Mixing
Graphing a complex equation
$begingroup$
Question: Identify (i.e., name of the shape) and graph the equation.
$|z-1|+|z+1|leq 2$
Work done so far:
Since $z$ is a complex number assume: $z=x+iy$
$$|x+iy-1|+|x+iy+1|leq 2$$
Since if $z=x+iy$, then $|z|=sqrt(x^2+y^2)$
$$sqrt{(x-1)^2+y^2}+sqrt{(x+1)^2+y^2}leq 2$$
After this I am completely lost. Please feel free to correct any errors in calculations and show me how to answer this sort of question.
complex-analysis
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
$endgroup$
add a comment |
$begingroup$
Question: Identify (i.e., name of the shape) and graph the equation.
$|z-1|+|z+1|leq 2$
Work done so far:
Since $z$ is a complex number assume: $z=x+iy$
$$|x+iy-1|+|x+iy+1|leq 2$$
Since if $z=x+iy$, then $|z|=sqrt(x^2+y^2)$
$$sqrt{(x-1)^2+y^2}+sqrt{(x+1)^2+y^2}leq 2$$
After this I am completely lost. Please feel free to correct any errors in calculations and show me how to answer this sort of question.
complex-analysis
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
$endgroup$
add a comment |
$begingroup$
Question: Identify (i.e., name of the shape) and graph the equation.
$|z-1|+|z+1|leq 2$
Work done so far:
Since $z$ is a complex number assume: $z=x+iy$
$$|x+iy-1|+|x+iy+1|leq 2$$
Since if $z=x+iy$, then $|z|=sqrt(x^2+y^2)$
$$sqrt{(x-1)^2+y^2}+sqrt{(x+1)^2+y^2}leq 2$$
After this I am completely lost. Please feel free to correct any errors in calculations and show me how to answer this sort of question.
complex-analysis
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
$endgroup$
Question: Identify (i.e., name of the shape) and graph the equation.
$|z-1|+|z+1|leq 2$
Work done so far:
Since $z$ is a complex number assume: $z=x+iy$
$$|x+iy-1|+|x+iy+1|leq 2$$
Since if $z=x+iy$, then $|z|=sqrt(x^2+y^2)$
$$sqrt{(x-1)^2+y^2}+sqrt{(x+1)^2+y^2}leq 2$$
After this I am completely lost. Please feel free to correct any errors in calculations and show me how to answer this sort of question.
complex-analysis
complex-analysis
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
asked Jan 24 at 1:29
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
483215
483215
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $(x-1)^2+y^2=r^2$ defines a circle with center $(1,0)$ and radius $r$, similarly for the second expression with center $(-1,0)$. You can also think of the first expression as the euclidean distance of a point $(x,y)$ from $(1,0)$ and similarly for the second with distance of a point $(x,y)$ from $(-1,0)$. You want the sum of these distances to be less or equal to 2. Let us pick the point $(0,0)$, it is equidistant from both centers and the sum of the distances is 2. I argue that due to the triangle inequality you cannot get a better distance, that is the only solutions is $(1-lambda)(-1,0) + lambda(1,0), lambda in [0,1]$ (it lies on a straight line between the two centers, imagine having a string and you can stretch it, moving away away on $y$ will only yield a greater distance). So to prove the convex linear combination, we set $y=0$ (since it will only increase the distance and thus violate the inequality if it's not 0). Then $sqrt{(x-1)^2} + sqrt{(x+1)^2} = 2|x|$, then $x in [-1,1]$.
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
$endgroup$
$begingroup$
Thank you!!!!!!!!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 1:56
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%2f3085314%2fgraphing-a-complex-equation%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
$begingroup$
Note that $(x-1)^2+y^2=r^2$ defines a circle with center $(1,0)$ and radius $r$, similarly for the second expression with center $(-1,0)$. You can also think of the first expression as the euclidean distance of a point $(x,y)$ from $(1,0)$ and similarly for the second with distance of a point $(x,y)$ from $(-1,0)$. You want the sum of these distances to be less or equal to 2. Let us pick the point $(0,0)$, it is equidistant from both centers and the sum of the distances is 2. I argue that due to the triangle inequality you cannot get a better distance, that is the only solutions is $(1-lambda)(-1,0) + lambda(1,0), lambda in [0,1]$ (it lies on a straight line between the two centers, imagine having a string and you can stretch it, moving away away on $y$ will only yield a greater distance). So to prove the convex linear combination, we set $y=0$ (since it will only increase the distance and thus violate the inequality if it's not 0). Then $sqrt{(x-1)^2} + sqrt{(x+1)^2} = 2|x|$, then $x in [-1,1]$.
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
$endgroup$
$begingroup$
Thank you!!!!!!!!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 1:56
add a comment |
$begingroup$
Note that $(x-1)^2+y^2=r^2$ defines a circle with center $(1,0)$ and radius $r$, similarly for the second expression with center $(-1,0)$. You can also think of the first expression as the euclidean distance of a point $(x,y)$ from $(1,0)$ and similarly for the second with distance of a point $(x,y)$ from $(-1,0)$. You want the sum of these distances to be less or equal to 2. Let us pick the point $(0,0)$, it is equidistant from both centers and the sum of the distances is 2. I argue that due to the triangle inequality you cannot get a better distance, that is the only solutions is $(1-lambda)(-1,0) + lambda(1,0), lambda in [0,1]$ (it lies on a straight line between the two centers, imagine having a string and you can stretch it, moving away away on $y$ will only yield a greater distance). So to prove the convex linear combination, we set $y=0$ (since it will only increase the distance and thus violate the inequality if it's not 0). Then $sqrt{(x-1)^2} + sqrt{(x+1)^2} = 2|x|$, then $x in [-1,1]$.
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
$endgroup$
$begingroup$
Thank you!!!!!!!!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 1:56
add a comment |
$begingroup$
Note that $(x-1)^2+y^2=r^2$ defines a circle with center $(1,0)$ and radius $r$, similarly for the second expression with center $(-1,0)$. You can also think of the first expression as the euclidean distance of a point $(x,y)$ from $(1,0)$ and similarly for the second with distance of a point $(x,y)$ from $(-1,0)$. You want the sum of these distances to be less or equal to 2. Let us pick the point $(0,0)$, it is equidistant from both centers and the sum of the distances is 2. I argue that due to the triangle inequality you cannot get a better distance, that is the only solutions is $(1-lambda)(-1,0) + lambda(1,0), lambda in [0,1]$ (it lies on a straight line between the two centers, imagine having a string and you can stretch it, moving away away on $y$ will only yield a greater distance). So to prove the convex linear combination, we set $y=0$ (since it will only increase the distance and thus violate the inequality if it's not 0). Then $sqrt{(x-1)^2} + sqrt{(x+1)^2} = 2|x|$, then $x in [-1,1]$.
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
$endgroup$
Note that $(x-1)^2+y^2=r^2$ defines a circle with center $(1,0)$ and radius $r$, similarly for the second expression with center $(-1,0)$. You can also think of the first expression as the euclidean distance of a point $(x,y)$ from $(1,0)$ and similarly for the second with distance of a point $(x,y)$ from $(-1,0)$. You want the sum of these distances to be less or equal to 2. Let us pick the point $(0,0)$, it is equidistant from both centers and the sum of the distances is 2. I argue that due to the triangle inequality you cannot get a better distance, that is the only solutions is $(1-lambda)(-1,0) + lambda(1,0), lambda in [0,1]$ (it lies on a straight line between the two centers, imagine having a string and you can stretch it, moving away away on $y$ will only yield a greater distance). So to prove the convex linear combination, we set $y=0$ (since it will only increase the distance and thus violate the inequality if it's not 0). Then $sqrt{(x-1)^2} + sqrt{(x+1)^2} = 2|x|$, then $x in [-1,1]$.
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
edited Jan 24 at 1:50
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
answered Jan 24 at 1:43
lightxbulblightxbulb
1,125311
1,125311
$begingroup$
Thank you!!!!!!!!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 1:56
add a comment |
$begingroup$
Thank you!!!!!!!!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 1:56
$begingroup$
Thank you!!!!!!!!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 1:56
$begingroup$
Thank you!!!!!!!!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 1:56
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%2f3085314%2fgraphing-a-complex-equation%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
