On the page 175 of the paper provided below, principle part of of a generating function $$f(z)=\frac{1}{2-e^z}$$
at its pole $z_0=\ln2$ is computed to be $$ f_p(z)=\frac{-1}{2(z-\ln2)}.$$
From my knowledge, the principle part is the part of the Laurant series which's input goes from $-\infty$ to $0$. Thus my question is how did the paper find such a compact form for the principle part, and how was it computed in the first place?
https://drive.google.com/file/d/1mvG22_40ZIvAr97xs8-UGX2TKOftzhsQ/view?usp=sharing
Thank you for the help in advance.
You have\begin{align}2-e^z&=-2 (z-\log (2))-(z-\log (2))^2-\frac{1}{3} (z-\log(2))^3+\cdots\\&=-2(z-\log(2))\left(1+\frac12(z-\log(2))+\frac16(z-\log(2))^2+\cdots\right)\end{align}and therefore$$\frac1{2-e^z}=-\frac1{2(z-\log(2))}\cdot\frac1{1+\frac12(z-\log(2))+\frac16(z-\log(2))^2+\cdots}.$$Since$$z\mapsto\frac1{1+\frac12(z-\log(2))+\frac16(z-\log(2))^2+\cdots}$$is an analytic function near $0$ which maps $\log(2)$ into $1$, it can be written (near $0$) as $1+a_1(z-\log(2))+a_2(z-\log(2))^2+\cdots$, and therefore\begin{align}\frac1{2-e^z}&=-\frac1{2(z-\log(2))}\cdot(1+a_1(z-\log(2))+a_2(z-\log(2))^2+\cdots)\\&=-\frac1{2(z-\log(2))}+a_1+a_2(z-\log(2))+\cdots\end{align}