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












1












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    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 $mathbb{C}setminusmathbb{R}_{leq 0}$.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:14












  • $begingroup$
    @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:22






  • 1




    $begingroup$
    No, it is not. The negative real axis and the origin are not in the image.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:23










  • $begingroup$
    @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:27










  • $begingroup$
    Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    $endgroup$
    – Sameer Baheti
    Dec 4 '18 at 9:31
















1












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    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 $mathbb{C}setminusmathbb{R}_{leq 0}$.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:14












  • $begingroup$
    @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:22






  • 1




    $begingroup$
    No, it is not. The negative real axis and the origin are not in the image.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:23










  • $begingroup$
    @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:27










  • $begingroup$
    Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    $endgroup$
    – Sameer Baheti
    Dec 4 '18 at 9:31














1












1








1





$begingroup$


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











$endgroup$




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 '18 at 1:42









Saad

19.7k92352




19.7k92352










asked Dec 4 '18 at 1:06









Cathy Cathy

15618




15618








  • 1




    $begingroup$
    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 $mathbb{C}setminusmathbb{R}_{leq 0}$.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:14












  • $begingroup$
    @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:22






  • 1




    $begingroup$
    No, it is not. The negative real axis and the origin are not in the image.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:23










  • $begingroup$
    @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:27










  • $begingroup$
    Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    $endgroup$
    – Sameer Baheti
    Dec 4 '18 at 9:31














  • 1




    $begingroup$
    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 $mathbb{C}setminusmathbb{R}_{leq 0}$.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:14












  • $begingroup$
    @Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:22






  • 1




    $begingroup$
    No, it is not. The negative real axis and the origin are not in the image.
    $endgroup$
    – Batominovski
    Dec 4 '18 at 1:23










  • $begingroup$
    @Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
    $endgroup$
    – Cathy
    Dec 4 '18 at 1:27










  • $begingroup$
    Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
    $endgroup$
    – Sameer Baheti
    Dec 4 '18 at 9:31








1




1




$begingroup$
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 $mathbb{C}setminusmathbb{R}_{leq 0}$.
$endgroup$
– Batominovski
Dec 4 '18 at 1:14






$begingroup$
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 $mathbb{C}setminusmathbb{R}_{leq 0}$.
$endgroup$
– Batominovski
Dec 4 '18 at 1:14














$begingroup$
@Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
$endgroup$
– Cathy
Dec 4 '18 at 1:22




$begingroup$
@Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
$endgroup$
– Cathy
Dec 4 '18 at 1:22




1




1




$begingroup$
No, it is not. The negative real axis and the origin are not in the image.
$endgroup$
– Batominovski
Dec 4 '18 at 1:23




$begingroup$
No, it is not. The negative real axis and the origin are not in the image.
$endgroup$
– Batominovski
Dec 4 '18 at 1:23












$begingroup$
@Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
$endgroup$
– Cathy
Dec 4 '18 at 1:27




$begingroup$
@Batominovski So the image is the whole plane except for origin and negative real axis? Why they're not in the image?
$endgroup$
– Cathy
Dec 4 '18 at 1:27












$begingroup$
Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:31




$begingroup$
Image is the half plane $big{zinmathbb{C},big|,text{Re}(z)>0big}$ because of what you have written in the question.($xgt 0$)
$endgroup$
– Sameer Baheti
Dec 4 '18 at 9:31










1 Answer
1






active

oldest

votes


















1












$begingroup$

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









$endgroup$













  • $begingroup$
    Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 2:52






  • 1




    $begingroup$
    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    $endgroup$
    – Timothy Hedgeworth
    Dec 4 '18 at 15:38










  • $begingroup$
    Sorry, I overlooked the negative part.
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 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',
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
});


}
});














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









1












$begingroup$

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









$endgroup$













  • $begingroup$
    Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 2:52






  • 1




    $begingroup$
    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    $endgroup$
    – Timothy Hedgeworth
    Dec 4 '18 at 15:38










  • $begingroup$
    Sorry, I overlooked the negative part.
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 22:20
















1












$begingroup$

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









$endgroup$













  • $begingroup$
    Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 2:52






  • 1




    $begingroup$
    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    $endgroup$
    – Timothy Hedgeworth
    Dec 4 '18 at 15:38










  • $begingroup$
    Sorry, I overlooked the negative part.
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 22:20














1












1








1





$begingroup$

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









$endgroup$



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 '18 at 1:31









Timothy HedgeworthTimothy Hedgeworth

1366




1366












  • $begingroup$
    Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 2:52






  • 1




    $begingroup$
    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    $endgroup$
    – Timothy Hedgeworth
    Dec 4 '18 at 15:38










  • $begingroup$
    Sorry, I overlooked the negative part.
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 22:20


















  • $begingroup$
    Only the origin isn't in the image, right ??@Timothy_Hedgeworth
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 2:52






  • 1




    $begingroup$
    Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
    $endgroup$
    – Timothy Hedgeworth
    Dec 4 '18 at 15:38










  • $begingroup$
    Sorry, I overlooked the negative part.
    $endgroup$
    – Anik Bhowmick
    Dec 4 '18 at 22:20
















$begingroup$
Only the origin isn't in the image, right ??@Timothy_Hedgeworth
$endgroup$
– Anik Bhowmick
Dec 4 '18 at 2:52




$begingroup$
Only the origin isn't in the image, right ??@Timothy_Hedgeworth
$endgroup$
– Anik Bhowmick
Dec 4 '18 at 2:52




1




1




$begingroup$
Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
$endgroup$
– Timothy Hedgeworth
Dec 4 '18 at 15:38




$begingroup$
Not quite, since setting $theta = -pi$ in $mathbb{R}^2$ gives the set ${(-r,0),|,rgeq 0}$ and this is not in the image!
$endgroup$
– Timothy Hedgeworth
Dec 4 '18 at 15:38












$begingroup$
Sorry, I overlooked the negative part.
$endgroup$
– Anik Bhowmick
Dec 4 '18 at 22:20




$begingroup$
Sorry, I overlooked the negative part.
$endgroup$
– Anik Bhowmick
Dec 4 '18 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.




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

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more