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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add a comment |
$begingroup$
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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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 $mathbb{C}setminusmathbb{R}_{leq 0}$.
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– Batominovski
Dec 4 '18 at 1:14
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@Batominovski I don't know complex number. Is the image the whole plane of $R^2$?
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– Cathy
Dec 4 '18 at 1:22
1
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No, it is not. The negative real axis and the origin are not in the image.
$endgroup$
– Batominovski
Dec 4 '18 at 1:23
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@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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
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.
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$begingroup$
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
add a comment |
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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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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