1. ## 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?

2. Originally Posted by JG89
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

3. 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.

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

4. Originally Posted by JG89
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.

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