Find Laurent Series for $f(z) = \frac{1}{z} \cdot \frac{1}{1+z^2}$ when $1 < |z| < \infty$ Q: Find the Laurent series that represents the following function when $1 < |z| < \infty$
\begin{align*}
  f(z) &= \frac{1}{z} \cdot \frac{1}{1+z^2} \\
\end{align*}
The answer is given:
\begin{align*}
  f(z) &= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{z^{2n+1}} \\
\end{align*}
I don't see how to do the problem to arrive at the answer.
BTW, to solve the Laurent series of the same function when $|z| < 1$, it's very straightforward:
Start with standard Maclaurin Series
\begin{align*}
  \frac{1}{1-z} = \sum\limits_{n=0}^\infty z^n \\
\end{align*}
Substitute $z$ with $-z^2$ to get:
\begin{align*}
  \frac{1}{1+z^2} = \sum\limits_{n=0}^\infty (-1)^n z^{2n} \\
\end{align*}
Multiply both sides by $\frac{1}{z}$:
\begin{align*}
  \frac{1}{z} \cdot \frac{1}{1+z^2} &= \sum\limits_{n=0}^\infty (-1)^n z^{2n-1} \\
  \frac{1}{z} \cdot \frac{1}{1+z^2} &= \frac{1}{z} + \sum\limits_{n=0}^\infty (-1)^{n+1} z^{2n+1} \\
\end{align*}
How do I solve the $1 < |z| < \infty$ version?

When $1 < |z| < \infty$ then $0 < \dfrac{1}{|z|} < 1$
$$\dfrac{1}{z}.\dfrac{1}{1+z^2}=\dfrac{1}{z^3}.\dfrac{1}{1+(\dfrac{1}{z})^2}=\\
\dfrac{1}{z^3}.\dfrac{1}{1-(-\dfrac{1}{z^2})}=\\
\dfrac{1}{z^3}(1+(-\dfrac{1}{z^2})+(-\dfrac{1}{z^2})^2+(-\dfrac{1}{z^2})^3+...)$$
