# does it come from cauchy integral formulae

Well I have to do the integral $$\int_{|z|=4}\frac{f(z) dz}{(z-2)(z-3)}$$

where $f(z)=\sin(\pi z^2)+\cos(\pi z^2)$

one breaks like: $$\int_{|z|=4}\frac{f(z) dz}{(z-2)(z-3)}$$ $$=\int_{|z-2|=1/2}\frac{f(z)/(z-3) dz}{(z-2)}$$ $$+\int_{|z-3|=1/2}\frac{f(z)/(z-2) dz}{(z-3)}$$

why it is true?Is there any easier method?

-
Recall the main trigonometric identity it holds even for complex arguments. –  Norbert Aug 25 '12 at 22:36
"Easier method"? Easier than that? I doubt it. –  DonAntonio Aug 25 '12 at 22:44
ok then, could any one tell me why we can break the integral like that? –  miosaki Aug 25 '12 at 22:45
If you "connect" the original path $\,|z|=4\,$ with those two paths $\,|z-2|=1/2\,,\,|z-3|=1/2\,$ via straight lines and you integrate forth and back those straight segments and around the new paths, the integral over each straight segment cancels out (both directions!) and you remain only with the integrals over the new paths, outside of which the integral equals zero as $\,\frac{1}{(z-2)(z-3)}\,$ is analytic.
Thus, the integral equals the sum of $$\oint_{|z-2|=1/2}\frac{dz/(z-3)}{z-2}=2\pi i\left.\left(\frac{1}{z-3}\right)\right|_{z=2}=-2\pi i$$ $$\oint_{|z-3|=1/2}\frac{dz/(z-2)}{z-3}=2\pi i\left.\left(\frac{1}{z-2}\right)\right|_{z=3}=2\pi i$$ All in all, we get that $$\oint_{|z|=4}\frac{dz}{(z-2)(z-3)}=-2\pi i+2\pi i =0$$