Exercise :
Let :
$$\frac{\cos(πz)}{e^{iπz/2} + 1} = \sum_{n=-\infty}^{\infty} a_nz^n$$
the Laurent Expansion of the function :
$$f(z) = \frac{\cos(πz)}{e^{iπz/2} + 1}$$
in the ring : $D=\{ z \in \mathbb C : 2 < |z| < 6 \}$ with centre the point $z_c = 0$.
Using the Laurent Theorem and the Residue Theorem, calculate the coefficients $a_n$, $\forall n \leq 0$.
Attempt :
$$e^{iπz/2} + 1 = 0 \Leftrightarrow e^{iπz/2} = -1 \Leftrightarrow e^{iπz/2} = e^{(2k+1)iπ} \Leftrightarrow z_k = 4k + 2, k\in \mathbb Z $$
So, we have the points : $\{z_0 =2, z_1 = 6, \dots\}.$ Only the point $z_0=2$ belongs inside the circle $|z| = r$ : $2<r<6$.
From the Laurent Theorem we have :
$$a_n = \frac{1}{2πi} \oint_{|z|=r} \frac{g(z)}{z^{n+1}}dz$$
where $g(z)$ we have the function :
$$g(z) = \frac{\cos(πz)}{e^{iπz/2} + 1}$$
First of all, we have :
$$a_0 = \oint_{|z|=r} \frac{g(z)}{z} dz= \oint_{|z|=r} \frac{\cos(πz)}{z(e^{iπz/2} + 1)}dz $$
Because the points : $0,2$ are simple poles of the function $h(z)= \frac{\cos(πz)}{z(e^{iπz/2} + 1)}$, from the Theorem of Residues we have :
$$a_0 = Res(h,0) + Res(h,2)$$
After that, if $n \leq -1$ then $ -(n+1) \geq 0$.
So, we have :
$$a_n = \oint_{|z|=r} \frac{z^{-(n+1)}\cos(πz)}{e^{iπz/2} + 1}dz$$
We see that $0$ is a (???) point (I do not know its name in English), so in this case only the point $z=2$ is inside the circle $|z| = r$ and it is a simple pole. So, from the Theorem of Residues, for $n\leq -1$, we have :
$$a_n = Res\Bigg(\frac{z^{-(n+1)}\cos(πz)}{e^{iπz/2} + 1},2\Bigg) , \forall n \leq 0$$
I would like to ask if my approach and solution is correct. I'm not sure about the ring thingy. If it's not correct, please point out my mistakes and provide me with a thorough explanation.