If $f$ is an entire function such that for each $\theta$, $|f(re^{i\theta})|\rightarrow \infty$ as $r\rightarrow \infty$

Here is a simpler example of a transcendental entire function $f : \mathbb{C} \rightarrow \mathbb{C}$ such that $f(r e^{i \theta}) \underset{r \rightarrow +\infty}{\longrightarrow} \infty$ for every $\theta \in \mathbb{R}$.

Consider the entire function $f : z \mapsto e^{z} +e^{-z} +z^{2}$.

It is obvious that $f$ is not polynomial - one can easily write its power series expansion or check that none of its derivatives is identically zero.

Furthermore, one can check that for all $\theta \in \mathbb{R}$, $f(r e^{i \theta}) \underset{r \rightarrow +\infty}{\longrightarrow} \infty$.
Indeed, let $\theta \in \mathbb{R}$.
Notice that since for all $z \in \mathbb{C}$, $f(z) = f(-z)$, we can assume that $\theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

• If $\theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$, then $$\left|f(r e^{i \theta})\right| \geq e^{r \cos(\theta)} - e^{-r \cos(\theta)} -r^{2} \underset{r \rightarrow +\infty}{\longrightarrow} +\infty$$

• If $\theta = \frac{\pi}{2}$, then $$\left|f(r e^{i \theta})\right| = |f(i r)| = \left|2 \cos(r) -r^{2}\right| \geq r^{2} -2 \underset{r \rightarrow +\infty}{\longrightarrow} +\infty$$