Discussion
Let \(f : \mathbf{C} \to \mathbf{C}\) be a non-constant analytic such that \(|f'(z)| \lt f(z)|\) for all \(z \in \mathbf{C}\). Then :
Answer: D
The conclusion is \(f'' = cf'\) for some constant \(c\).
Reading the condition as \(|f'(z)| \lt |f(z)|\): if \(f(z_{0}) = 0\) the inequality would say \(|f'(z_{0})| \lt 0\), impossible, so \(f\) never vanishes. Then \(g = f'/f\) is entire and \(|g| \lt 1\), so by Liouville's theorem \(g\) is a constant \(c\). Thus \(f' = cf\), and differentiating, \(f'' = cf'\).
In fact \(f = Ae^{cz}\) with \(c \ne 0\) (as \(f\) is non-constant). So \(1/f\) is analytic, \(f\) is not a polynomial, and \(f' = cAe^{cz}\) never vanishes, which rules out the other options.
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