I want to compute the integrals $\int_{|z|=r}\frac{dz}{(z+1-i)(z-3-2i)^{4}} $ where $r\in {1,2,3} $ with Cauchy's integral formula but I'm not sure if I did it correctly.
I started by looking where the denominator is undefined which is $z=-1+i$ or $z=3+2i $ then looking at the integral for each $r$. $$\int_{|z|=1}\frac{dz}{(z+1-i)(z-3-2i)^{4}} $$ It is analytic everywhere inside the circle so the integral should be $0$ due to Cauchy's theorem.
$$\int_{|z|=2}\frac{dz}{(z+1-i)(z-3-2i)^{4}} $$ It fails to be analytic at the point $z=-1+i$ so I put $$f(z)=\dfrac{1}{(z-3-2i)^{4}} $$ and use Cauchy's integral formula: $$f(z_0)=\dfrac{1}{2\pi i}\int \dfrac{f(z)}{z-z_0}dz $$ $$2\pi i\bigg(\dfrac{1}{(z-3-2i)^{4}}\bigg)\bigg|_{z=-1+i}=\int_{|z|=2}\dfrac{f(z)}{(z+1-i)}dz $$ $$I=2\pi i\bigg(\dfrac{1}{(-1+i-3-2i)^4}\bigg) $$ $$I=2\pi i\bigg(\dfrac{1}{(-4-i)^4}\bigg) $$
It should also be the same for $$\int_{|z|=3 }.....dz $$
Since it only fails to be analytic at the point $z=-1+i$. Is this correct or did I mess something up?
Thanks in advance!