Correct answer: (C) Every such entire f is zero-free.
Explanation
Define entire $g(z)=\overline{f(\bar z)}$. On the real axis f(z)g(z)=|f(z)|²=1. The identity theorem gives fg≡1 on ℂ, so f has no zeros. **C**. The function e^(iz) is a nonconstant entire example, refuting B; a polynomial with bounded modulus on the whole real axis must be constant.
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