Evaluate the integral $$\int_{|z|=1} \frac{\tan(z)}{z^3}dz, \text{ in counterclockwise orentation.}$$
My approach
There is only one pole inside $|z|=1$,$z=0$, a pole of order 2.
Using Residue theorem, $$\int_{|z|=1} \frac{\tan(z)}{z^3}dz =2\pi i * \text{ sum of residue inside the contours}$$
$\text{Res}(f(z),z=0) = \lim_{z \to0} \frac{d}{dz}(z-0)^2 \frac{\tan(z)}{z^3}=0$
Adding all residues gives $$\int_{|z|=1} \frac{\tan(z)}{z^3}dz=0$$
My question: Are my calculations right?