Expanding Laurent series Expand the following Laurent series centered at the origin. Indicate the annulus of convergence. $$\frac{1}{(z^2+1)(z-2)}$$
I am attempting to decompose using partial fractions as I did on previous problems but I believe the $(z^2+1)$ is giving me problems. I have the numerator as $c_1z+c_2$ but i do not believe i am reaching a proper decomposition. 


The function
\begin{align*}
 f(z)&=\frac{1}{(z^2+1)(z-2)}\\
&=-\frac{1+2i}{10}\frac{1}{z+i}+\frac{1-2i}{10}\frac{1}{z-i}+\frac{1}{5}\frac{1}{z-2}
\end{align*}
   has three simple poles at $+i,-i$ and $2$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $\pm i$ and $2$ and see they determine three regions.
\begin{align*}
 |z|<1,\qquad\quad
 1<|z|<2,\qquad\quad
 2<|z|
 \end{align*} 
  
  
*
  
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the poles $\pm i$ at the boundary of the disc. In the interior of this disc all three fractions with poles $\pm i$ and $2$  admit a representation as power series at $z=0$.
  
*The second region $1<|z|<2$ is the annulus with center $0$, inner radius $1$ and outer radius $2$. Here we have a representation of the fraction with poles $\pm i$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $2$ admits a representation as power series.
  
*The third region $|z|>2$ containing all points outside the disc with center $0$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}

We can now obtain the Laurent expansion of $f(x)$ at $z=0$ for all three regions
  
  
*
  
*Region 1: $|z|<1$
  
  
  \begin{align*}
f(z)&=-\frac{1+2i}{10}\frac{1}{z+i}+\frac{1-2i}{10}\frac{1}{z-i}+\frac{1}{5}\frac{1}{z-2}\\
&=-\frac{1+2i}{10}\sum_{n=0}^\infty \frac{1}{i^{n+1}}(-z)^n+\frac{1-2i}{10}\sum_{n=0}^\infty \frac{1}{(-i)^{n+1}}(-z)^n
+\frac{1}{5}\sum_{n=0}^{\infty}\frac{1}{(-2)^{n+1}}(-z)^n\\
&=\frac{1}{10}\sum_{n=0}^{\infty}\left((-2+i)i^n+(2+i)(-i)^n-\frac{1}{2^n}\right)z^n
\end{align*}

The Laurent expansion for the other regions can be calculated similarly.
