$f$ entire function with $|f(z)| \leq 1 + |\Im z|$, is $f$ constant?

Question

Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be entire such that for any $z \in \mathbb{C}$, $|f(z)| \leq 1 + |\Im(z)|$. Is $f$ constant?

I think it is linear and non-constant: For $a_{n}$ the $n$-th coefficient in the power series of $f$ about the origin [and $C_{R} := \partial D(0, R)$], we have

$$|a_{n}| \leq \frac{1}{2\pi} \int_{C_{R}} \frac{1 + |\Im(z)|}{|z|^{n+1}} dz \leq \frac{1}{2\pi} \int_{0}^{2\pi} R \cdot \frac{1 + R|\sin t|}{R^{n+1}} dt = \frac{1}{R^{n}} + \frac{2}{\pi R^{n-1}}.$$

Letting $R \rightarrow \infty$, $|a_{n}|$ will be $0$ when $n > 1$. Since I've only bounded above, I can't say anything about $a_{1}$. On the other hand, I can't find an example of a non-constant $f$ satisfying the hypothesis.

Answer 1

$$ |f(z)| \leq 1 + |\Im(z)| \le 1 + |z| $$ implies that $f$ is (at most) linear, i.e. $$ f(z) = a_0 + a_1 z $$ with complex constants $a_0, a_1$. Now choose $z = x \in\Bbb R$ with $x\to \infty$ to conclude that $a_1=0$ and $f$ is constant.

Answer 2

One general tool that applies to such situations is the Phragmen–Lindelöf theorem, which extends the maximum modulus principle to unbounded domains as long as the function in question doesn't grow too fast. In this case, we have $|f(z)| \le 1$ on the boundary of the upper half plane (namely the real axis), and the growth bound is easily enough to apply Phragmen–Lindelöf, whereby $|f(z)| \le 1$ on the entire upper half plane. The same applies to the lower half plane, and so $|f(z)|\le1$ everywhere, hence is constant.