Zeroes about entire function Given an entire function $f(z)$ which satisfies $|f(z)|=1, \forall z \in \mathbb{R}$, the problem asks to show that there exists an entire function $g(z)$ such that $f(z)=\exp(g(z))$.
The only thing need to show is that $f(z)$ admits no zeros on $\mathbb{C}$ so that we can define $g(z)$ by a standard single-valued branch argument of logarithm. But does the assumption $|f(z)|=1, \forall z \in \mathbb{R}$ is so strong that entire functions satisfying this cannot have zeroes? Intuitive we can take an example of $f(z)=\exp(iz)$ where $\infty$ is its essential singularity so it is hard to expect we can turn the real line into the boundary of unit circle and then use somekind of Maximum Principle, etc. Basically I do not get the picture of what the assumption is talking about.

Consider the function $h(z) = \overline{f(\overline{z})}$. That is an entire function too, and hence so is $k(z) = f(z)\cdot h(z)$. On the real line, you have
$$k(x) = f(x)\cdot h(x) = f(x) \overline{f(x)} = \lvert f(x)\rvert^2 = 1,$$
hence $k \equiv 1$. That guarantees that $f$ has no zeros.
