Find number of zeros of $\sin \pi x$ on a domain $D=\{|z-3-4i|<6\}$. I am going to take written exam in complex analysis in a week. 
Among sample problems I found the following one:
Find $\lim \limits_{n \to \infty}N_{P_n}(D)$ where $N$ is number of zeros of $P_n$ on a  domain, $D=\{|z-3-4i|<6\}$ and $$P_n = \sum_{k=0}^n (-1)^k \frac{(\pi z)^{2k+1}}{(2k+1)!}$$
Update: as it was pointed out by Wauzl and  Arnaud D. $\lim \limits_{n \to \infty}N_{P_n}(D)$ is in fact $\sin \pi x$. I checked another samples, and they are all in fact have the form $\sin ax$ or $\cos ax$ or $e^{ax}$. So I have to be able to find 
number of zeros of these functions in $D$.
Now I guess I have to use Rouché_theorem. $P$ is actually sum of exponents.
Unfortunately, I do not see how to apply the theorem here since $\sin \pi x$ has no polynomial part.
Any ideas?
Thanks a lot for your help and advices!

Notice that $\sin(\pi x)$ only has zeroes when $x$ is a real integer (source), so you just need to figure out how many real integers are in the open disk you've mentioned.
Similarly for $\cos(\pi x)$ have a zero only when $x+\frac12$ is an integer, so you need to check how many of these are in the disk.
For $e^{ax}$ just notice that the exponential function has no zeroes whatsoever.
Now there's a potential pitfall in that the polynomials, that are used to approximate the functions, always have some roots. However when approximating for example the exponential function with taylor series, the roots are moved further and further away from the origin each iteration, so after some number of iterations, the polynomial will be close enough to the exponential function that there really are no roots in the disk $D$.
