Showing that this meromorphic function is bounded

Question

I currently have a function $f = a + ib$ that is meromorphic such that there is some $Q \in \mathbb{R}$ such that for all $z \in \mathbb{C},$ $a \geq Q.$ I want to show that this implies that $f$ is bounded. The reason for this is because if $f$ is bounded, I can show that the singularities in $f$ are removable.

I was under the impression that I should take advantage of the fact that $f$ is the quotient of two holomorphic functions, although when I take

$$f = \frac{g}{h} = \frac{c + id}{e + if}$$ $$\implies (u + iv)(e+if) = ue +iuf + ive - vf = c + id$$ $$\implies c + id = (ue - vf) + i(uf + ve) \geq (Qe - vf) + i(Qf + ve).$$

However, now I am not sure where to go to show $f$ is bounded. Any suggestions?

Answer 1

Take $g(z) = {1 \over f(z) +1-Q} $, then $g$ is meromorphic and $|g(z)| \le 1$.

Since $f$ is analytic except for poles, so is $g$, and since $g$ is bounded, any singularities are removable, hence $g$ is entire and hence a constant (thanks to Alex M. for clarifying my muddled thinking). Since $f$ takes some value in $\mathbb{C}$, we must have $g(0) \neq 0$.

Hence $f(z) = {1 \over g(0)} +Q-1$, hence constant.

Answer 2

Hint: For which regions in $\mathbb{C}$ is the function $z \mapsto e^z$ bounded.

Solution: If your requirements are fulfilled, then $f$ is already constant. Look at the map $z \mapsto e^{-f(z)}$. Since $\mathfrak{R}(f)\geq Q$ we have that $| e^{-f(z)}| = |e^{-\mathfrak{R}(f(z))}| \leq e^{-Q}$.

Thus the map $z \mapsto e^{-f(z)}$ is bounded and defined on the whole of $\mathbb{C}$. By Liouville it is constant, in particular bounded.