So I have two questions. 1) Evaluate $$ \oint\limits_{|z|=1} \dfrac{\cos(\pi z^2)}{(z-2)(z-4)^3} dz$$ and 2) Evaluate $$ \oint\limits_{|z|=6} \dfrac{\cos(\pi z^2)}{(z-2)(z-4)^3} dz.$$ Now I know the integrand is analytic when $ z\neq 2, 4$ so for 1) the integrand is analytic on $|z|=1$, do we not use cauchy's integral then?
I'm just confused because I only really understand the case for when one of the poles makes the integrand non-analytic. That's why for 2) I'm stuck as well because both $z= 2, 4 $ are included in $ |z|=6 $. We haven't learned the residue theorem yet either. Can anyone help me out?