Integrate $\int{ \frac{\sin(\pi z^2)+\cos(\pi z^2)}{\{(z-1)(z-2)\}^{4}} dz }$

Question

I want to evaluate $$\int{ \frac{\sin(\pi z^2)+\cos(\pi z^2)}{\{(z-1)(z-2)\}^{4}} dz }.$$

This is the contour integration I came across. I know Cauchy's integral formula and Cauchy's integral formula for higher derivatives. First I separate $\cos$ and $\sin$ terms. Then I cosider $f(z)= \frac{\sin(\pi z^2)}{(z-1)^4}$ and z.$=-2$ and then I use Cauchy's formula somehow. Am I right in my approach? If not, how do I proceed?

Answer 1

Expand $$ \frac{1}{\{(z-1)(z-2)\}^4} $$ in terms of partial fractions.

http://www4c.wolframalpha.com/Calculate/MSP/MSP21321ie0f5d72de38918000019cf96d2g3bad01f?MSPStoreType=image/gif&s=41&w=513.&h=78.

Then multiply the numerator and integrate each term using the Cauchy integral formula or the Cauchy Goursat theorem, depending on which of the points $z =1 $ and $z = 2$ are enclosed by your contour.