Discussion
Consider the function \(f : \mathbf{R}^{2} \to \mathbf{R}^{2}\) defined as \(f(x, y) = (x^{2} - y^{2}, 2xy)\). Then the image of the ball \(B(P, \frac{1}{2})\), where P = (1, 0) is the centre and \(\frac{1}{2}\) is radius, is :
Answer: B
The image is an open set.
In complex notation \(f(z) = z^{2}\), since \((x + iy)^{2} = x^{2} - y^{2} + 2ixy\). The ball centred at 1 with radius 1/2 does not contain 0, so \(f'(z) = 2z \neq 0\) there. A non-constant analytic function is an open map (open mapping theorem), so the image of an open ball is open.
It is bounded, since \(|z| \lt 3/2\) gives \(|z^{2}| \lt 9/4\), so D is wrong. An open, non-empty, bounded proper subset of the plane cannot be closed, which rules out A and C.
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