<s>[INST] Laurent series expansion, can one simplify this? I have to expand $f(z)=\frac{z-1}{(z^2+1)z}$ in an annulus $R(i,1,2)$.

$$f(z)=\frac{1}{z-i}\frac{1}{z+i}-\frac{1}{z-i}\Big(\frac{i}{z+i}-\frac{i}{z}\Big)$$

$$\frac{1}{z-i}\frac{1}{z+i}=\frac{1}{z-i}\cdot\frac{1}{2i}\cdot\frac{1}{1-(-(z-i)/2i)}=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^n$$

$$\frac{i}{z+i}=\frac{1}{2}\cdot \frac{1}{1-(-(z-i)/2i)}=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n}\frac{1}{2}(z-i)^n$$

$$\frac{i}{z}=\frac{i}{z-i}\cdot\frac{1}{1-(-i/(z-i))}=\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+1}$$

Finally we got:
$$f(z)=\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^n-\sum_{n=0}^{\infty}(-1)^n (\frac{1}{2i})^{n+1}(z-i)^{n-1}+\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+2}=\frac{1}{2i}\cdot\frac{1}{z-i}+\sum_{n=1}^{\infty}\Big( (-1)^{n-1}(\frac{1}{2i})^{n}-(-1)^n (\frac{1}{2i})^{n+1}\Big)(z-i)^{n-1}+\sum_{n=0}^{\infty}(-1)^n i^{n+1}(\frac{1}{z-i})^{n+2}$$
My questions are: is this expansion correct? And can one simplify it?
 [/INST] From partial fractions, we have that
$$
f(z) = \frac{-1}{z} + \frac{1}{2}\frac{1+i}{z+i} + \frac{1}{2}\frac{1-i}{z-i}
$$
For the Laurent series in annulus, I obtained:
\begin{align}
\frac{-1}{z} + \frac{1}{2}\frac{1+i}{z+i} + \frac{1}{2}\frac{1-i}{z-i}
&= \frac{-1}{z-i}\frac{1}{1+\bigl(\frac{i}{z-i}\bigr)} + \frac{1}{4i}\frac{1+i}{1+\bigl(\frac{z -i}{2i}\bigr)}+\frac{1}{2}\frac{1-i}{z-i}\\
&= \frac{1}{2}\frac{1-i}{z-i}+\sum_{n=0}^{\infty}\biggl[i^n\frac{(-1)^{n+1}}{(z-i)^{n+1}}+\frac{1+i}{4i}\Bigl(\frac{i}{2}\Bigr)^n(z-i)^n\biggr]
\end{align}
I checked the results on Mathematica as well and it checked out. However, Wolfram online wouldn't compute it, so I attached a screen shot of my results.

</s>