Finding the types of singularities of $\oint \frac{\sin(\pi \cdot z)}{(z-1)^2}$

Question

I want to find the types of singularities of $$\oint \frac{\sin(\pi \cdot z)}{(z-1)^2}$$ the point is $z=1$
I know that: $$f(z)=\frac{p(z)}{q(z)},q(a)=0,p(a)\neq 0,p(z)$$ so $p(z)$ analytic in $a$ and $a$ the point in order $m$ if $q'(a)=0$,$\dots $,$q^{(m-1)}$
I would like to get some hints.
Thanks!

Answer 1

Note that

$$\lim_{z\to 1} \frac{\sin{\pi z} - \sin{\pi}}{z-1} = \pi \cos{\pi} = -\pi$$

(i.e., the derivative of $\sin{\pi z}$ at $z=1$). This is independent of direction as $\sin{\pi z}$ is analytic. Thus, the integrand has a simple pole, not a double pole, at $z=1$, and the integral (again, assuming that the contour fully surrounds $z=1$ and is taken in a positive sense) is given by the formula

$$\oint_C dz \frac{f(z)}{z-1} = i 2 \pi f(1)$$

where $f(z) = \sin{\pi z}/(z-1)$.