Show that there cannot be an entire function that satisfy $|z+(\cos (z)-1)f(z)|\leq 7$

Question

Show that there cannot be an entire function $f(z)$ that satisfy $|z+(\cos (z)-1)f(z)|\leq 7$.

I thought about showing somehow that $f(z)$ is bounded and by Liouville's theorem it is constant, then the inequality does not hold for all $z\in \mathbb{C}$.

But I'm not sure how to do it or if it is right.

Answer 1

Suppose $f$ is entire. Then $g(z) = z + ( \cos z - 1 ) f(z)$ is entire also. By Liousville's theorem, the given inequality $|g(z)| \leq 7$ implies that $g$ must be a constant, say $g(z) = c$. But $g(z) = c$ implies that $f(z) = \frac{c - z}{\cos z - 1}$, and this is a problem, for $\cos z - 1 = 0 \iff z \in \{ 2\pi n \;|\; n \in \mathbb{Z} \}$. That is, $f$ has poles at $\{ 2\pi n \;|\; n \in \mathbb{Z} \}$, contradicting that it is entire.

Answer 2

If such a function existed, $z+\bigl(\cos(z)-1\bigr)f(z)$ would be a constant $k$, by Liouville's theorem. Actually, it would be $0$, since$$k=0+\bigl(\cos(0)-1\bigr)f(0)=0.$$So,$$0=2\pi+\bigl(\cos(2\pi)-1\bigr)f(2\pi)=2\pi,$$which is absurd.