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











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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







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  • 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






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share|cite|improve this question













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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
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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











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up vote
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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


















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