$f(z)=\frac{\sin^{2014}(z) e^{\frac{1}{(z-10)^2}}}{(z-2i)^8(z-2014\pi)^2}$ Find the type of singularity $z=2014\pi$

Question

$$f(z)=\frac{\sin^{2014}(z) e^{\frac{1}{(z-10)^2}}}{(z-2i)^8(z-2014\pi)^2}$$ Find the type of singularity $z=2014\pi$?
My attempt:

I try put and see what is $\lim_{z \rightarrow 2014\pi} |f(z)|$ but I get $\frac{0}{0}$ which isn't determined. And I also see that isn't idea to write Laurent series of $f$ at $z=2014\pi$ beacause it is too much job.

Answer 1

It has a removable singularity there, since\begin{align}\lim_{z\to2014\pi}\frac{\sin^{2014}(z)e^{1/(z-10)^2}}{(z-2i)^8(z-2014\pi)^2}&=\lim_{z\to2014\pi}\left(\frac{\sin(z)-\overbrace{\sin(2014\pi)}^{\phantom{0}=0}}{z-2014\pi}\right)^2\times\frac{\sin^{2012}(z)e^{1/(z-10)^2}}{(z-2i)^8}\\&=1\times0\\&=0.\end{align}