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Math Help - Image of set

  1. #1
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    Image of set

    Let  f(x,y) = (x^2 - y^2, 2xy) where f is defined on the open set  A = \{ (x,y) \in \mathbb{R}^2 : x > 0 \} .

    Question: What is the set f(A)?

    I'm guessing they're asking me to describe f(A) using set notation. But I don't know how to do this. I've tried plotting a set of points to get an idea what f(A) would look like on the plane, but I can't get any idea whatsoever.

    Any help or ideas?
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  2. #2
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    Quote Originally Posted by JG89 View Post
    Let  f(x,y) = (x^2 - y^2, 2xy) where f is defined on the open set  A = \{ (x,y) \in \mathbb{R}^2 : x > 0 \} .

    Question: What is the set f(A)?

    I'm guessing they're asking me to describe f(A) using set notation. But I don't know how to do this. I've tried plotting a set of points to get an idea what f(A) would look like on the plane, but I can't get any idea whatsoever.

    Any help or ideas?


    Hint: prove that for any (a,b)\in\mathbb{R}^2 there exist x,y\in\mathbb{R} s.t. 2xy=b\,\,and\,\,x^2-y^2=a . This solves your problem.

    Tonio
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  3. #3
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    Tonio, this isn't true. Consider the origin (0,0). If there is x, y such that  2xy = 0, x^2 - y^2 = 0 then we must have  y = 0 in order for  2xy = 0 to hold, since x > 0. Thus  x^2 - y^2 = x^2 > 0 . So f will never map to the origin.

    How about this:

    Think about any point (x,y) in A as the complex number  x + iy = re^{i \theta} where  - \pi / 2 < \theta < \pi / 2 .Then  f(x+iy) = (x+iy)^2 = (re^{i\theta})^2 = r^2 e^{i2\theta} . Let  r' = r^2, 2\theta = \phi . Then  r^2 e^{i 2 \theta} = r' e^{i \phi} where  - \pi < \phi < \pi .

    Thus  f(A) = \{(x,y) \in \mathbb{R}^2 : (x,y) \neq (x, 0) for x <= 0 \}
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  4. #4
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    Quote Originally Posted by JG89 View Post
    Tonio, this isn't true. Consider the origin (0,0). If there is x, y such that  2xy = 0, x^2 - y^2 = 0 then we must have  y = 0 in order for  2xy = 0 to hold, since x > 0. Thus  x^2 - y^2 = x^2 > 0 . So f will never map to the origin.

    How about this:

    Think about any point (x,y) in A as the complex number  x + iy = re^{i \theta} where  - \pi / 2 < \theta < \pi / 2 .Then  f(x+iy) = (x+iy)^2 = (re^{i\theta})^2 = r^2 e^{i2\theta} . Let  r' = r^2, 2\theta = \phi . Then  r^2 e^{i 2 \theta} = r' e^{i \phi} where  - \pi < \phi < \pi .

    Thus  f(A) = \{(x,y) \in \mathbb{R}^2 : (x,y) \neq (x, 0) for x <= 0 \}

    Oops! I didn't see the condition x>0 ....jeje

    Tonio
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