all circles positively oriented. Use Cauchy's Integration Formulas to compute each of the following integrals. HINT: use partial fractions where appropriate.

a) $\displaystyle \int_{|z-2|=2} \frac{cosz}{(z-3)^2(z+1)}dz$

b) $\displaystyle \int_{|z+2|=1} \frac{sinz}{z^2(z-1)^2}dz$

c) $\displaystyle \int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz$ , t>0

The Cauchy integral formula : $\displaystyle \int_C \frac{f(z)dz}{z-z_o}=2\pi i f(z_o)$

NOW FOR c) in the denominator -1 is interior to C so i go as follows,

$\displaystyle \int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz=\int_{|z+2|=2} \frac{ze^{zt}}{(z-(-1))^{2+1}}dz$ so n=2 in the extension of Cauchy's integral formula given as $\displaystyle \int_C \frac{f(z)dz}{(z-z_0)^{n+1}}=\frac{2\pi i}{n!}f^{(n)}(z_0)$

so i let f(z) = $\displaystyle ze^{zt}$ then f'(z) = $\displaystyle e^{zt}+zte^{zt}$ and f ''(z) = $\displaystyle 2te^{zt}+zt^2e^{zt} = e^{zt}(2t+zt^2)$

hence f '' (-1) =$\displaystyle e^{-t}(2t-t^2) $and then by the extended Cauchy integral formula i get

$\displaystyle \int_{|z+2|=2} \frac{ze^{zt}}{(z+1)^3}dz = \frac{2\pi i}{2} e^{-t}(2t-t^2)= \pi i e^{-t}(2t-t^2)$ , is this correct ?

OK, things i don't know :

Now in a) neither 3 nor -1 are interior to C so neither of them are my Zo point. The same goes for b) in the denominator neither 0 nor 1 are interior to C, so where do i find Zo ? Now if the answer is to use partial fraction decomposition, my attempt :

for a) $\displaystyle \frac{cosz}{(z-3)^2(z+1)}= \frac{A}{z-3}+\frac{B}{(z-3)^2}+\frac{C}{z+1}$

$\displaystyle cosz = A(z-3)(z+1)+B(z+1)+C(z-3)^2$

$\displaystyle cosz=z^2(A+C)+z(-2A+B-6C)+(-3A+B+9C)$ and now what ? How do i equate anything??? I have the same confusion for b). I aprreciate any help.