Laurent expansion of $\frac{1}{1+z^2}$ in $A:= \{z \in \mathbb{C} : | z - z_0 | \gt |z_0 + i|, Im(z_0) \gt 0\}$ I need to find the Laurent expansion of $f(z) := \dfrac{1}{1+z^2}$ in the set $A:= \{z \in \mathbb{C} : | z - z_0 | \gt |z_0 + i|, Im(z_0) \gt 0\}$. I've drawn a picture of this:

I know that if $r \lt R$ and $f \in \mathcal{O}\left(K_{r,R}(z_0)\right)$, where $K_{r,R}(z_0) = \{z \in \mathbb{C}:r <|z-z_0| < R \}$, then one can find two functions $f_N(z) \in \mathcal{O}\left(B_R(z_0)\right) \text{ and } f_H \in \mathcal{O} \left(\mathbb{C} \backslash \overline{B_r(z_0)}\right)$ such that 
$$f(z) = f_N (z) + f_H(z) \quad \forall z \in K_{r,R} (z_0)$$
In my case I've chosen $r:= |z_0+i|$ and $R = \infty$. To evaluate the Laurent expansion I've decomposed $f(z)$ in this way:
$$f(z) = \dfrac{1}{1+z^2} = \dfrac{1}{2i}\dfrac{1}{ (z-i)} - \dfrac{1}{2i}\dfrac{1}{(z+i)}$$
Now i noticed that: 
• $\dfrac{1}{2i}\dfrac{1}{ (z-i)} - \dfrac{1}{2i}\dfrac{1}{(z+i)} \notin \mathcal{O}\left(B_R(z_0)\right)$
• $\dfrac{1}{2i}\dfrac{1}{ (z-i)} \notin \mathcal{O}\left(B_R(z_0)\right)$
• $- \dfrac{1}{2i}\dfrac{1}{(z+i)} \notin \mathcal{O}\left(B_R(z_0)\right)$
Hence edit: there are some functions for which the secondary expansion $f_N(z) \equiv 0$ I've then tried to expand the function with the help of the geometric series, like I've seen in an example in the lecture:
$$- \dfrac{1}{2i}\dfrac{1}{(z+i)} = \dfrac{-1}{2i} \dfrac{1}{z-z_0} \dfrac{1}{1+ \frac{z_0+i}{z-z_0}} \qquad (1)$$
Now we see from the picture that $\mid \frac{z_0+i}{z-z_0} \mid < 1 \quad \forall z \in A$
$$(1) = \dfrac{-1}{2i} \dfrac{1}{z-z_0} \sum_{\nu \ge 0} (-1)^{\nu}\left(\dfrac{z_0 + i}{z-z_0}\right)^{\nu} = \dfrac{-1}{2i} \sum_{\nu \ge 0} \dfrac{(-1)^{\nu}\left(z_0 + i\right)^{\nu}}{\left(z-z_0\right)^{\nu +1}}$$
The same for the other part:
$$\dfrac{1}{2i}\dfrac{1}{(z-i)} = \dfrac{1}{2i} \dfrac{1}{z-z_0} \dfrac{1}{1- \frac{i-z_0}{z-z_0}} \qquad (2)$$
Since $\mid \frac{i-z_0}{z-z_0} \mid \lt 1 \quad \forall z \in A$
$$(2) = \dfrac{1}{2i} \sum_{\nu \ge 0} \dfrac{\left(i-z_0\right)^{\nu}}{\left(z-z_0\right)^{\nu + 1}}$$
hence the Laurent expansion in $K_{r,\infty}(z_0)$ is
$$f_H(z) = \dfrac{1}{2i} \sum_{\nu \ge 0} \dfrac{\left(i-z_0\right)^{\nu}}{\left(z-z_0\right)^{\nu + 1}} + \dfrac{-1}{2i} \sum_{\nu \ge 0} \dfrac{(-1)^{\nu}\left(z_0 + i\right)^{\nu}}{\left(z-z_0\right)^{\nu +1}}$$
$$ edit: f(z) = \frac{1}{2i} \sum_{\nu = 0}^\infty \frac{\left((i-z_0)^\nu + (-1)^{\nu+1}(i+z_0)^\nu\right)}{(z-z_0)^{(\nu+1)}}.$$
Is this right? I'm not very sure about the choice of $R = \infty$ and the fact that $f_N(z) \equiv 0$
 A: First things first: Yes, it's correct. Good work.
With one caveat: if $z_0 = i$, your expansion of $(2)$ is unnecessarily complicated, it's plain $\frac{1}{2i} \frac{1}{z-i}$ then. But, writing it as a series with infinitely many terms (all but the first of which turn out to be $0$) keeps the development uniform (no special case for $z_0 = i$), and is therefore good. It may, however, be good to say a word about that (a parenthetical remark à la "for uniformity of the exposition, no special treatment of $z_0 = i$"). But it's not necessary to add such a remark.
A Laurent expansion is always valid for the largest annulus not containing any singularity of the function, and since $f$ has only the two poles at $i$ and $-i$ as singularities, a Laurent expansion valid in any point $z$ with $\lvert z-z_0\rvert > \lvert z_0+i\rvert$ automatically holds on the whole complement of the disk $\{z : \lvert z-z_0\rvert \leqslant \lvert z_0+i\rvert\}$, so you don't "choose" $R=\infty$, that drops out automatically.
A minor cosmetic point: it is desirable to write the representation of $f$ as a single series, combining the two parts into one,
$$f(z) = \frac{1}{2i} \sum_{\nu = 0}^\infty \left((i-z_0)^\nu + (-1)^{\nu+1}(i+z_0)^\nu\right)(z-z_0)^{-(\nu+1)}.$$
One point that I found irritating was

Now I noticed that [some functions are not in $\mathcal{O}(B_R(z_0))$]. Hence $f_N(z) \equiv 0$.

That conclusion is valid because of the special type of the functions involved here, but in general, a function has nontrivial primary and secondary parts. Here, you have two functions of the form $(z-a)^k$ (times a constant), and such functions always contribute only to either the principal or the secondary part, not to both. But composite functions generally contribute to both parts.
