Discussion
Let \(f : \mathbb{C} \to \mathbb{C}\) be an entire function with \(f(0) = 1\), \(f(1) = 2\) and \(f'(0) = 0\). If there exists \(M \gt 0\) such that \(|f''(z)| \le M\) for all \(z \in \mathbb{C}\), then \(f(2) =\)
Answer: B
\(f(2) = 5\).
Since \(f\) is entire, \(f''\) is entire too, and it is bounded by \(M\) on the whole plane. By Liouville's theorem \(f''\) is constant, so \(f\) is a polynomial of degree at most 2: \(f(z) = az^{2} + bz + c\).
\[ \begin{aligned} f(0) = 1 &\Rightarrow c = 1 \\ f'(0) = 0 &\Rightarrow b = 0 \\ f(1) = 2 &\Rightarrow a + 1 = 2,\ a = 1 \end{aligned} \]
So \(f(z) = z^{2} + 1\) and \(f(2) = 5\). The value is real because every coefficient is real, which rules out \(2 + 5i\) and \(5 + 2i\); 2 is what you get by taking \(f\) linear and ignoring the quadratic term.
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