Solution:

Let

$ \begin{aligned}I=\oint_{c} \frac{\sin \pi z^{2}+\cos \pi z^{2}}{(z-1)(z-2)^2} d z \end{aligned}$

Poles of the integrand are given by putting the denominator equal to zero.

$$ (z-1)(z-2)^2=0 \Rightarrow z=1,2 $$

The integrand has two poles at $z=1,2$.

both the poles $z=1$, and $z=2$.

$\begin{aligned} \oint_{c} \frac{\sin \pi z^{2}+\cos \pi z^{2}}{(z-1)(z-2)^2} d z =& \oint \frac{f(z)dz}{(z-1)(z-2)^2} \end{aligned} $

Using partial fraction

$\begin{aligned} \frac{1}{(z-1)(z-2)^2} =\frac{A}{(z-1)}+\frac{B}{(z-2)} &+\frac{C}{(z-2)^2} \end{aligned}$

By solving it we get,

$A=-1,B=\frac{1}{2},C=\frac{1}{2}$

$ \begin{aligned}I= &\oint{\frac{-1f(z)dz}{(z-1)}+\frac{f(z)dz}{2(z-2)}+\frac{f(z)dz}{2(z-2)^2}} \\ I=I_1+I_2+I_3 \end{aligned} $

Applying Cauchy's integral formulae

$$\begin {aligned} \oint \frac{f(z)dz}{z-a} &=2 \pi if(a) \\ \oint \frac{f(z)dz}{(z-a)^2} &=2 \pi if'(a) \\ \end{aligned} \space $$

$\begin {aligned} I_1 &=\oint \frac{-1f(z)dz}{(z-1)} \\ &=-2 \pi i[sin \pi + cos \pi ]\\ &=2 \pi i \end{aligned}$

$\begin {aligned} I_2 &=\oint \frac{f(z)dz}{2(z-2)} \\ &=2 \times (2\pi i[sin 4\pi + cos4\pi ])\\ &=4 \pi i \end{aligned}$

$\begin {aligned} I_3 &=\oint \frac{f(z)dz}{2(z-2)^2} \\ &=2 \times {2\pi i[4\pi (cos 4\pi - sin4\pi ) ]}\\ &=16 \pi^2 i \end{aligned} $

$\begin {aligned} I &=\oint_{c} \frac{\sin \pi z^{2}+\cos \pi z^{2}}{(z-1)(z-2)^2} d z \\ &= 2 \pi i+4 \pi i+16 \pi^2 i \\ &= 2 \pi i(3+8\pi ) \end{aligned} $