I'm learning about different types of singularities in complex analysis, and stumbled upon a problem with finding what order of pole $f(z)=\frac{\sin(z)}{z^3}$ have at $z_0=0$.
My solution is
$$\lim_{z \to 0} ~(z-z_0)^3 \frac{\sin(z)}{z^3} = \sin(0)$$
which according to me should give me a pole of order 3, however, the correct answer is 2. Why so? Would anyone mind explain?
If we let $g(z)=\frac {\sin z}{z}$, then
$$f(z)=\frac{g(z)}{z^2}$$
Can you show $z=0$ in $g(z)$ is actually a removable singularity? Hence, $z=0$ is a second order pole for $f(z)$.