Discussion
Which of the following continuous functions is not an open mapping from \(\mathbf{C}\) to \(\mathbf{C}\) ?
Answer: B
The function that is not an open mapping is \(f(z) = |z|\).
By the open mapping theorem, every non-constant analytic function on a domain is open. \(e^{z^{2}}\), \(\sin z + \cos z^{2}\) and the series \(1 - z^{2}/2! + z^{4}/4! - \cdots = \cos z\) are non-constant entire functions, so they are open maps.
\(|z|\) is continuous but takes only real, non-negative values. The image of any open disc is a subset of the real line, which contains no open set of \(\mathbf{C}\), so the map is not open. (It is also not analytic, so the theorem does not apply.)
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