Compute integrals using Cauchy Integral Formula

Question

The integrals:

1. $\int_{|z+1|=1}\frac{1}{(z+1)(z-1)^4} dz $

2. $\int_{|z|=2}\frac{\cos(z)}{(z+i)} dz $

idea:

As I mentioned in the title I want to use the Cauchy integral formula:

$f(a) = \frac{1}{2\pi i }\int_{|z- z_0|=r} \frac{f(z)}{z-a} dz $

So I believe that $f$ is:

1. $\frac{1}{(z-1)^4}$ and $ a=-1$

2. $ \cos(z) $ and $ a = -i $

So I would get:

1. $\int_{|z+1|=1}\frac{1}{(z+1)(z-1)^4} dz = \int_{|z-(-1)|=1}\frac{1}{(z+1)(z-1)^4} dz = \int_{|z-(-1)|=1}\frac{f(z)}{(z-(-1))} dz = 2\pi i f(a) = 2\pi i \frac{1}{((-1)-1)^4} = \frac{\pi i }{8} $

2. $\int_{|z|=2}\frac{\cos(z)}{(z+i)} dz=\int_{|z-0|=2}\frac{\cos(z)}{(z+i)} dz =\int_{|z-0|=2}\frac{f(z)}{(z-(-i))} dz = 2\pi i f(a) = 2 \pi i \cos(-i) $

is this correct?