What is the image of $f: mathbb{R}^2 rightarrow mathbb{R}^2$ $f(x, y) = (x^2-y^2, 2xy)$ with $x>0$?











up vote
1
down vote

favorite












What I have now:



Let $x=rcostheta, y=rsintheta$, with $r>0, -dfrac{pi}{2}<theta <dfrac{pi}{2}$, and then$$f(x, y)=(r^2{costheta}^2- r^2{sintheta}^2, 2costheta sintheta)=(r^2cos2theta, r^2sin2theta)$$
I am not sure where to go beyond those steps. It seems that it should be circles with radius $r^2$, and centred at $(0,0)$. Any help, thanks ahead!






calculus







share|cite|improve this question




















  • 1




    If you know complex numbers, then identify $(x,y)inmathbb{R}^2$ with $x+ytext{i}inmathbb{C}$. Then, $f(z)=z^2$ for all $zinmathbb{C}$. Consequently, the image of the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ is $mathbb{C}setminusmathbb{R}_{leq 0}$.
    – Batominovski
    Dec 4 at 1:14












  • @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    – Cathy
    Dec 4 at 1:22






  • 1




    No, it is not. The negative real axis and the origin are not in the image.
    – Batominovski
    Dec 4 at 1:23










  • @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    – Cathy
    Dec 4 at 1:27










  • Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    – Sameer Baheti
    Dec 4 at 9:31















up vote
1
down vote

favorite












What I have now:



Let $x=rcostheta, y=rsintheta$, with $r>0, -dfrac{pi}{2}<theta <dfrac{pi}{2}$, and then$$f(x, y)=(r^2{costheta}^2- r^2{sintheta}^2, 2costheta sintheta)=(r^2cos2theta, r^2sin2theta)$$
I am not sure where to go beyond those steps. It seems that it should be circles with radius $r^2$, and centred at $(0,0)$. Any help, thanks ahead!






calculus







share|cite|improve this question




















  • 1




    If you know complex numbers, then identify $(x,y)inmathbb{R}^2$ with $x+ytext{i}inmathbb{C}$. Then, $f(z)=z^2$ for all $zinmathbb{C}$. Consequently, the image of the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ is $mathbb{C}setminusmathbb{R}_{leq 0}$.
    – Batominovski
    Dec 4 at 1:14












  • @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    – Cathy
    Dec 4 at 1:22






  • 1




    No, it is not. The negative real axis and the origin are not in the image.
    – Batominovski
    Dec 4 at 1:23










  • @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    – Cathy
    Dec 4 at 1:27










  • Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    – Sameer Baheti
    Dec 4 at 9:31













up vote
1
down vote

favorite









up vote
1
down vote

favorite











What I have now:



Let $x=rcostheta, y=rsintheta$, with $r>0, -dfrac{pi}{2}<theta <dfrac{pi}{2}$, and then$$f(x, y)=(r^2{costheta}^2- r^2{sintheta}^2, 2costheta sintheta)=(r^2cos2theta, r^2sin2theta)$$
I am not sure where to go beyond those steps. It seems that it should be circles with radius $r^2$, and centred at $(0,0)$. Any help, thanks ahead!






calculus







share|cite|improve this question















What I have now:



Let $x=rcostheta, y=rsintheta$, with $r>0, -dfrac{pi}{2}<theta <dfrac{pi}{2}$, and then$$f(x, y)=(r^2{costheta}^2- r^2{sintheta}^2, 2costheta sintheta)=(r^2cos2theta, r^2sin2theta)$$
I am not sure where to go beyond those steps. It seems that it should be circles with radius $r^2$, and centred at $(0,0)$. Any help, thanks ahead!






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 at 1:42









user302797

19.4k92252




19.4k92252










asked Dec 4 at 1:06









Cathy

13217




13217








  • 1




    If you know complex numbers, then identify $(x,y)inmathbb{R}^2$ with $x+ytext{i}inmathbb{C}$. Then, $f(z)=z^2$ for all $zinmathbb{C}$. Consequently, the image of the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ is $mathbb{C}setminusmathbb{R}_{leq 0}$.
    – Batominovski
    Dec 4 at 1:14












  • @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    – Cathy
    Dec 4 at 1:22






  • 1




    No, it is not. The negative real axis and the origin are not in the image.
    – Batominovski
    Dec 4 at 1:23










  • @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    – Cathy
    Dec 4 at 1:27










  • Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    – Sameer Baheti
    Dec 4 at 9:31














  • 1




    If you know complex numbers, then identify $(x,y)inmathbb{R}^2$ with $x+ytext{i}inmathbb{C}$. Then, $f(z)=z^2$ for all $zinmathbb{C}$. Consequently, the image of the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ is $mathbb{C}setminusmathbb{R}_{leq 0}$.
    – Batominovski
    Dec 4 at 1:14












  • @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    – Cathy
    Dec 4 at 1:22






  • 1




    No, it is not. The negative real axis and the origin are not in the image.
    – Batominovski
    Dec 4 at 1:23










  • @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    – Cathy
    Dec 4 at 1:27










  • Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    – Sameer Baheti
    Dec 4 at 9:31








1




1




If you know complex numbers, then identify $(x,y)inmathbb{R}^2$ with $x+ytext{i}inmathbb{C}$. Then, $f(z)=z^2$ for all $zinmathbb{C}$. Consequently, the image of the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ is $mathbb{C}setminusmathbb{R}_{leq 0}$.
– Batominovski
Dec 4 at 1:14






If you know complex numbers, then identify $(x,y)inmathbb{R}^2$ with $x+ytext{i}inmathbb{C}$. Then, $f(z)=z^2$ for all $zinmathbb{C}$. Consequently, the image of the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ is $mathbb{C}setminusmathbb{R}_{leq 0}$.
– Batominovski
Dec 4 at 1:14














@Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
– Cathy
Dec 4 at 1:22




@Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
– Cathy
Dec 4 at 1:22




1




1




No, it is not. The negative real axis and the origin are not in the image.
– Batominovski
Dec 4 at 1:23




No, it is not. The negative real axis and the origin are not in the image.
– Batominovski
Dec 4 at 1:23












@Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
– Cathy
Dec 4 at 1:27




@Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
– Cathy
Dec 4 at 1:27












Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
– Sameer Baheti
Dec 4 at 9:31




Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
– Sameer Baheti
Dec 4 at 9:31










1 Answer
1






active

oldest

votes

















up vote
1
down vote













What you have done is fine. You have shown image is the set ${(r^2cos 2theta,r^2sin2theta),|, r >0, -pi/2 < theta < pi/2} = {(r^2cos phi,r^2sinphi),|, r >0, -pi < theta < pi}$. The image, however, can be simplified.



The image is `most' of $mathbb{R}^2$ since $mathbb{R}^2 = {(rcos phi,rsinphi),|, r geq 0, -pi leq theta < pi}$. Now ask yourself, what elements are in $mathbb{R}^2$ that aren't in the image.






share|cite|improve this answer





















  • Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    – Anik Bhowmick
    Dec 4 at 2:52






  • 1




    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    – Timothy Hedgeworth
    Dec 4 at 15:38










  • Sorry, I overlooked the negative part.
    – Anik Bhowmick
    Dec 4 at 22:20











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',
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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024970%2fwhat-is-the-image-of-f-mathbbr2-rightarrow-mathbbr2-fx-y-x2%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








up vote
1
down vote













What you have done is fine. You have shown image is the set ${(r^2cos 2theta,r^2sin2theta),|, r >0, -pi/2 < theta < pi/2} = {(r^2cos phi,r^2sinphi),|, r >0, -pi < theta < pi}$. The image, however, can be simplified.



The image is `most' of $mathbb{R}^2$ since $mathbb{R}^2 = {(rcos phi,rsinphi),|, r geq 0, -pi leq theta < pi}$. Now ask yourself, what elements are in $mathbb{R}^2$ that aren't in the image.






share|cite|improve this answer





















  • Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    – Anik Bhowmick
    Dec 4 at 2:52






  • 1




    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    – Timothy Hedgeworth
    Dec 4 at 15:38










  • Sorry, I overlooked the negative part.
    – Anik Bhowmick
    Dec 4 at 22:20















up vote
1
down vote













What you have done is fine. You have shown image is the set ${(r^2cos 2theta,r^2sin2theta),|, r >0, -pi/2 < theta < pi/2} = {(r^2cos phi,r^2sinphi),|, r >0, -pi < theta < pi}$. The image, however, can be simplified.



The image is `most' of $mathbb{R}^2$ since $mathbb{R}^2 = {(rcos phi,rsinphi),|, r geq 0, -pi leq theta < pi}$. Now ask yourself, what elements are in $mathbb{R}^2$ that aren't in the image.






share|cite|improve this answer





















  • Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    – Anik Bhowmick
    Dec 4 at 2:52






  • 1




    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    – Timothy Hedgeworth
    Dec 4 at 15:38










  • Sorry, I overlooked the negative part.
    – Anik Bhowmick
    Dec 4 at 22:20













up vote
1
down vote










up vote
1
down vote









What you have done is fine. You have shown image is the set ${(r^2cos 2theta,r^2sin2theta),|, r >0, -pi/2 < theta < pi/2} = {(r^2cos phi,r^2sinphi),|, r >0, -pi < theta < pi}$. The image, however, can be simplified.



The image is `most' of $mathbb{R}^2$ since $mathbb{R}^2 = {(rcos phi,rsinphi),|, r geq 0, -pi leq theta < pi}$. Now ask yourself, what elements are in $mathbb{R}^2$ that aren't in the image.






share|cite|improve this answer












What you have done is fine. You have shown image is the set ${(r^2cos 2theta,r^2sin2theta),|, r >0, -pi/2 < theta < pi/2} = {(r^2cos phi,r^2sinphi),|, r >0, -pi < theta < pi}$. The image, however, can be simplified.



The image is `most' of $mathbb{R}^2$ since $mathbb{R}^2 = {(rcos phi,rsinphi),|, r geq 0, -pi leq theta < pi}$. Now ask yourself, what elements are in $mathbb{R}^2$ that aren't in the image.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 at 1:31









Timothy Hedgeworth

1366




1366












  • Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    – Anik Bhowmick
    Dec 4 at 2:52






  • 1




    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    – Timothy Hedgeworth
    Dec 4 at 15:38










  • Sorry, I overlooked the negative part.
    – Anik Bhowmick
    Dec 4 at 22:20


















  • Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    – Anik Bhowmick
    Dec 4 at 2:52






  • 1




    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    – Timothy Hedgeworth
    Dec 4 at 15:38










  • Sorry, I overlooked the negative part.
    – Anik Bhowmick
    Dec 4 at 22:20
















Only the origin isn't in the image, right ??@Timothy_Hedgeworth
– Anik Bhowmick
Dec 4 at 2:52




Only the origin isn't in the image, right ??@Timothy_Hedgeworth
– Anik Bhowmick
Dec 4 at 2:52




1




1




Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
– Timothy Hedgeworth
Dec 4 at 15:38




Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
– Timothy Hedgeworth
Dec 4 at 15:38












Sorry, I overlooked the negative part.
– Anik Bhowmick
Dec 4 at 22:20




Sorry, I overlooked the negative part.
– Anik Bhowmick
Dec 4 at 22:20


















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024970%2fwhat-is-the-image-of-f-mathbbr2-rightarrow-mathbbr2-fx-y-x2%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more