Let $\rho(x)$ be a holomorphic function on a disk $D \subseteq \mathbb{C}$ with the property that $\rho(x) \notin \mathbb{N^*} = \{1,2,\dots\}$ on $D$.
Prove the following: There exists an $R$ sufficiently small so that in $D_R = \{x : |x| \leq R \}$, $$ |m-\rho(x)| \geq\sigma m, \qquad m=1,2,\dots$$ for some $\sigma>0$.
The requirement can be written as $|1-{\rho(x) \over m} | \ge \sigma$, for $m=1,2...$.
By continuity, $\rho$ is bounded on the compact disk $\bar{B}(0,1)$ and so for some $M$ we have $|1-{\rho(x) \over m} | \ge {1 \over 2}$ for all $ x \in \bar{B}(0,1)$ and $m \ge M$.
Now let $\phi(x) = \max(|1-{\rho(x) \over 1}|,...,|1-{\rho(x) \over M-1} |)$ and note that $\phi(0) >0$ and $\phi$ is continuous.
Hence there is some $\delta>0$ such that $\phi(x) \ge {\phi(0) \over 2}$ for all $x$ satisfying $|x| < \delta$.
Now pick $R = {\delta \over 2}$ and $\sigma = \min({\phi(0) \over 2}, {1 \over 2})$.