Showing ${Mf(z) \over 10M - f(z)}$ is bounded if $\text{Re}(f(z)) < M$ Data: Let $f(z)$ be analytic on $\Omega - \{a\}$.  Suppose further that $\text{Re}(f(z)) < M$ for some $M \in \mathbb{R}_{> 0}$.
Define $g(z)$ as follows:
$$
g(z) = {Mf(z) \over 10M - f(z)}
$$
Since $\text{Re}(f(z)) < M$, we have that the denominator of $g(z)$ is never zero on $\Omega$ so that $g(z)$ is analytic on $\Omega$.
Question: How do we show that $g(z)$ is bounded by a constant?
Attempt:
$$
|g(z)| = \left|{M f(z) \over 10M - f(z)}\right| = {|Mf(z)| \over |10M - f(z)|} \le {|Mf(z)| \over |9M|} \le {1 \over 9} |f(z)|
$$
which doesn't quite do the trick.

@Robert Israel
We have that
$$
|g(z)| =  {|Mf(z)| \over |10M - f(z)|}  = \underbrace{|-M| + {|10M^2| \over |10M - f(z)|}}_{\text{via (*) below}} \le {10M^2 \over 9M} + M = {10 \over 9} M + M < 3M 
$$
via 
\begin{equation*}
\tag{*}
{M f(z) \over 10M - f(z)} = {-M(10M - f(z)) + 10M^2 \over 10M - f(z)} =  -M + {10M^2 \over 10M - f(z)}
\end{equation*}
However, I have independent reason to believe this bound can actually be made less than $M$ (as opposed to $3M$ below).  How to show this?
