$$f(z)=\cot z - \frac{1}{z}=\frac{z \cos z - \sin z}{z ~ \sin z}$$ having poles at $z=n \pi$ with residue $1$ at each pole. Also $f(z)$ has a removable singularity at $z=0.$ All these are asked to calculate and I've done that. If $C_N$ is the circle enclosing all the poles having the center at the origin and radius $R_N,$ how can I show that $f$ is bounded on $C_N$ for a suitable $R_N$ ?
Any help is much appreciated.
Edit: 
Note that
Choose $R_N = (\pi/2)(2N + 1)$ so that $C_N$ does not intersect the poles. We have $$\vert f(z) \vert \leq \vert \cot z \vert + \vert 1/z \vert.$$ The $1/z$ part can be made arbitrarily small by choosing a sufficiently large $R_N$. You can show $\vert cot z \vert \leq \vert \coth y \vert$, where $y = \textrm{Im}(z)$. Now if $\vert y \vert > \vert y' \vert$ then $\vert \coth y \vert < \vert \coth y' \vert$, so we can bound $\cot z$ on the parts of $C_N$ where $\textrm{Im}(z) \geq y$ for some chosen $y > 0$. The problem is that $\vert \coth y \vert$ approaches $\infty$ as $y \to 0$. However, this is easy to circumvent because $\cot z$ is periodic. Simply choose a closed square centered at $\pi/2$ which avoids the poles $0$ and $\pi$, then take the max of $\vert \cot z \vert$ on this square. By periodicity, this bound will work near the real points of $C_N$ for any $N \geq 0$.