# Showing that this meromorphic function is bounded

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?

• Concerning your own work, remember that you cannot use inequalities when dealing with complex numbers, so writing $c + d \Bbb i \ge (Qe-vf) + (Qf+ve)\Bbb i$ doesn't make sense. Also, using $f$ as both a function and as the imaginary part of the denominator is confusing. – Alex M. Dec 7 '16 at 14:51

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.

• Note that the map above is indeed bounded, but do you not need $f$ to be analytic (as opposed to meromorphic) to apply Liouville? Furthermore, even if $|e^z| \le M$ for $z \in A$, we cannot conclude that $A$ is bounded since $|e^z | = e^{z+it}|$ for all $t$. – copper.hat Dec 6 '16 at 16:16
• (I added a +1 to counteract the silent downvoter.) – copper.hat Dec 6 '16 at 16:17
• @copper.hat: Indeed, the argument is a bit sloppy but essentially correct. If $z_0$ is a pole for $f$, can it also be one for $\Bbb e ^{-f(z)}$? It can't, because $\Bbb e ^{-f(z)}$ is holomorphic on some neighbourhood of $z_0$ (because $f$ is so) and bounded (as shown above), so by Riemann's theorem $z_0$ is in fact a removable singularity, therefore not a pole for $\Bbb e ^{-f(z)}$. This shows that $\Bbb e ^{-f(z)}$ is in fact extendable to a holomorphic function, to which Liouville's theorem may then be applied. – Alex M. Dec 7 '16 at 15:02
• @AlexM.: How does boundedness of $f$ follow from boundedness of $z \mapsto e^{-f(x)}$? – copper.hat Dec 7 '16 at 15:16
• @copper.hat: No, the boundedness and entireness of $\Bbb e ^{-f(z)}$ imply that $\Bbb e ^{-f(z)}$ is a constant. This does not immediately imply that $f$ is constant too, but since outside of its poles (which form a set of isolated points) $f$ is continuous, it is easy to show that it is constant too. – Alex M. Dec 7 '16 at 15:22

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.