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







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$endgroup$








  • 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






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$begingroup$

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











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1 Answer
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$begingroup$

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


















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