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**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.

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 \} $