Discussion
Let \(f\) be a non-constant analytic function on the unit disc. Then which of the following is impossible ?
Answer: D
The impossible case is \(f(G) = \{z : |z| = 1\}\) (G being the unit disc).
By the open mapping theorem, a non-constant analytic function maps open sets to open sets, so \(f(G)\) is open. The unit circle contains no open disc, so it cannot be the image.
The other images are possible: \(f(z) = z + 2\) maps the disc into \(\mathrm{Re}(z) \ge 1\); \(f(z) = 2z\) maps it onto \(|z| \lt 2\); and the open set \(|z| \gt 1\) is also the image of a suitable analytic map (for instance \(e^{g(z)}\) with \(g\) mapping the disc onto the half plane \(\mathrm{Re} \gt 0\)).
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