Determineing the largest number such that the Laurent series of converges for a trig function. Question
How to determine the largest number $R$ such that the Laurent series of
$$f(z)= \dfrac{2sin(z)}{z^2-4} + \dfrac{cos(z)}{z-3i}$$
about $z=-2$ converges for $0<|z+2|<R$?
Attempt :
Its my understanding that with laurent series you have to manipulate the questions in a such way that you can use maclaurin series, for example in this case it would be: 

$\cos (x) = \sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$ 
$\sin (x) = \sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ and 
$\dfrac{1}{1-(x)} = \sum_{n=0}^\infty x^{n}$  $\text{where} \left| x \right| < 1$

But why does $|z+2|$ matter, and what is the meaning of the lowest number, and how do i calculate the series?

The radius of convergence of the Laurent series of a meromorphic function about one of its poles is equal to the distance to the nearest neighboring pole.
In the case of interest, $f(z)=\frac{2\sin(z)}{z^2-4}+\frac{\cos(z)}{z-3i}$, and there are three poles; $z=-2$, $z=3i$, and $z=2$.  
The distances between the pole at $z=-2$ and the other poles are $|-2-2|=4$, and $|-2-3i|=\sqrt{13}<4$.  
Therefore, the Laurent series of $f(z)$ around $z=-2$ has a radius of convergence $R=\sqrt{13}$.
