Results 1 to 4 of 4

Thread: Image of set

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    130

    Image of set

    Let $\displaystyle f(x,y) = (x^2 - y^2, 2xy) $ where f is defined on the open set $\displaystyle 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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by JG89 View Post
    Let $\displaystyle f(x,y) = (x^2 - y^2, 2xy) $ where f is defined on the open set $\displaystyle 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 $\displaystyle (a,b)\in\mathbb{R}^2$ there exist $\displaystyle x,y\in\mathbb{R}$ s.t. $\displaystyle 2xy=b\,\,and\,\,x^2-y^2=a$ . This solves your problem.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2009
    Posts
    130
    Tonio, this isn't true. Consider the origin (0,0). If there is x, y such that $\displaystyle 2xy = 0, x^2 - y^2 = 0 $ then we must have $\displaystyle y = 0 $ in order for $\displaystyle 2xy = 0 $ to hold, since x > 0. Thus $\displaystyle 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 $\displaystyle x + iy = re^{i \theta} $ where $\displaystyle - \pi / 2 < \theta < \pi / 2 $.Then $\displaystyle f(x+iy) = (x+iy)^2 = (re^{i\theta})^2 = r^2 e^{i2\theta} $. Let $\displaystyle r' = r^2, 2\theta = \phi $. Then $\displaystyle r^2 e^{i 2 \theta} = r' e^{i \phi} $ where $\displaystyle - \pi < \phi < \pi $.

    Thus $\displaystyle f(A) = \{(x,y) \in \mathbb{R}^2 : (x,y) \neq (x, 0) for x <= 0 \} $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by JG89 View Post
    Tonio, this isn't true. Consider the origin (0,0). If there is x, y such that $\displaystyle 2xy = 0, x^2 - y^2 = 0 $ then we must have $\displaystyle y = 0 $ in order for $\displaystyle 2xy = 0 $ to hold, since x > 0. Thus $\displaystyle 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 $\displaystyle x + iy = re^{i \theta} $ where $\displaystyle - \pi / 2 < \theta < \pi / 2 $.Then $\displaystyle f(x+iy) = (x+iy)^2 = (re^{i\theta})^2 = r^2 e^{i2\theta} $. Let $\displaystyle r' = r^2, 2\theta = \phi $. Then $\displaystyle r^2 e^{i 2 \theta} = r' e^{i \phi} $ where $\displaystyle - \pi < \phi < \pi $.

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

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

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pre-Image / Image proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Mar 9th 2012, 02:30 AM
  2. Image of T
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 23rd 2010, 12:44 PM
  3. Replies: 1
    Last Post: Oct 27th 2009, 12:09 AM
  4. image set
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: Oct 26th 2009, 02:59 PM
  5. Replies: 3
    Last Post: Sep 16th 2009, 04:17 PM

Search Tags


/mathhelpforum @mathhelpforum