{-1}(z)$, because if $w$ is a point in $T(A)$, then $T^{-1}(w)in A$. I cannot get anything decent from this, but honestly I don't know if I'm proceeding correctly.
$T^{-1}(z)=frac{z+i}{z-i}$, so if $T^{-1}(w)in A$, we have $left|frac{w+i}{w-i}right|>1$... but I can't get anywhere decent from this.
Here are some calculations: let $w=x+iy$,
$$left|frac{w+i}{w-i}right|>1 Leftrightarrow left|frac{(w+i)(bar{w}+i)}{(w-i)(bar{w}+i)}right|>1 Leftrightarrow left|frac{x^2+y^2-1+2ix}{x^2+y^2-2y+1}right|>1$$
which eventually leads me to the inequality $y(x^2+(y-1)^2)>0$, which doesn't seem to be a reasonable answer for this kind of problem.
complex-analysis
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
$endgroup$
Let $T(z)=ifrac{z+1}{z-1}$, and $A={zinmathbb{C}:|z|>1}cup{infty}$.
How does one find $T(A)$? I tried finding $T^{-1}(z)$, because if $w$ is a point in $T(A)$, then $T^{-1}(w)in A$. I cannot get anything decent from this, but honestly I don't know if I'm proceeding correctly.
$T^{-1}(z)=frac{z+i}{z-i}$, so if $T^{-1}(w)in A$, we have $left|frac{w+i}{w-i}right|>1$... but I can't get anywhere decent from this.
Here are some calculations: let $w=x+iy$,
$$left|frac{w+i}{w-i}right|>1 Leftrightarrow left|frac{(w+i)(bar{w}+i)}{(w-i)(bar{w}+i)}right|>1 Leftrightarrow left|frac{x^2+y^2-1+2ix}{x^2+y^2-2y+1}right|>1$$
which eventually leads me to the inequality $y(x^2+(y-1)^2)>0$, which doesn't seem to be a reasonable answer for this kind of problem.
complex-analysis
complex-analysis
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
asked Jan 25 at 1:25
AstlyDichrarAstlyDichrar
42248
42248
$begingroup$
Wait... does this mean $T(A)={zinmathbb{C}:text{Im}(z)>0, zneq i}$?
$endgroup$
– AstlyDichrar
Jan 25 at 1:52
add a comment |
$begingroup$
Wait... does this mean $T(A)={zinmathbb{C}:text{Im}(z)>0, zneq i}$?
$endgroup$
– AstlyDichrar
Jan 25 at 1:52
$begingroup$
Wait... does this mean $T(A)={zinmathbb{C}:text{Im}(z)>0, zneq i}$?
$endgroup$
– AstlyDichrar
Jan 25 at 1:52
$begingroup$
Wait... does this mean $T(A)={zinmathbb{C}:text{Im}(z)>0, zneq i}$?
$endgroup$
– AstlyDichrar
Jan 25 at 1:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It takes $mid zmid=1$ to the line through $1$ and $0$. That is, the $x$-axis.
To check which half-plane, use a test point. $0to -i$. $therefore $ it's the upper half-plane.
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
So, it is always sufficient to just check where $T$ sends the boundary of $A$ to and a single point in $A$ if we're dealing with bilinear transformations?
$endgroup$
– AstlyDichrar
Jan 25 at 1:59
$begingroup$
For a Möbius transformation this strategy will work. They take generalized circles to generalized circles. Notice $1toinfty$.
$endgroup$
– Chris Custer
Jan 25 at 2:02
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%2f3086593%2ffinding-image-of-a-set-under-bilinear-transformation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It takes $mid zmid=1$ to the line through $1$ and $0$. That is, the $x$-axis.
To check which half-plane, use a test point. $0to -i$. $therefore $ it's the upper half-plane.
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
So, it is always sufficient to just check where $T$ sends the boundary of $A$ to and a single point in $A$ if we're dealing with bilinear transformations?
$endgroup$
– AstlyDichrar
Jan 25 at 1:59
$begingroup$
For a Möbius transformation this strategy will work. They take generalized circles to generalized circles. Notice $1toinfty$.
$endgroup$
– Chris Custer
Jan 25 at 2:02
add a comment |
$begingroup$
It takes $mid zmid=1$ to the line through $1$ and $0$. That is, the $x$-axis.
To check which half-plane, use a test point. $0to -i$. $therefore $ it's the upper half-plane.
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
So, it is always sufficient to just check where $T$ sends the boundary of $A$ to and a single point in $A$ if we're dealing with bilinear transformations?
$endgroup$
– AstlyDichrar
Jan 25 at 1:59
$begingroup$
For a Möbius transformation this strategy will work. They take generalized circles to generalized circles. Notice $1toinfty$.
$endgroup$
– Chris Custer
Jan 25 at 2:02
add a comment |
$begingroup$
It takes $mid zmid=1$ to the line through $1$ and $0$. That is, the $x$-axis.
To check which half-plane, use a test point. $0to -i$. $therefore $ it's the upper half-plane.
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
$endgroup$
It takes $mid zmid=1$ to the line through $1$ and $0$. That is, the $x$-axis.
To check which half-plane, use a test point. $0to -i$. $therefore $ it's the upper half-plane.
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
answered Jan 25 at 1:57
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
So, it is always sufficient to just check where $T$ sends the boundary of $A$ to and a single point in $A$ if we're dealing with bilinear transformations?
$endgroup$
– AstlyDichrar
Jan 25 at 1:59
$begingroup$
For a Möbius transformation this strategy will work. They take generalized circles to generalized circles. Notice $1toinfty$.
$endgroup$
– Chris Custer
Jan 25 at 2:02
add a comment |
$begingroup$
So, it is always sufficient to just check where $T$ sends the boundary of $A$ to and a single point in $A$ if we're dealing with bilinear transformations?
$endgroup$
– AstlyDichrar
Jan 25 at 1:59
$begingroup$
For a Möbius transformation this strategy will work. They take generalized circles to generalized circles. Notice $1toinfty$.
$endgroup$
– Chris Custer
Jan 25 at 2:02
$begingroup$
So, it is always sufficient to just check where $T$ sends the boundary of $A$ to and a single point in $A$ if we're dealing with bilinear transformations?
$endgroup$
– AstlyDichrar
Jan 25 at 1:59
$begingroup$
So, it is always sufficient to just check where $T$ sends the boundary of $A$ to and a single point in $A$ if we're dealing with bilinear transformations?
$endgroup$
– AstlyDichrar
Jan 25 at 1:59
$begingroup$
For a Möbius transformation this strategy will work. They take generalized circles to generalized circles. Notice $1toinfty$.
$endgroup$
– Chris Custer
Jan 25 at 2:02
$begingroup$
For a Möbius transformation this strategy will work. They take generalized circles to generalized circles. Notice $1toinfty$.
$endgroup$
– Chris Custer
Jan 25 at 2:02
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%2f3086593%2ffinding-image-of-a-set-under-bilinear-transformation%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
$begingroup$
Wait... does this mean $T(A)={zinmathbb{C}:text{Im}(z)>0, zneq i}$?
$endgroup$
– AstlyDichrar
Jan 25 at 1:52