# Laurent series of $\frac{1}{(z+2)}\frac{1}{(e^{i\pi z /2}+1)}$

I'm re-learning this stuff. It is a little confusing to me how to obtain the Laurent expansion of exponential functions in denominator.

I have to determine the Laurent series of $$\frac{1}{(z+2)}\frac{1}{(e^{i\pi z /2}+1)}$$ centered at $$z_0=−2$$ that converges at $$0<|z+2|<4$$.

The "centered" part is clear to me, but what does it mean that converges at $$0<|z+2|<4$$?

I know that I have to find the series of each term centered at $$z_0=−2$$. But when I try to do that I simply get some strange series that doesn't match with anything $$\frac{1}{(z+2)}\cdot\sum_{m=0}^{\infty}\left[ (-1)^m \left(\sum_{n=0}^{\infty}\frac{1}{n!}\left[\frac{\pi(z+2)}{2}-1 \right]^n \right)^m\right]$$

Second question. Can I make another Laurent expansion for $$4<|z+2|<8$$? How should it be?

In general, the "region of convergence" of a Laurent series $$\sum_{n = -\infty}^{\infty}a_nz^n$$ about $$z = 0$$ is an annulus with inner radius $$r_0 = \limsup_{n \to \infty}|a_{-n}|^{1/n} \in [0, \infty]$$ and outer radius $$r_1 = \frac{1}{\limsup_{n \to \infty}|a_{n}|^{1/n}} \in [0, \infty]$$. The formulas given here for $$r_0$$ and $$r_1$$ can be derived using the root test. This means that the region of convergence of the Laurent series is $$A_{r_0, r_1}(0) = \{z \in \mathbb{C} : r_1 < |z| < r_2\}$$.
Conversely, if $$f$$ is holomorphic on some annulus $$A = A_{r_1, r_2}(0)$$, then on $$A$$, $$f$$ can be written as a Laurent series around $$0$$. This converse is a consequence of the Cauchy integral formula. Of course, you can translate coordinates and expand around any $$z_0 \in \mathbb{C}$$.
For your problem, your function is holomorphic on $$A_{0, 4}(-2) = \{z \in \mathbb{C} : 0 < |z + 2| < 4\}$$, so you automatically know that it has a Laurent series there. You just need to find the coefficients. The $$\frac{1}{z + 2}$$ factor is fine, so you just need the Laurent series for $$\frac{1}{\exp(i\pi z/2) + 1}$$. This function is not holomorphic around $$z = -2$$; it has a pole of order $$1$$ there. I would compute the Laurent series by first factoring out the pole to reduce this to computing an ordinary power series, and then using the formula for the quotient of two series (Dividing an infinite power series by another infinite power series).