Complex integral with pole

Question

I am doing GRE problems to prepare for the subject exam. My complex analysis is a bit rusty, can someone give me a hint here:

$$I=\oint_C \frac{\cos z}{z(z-\pi)} dz$$

where $C$ is the circle $\vert z -1 \vert = 2$.

Am I correct in my following intuition.

First we are considering the circle with radius $\sqrt{2}$ centered at $(1,0)$ on the complex plane correct? And we have two points to consider as poles $z=0,\pi$ but one of them is outside the circle correct? Namely, $z=\pi$ is outside and only the origin is inside it. So Do I calculate the residue at the one pole I have? Also if they were both outside, it would be zero correct?

Answer 1

So $I$ has two singularities to consider, at $z=0,\pi$. By the residue theorem we know that

$$\oint_C f(z) = 2 \pi i \sum_{k=1}^m \eta(C;a_k) \text{Res} (f;a_k)$$

Where $\eta$ is winding number. Since the winding number for $z=\pi$ is zero, we need to only compute the residue at $z=0$ which has winding number $1$. Then we get

$$\oint_C f(z) = 2 \pi i (\frac{-1}{\pi})=-2i$$

as needed.

Answer 2

You can apply Cauchy's integral formula to $f(z)=\frac{\cos z}{z-\pi}$, getting\begin{align}\oint_{|z-1|=2}\frac{\cos z}{z(z-\pi)}\,\mathrm dz&=\oint_{|z-1|=2}\frac{f(z)}z\,\mathrm dz\\&=2\pi if(0)\\&=-\frac{2\pi i}\pi\\&=-2i.\end{align}