Discussion
Consider the following condition on a function \(f : \mathbb{C} \to \mathbb{C}\)
\(|f(z)| = 1\) for all \(z \in \mathbb{C}\) such that \(\text{Im}(z) = 0\). (P)
Which one of the following is correct?
Answer: C
The correct statement is Every entire function \(f\) satisfying (P) has no zeroes in \(\mathbb{C}\).
Let \(g(z) = f(z)\overline{f(\overline{z})}\). The function \(z \mapsto \overline{f(\overline{z})}\) is entire, so \(g\) is entire, and for real \(z\), \(g(z) = |f(z)|^{2} = 1\). An entire function equal to 1 on the real line is identically 1, so \(f(z)\overline{f(\overline{z})} = 1\) everywhere and \(f\) can never vanish.
This rules out the option with infinitely many zeroes. \(f(z) = e^{iz}\) satisfies (P) and is not constant, so 'every such function is constant' is false. A non-constant polynomial has \(|p(x)| \to \infty\) as \(x \to \infty\) along the real axis, so no non-constant polynomial satisfies (P).
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