Laurent expansion of $f(z)=\frac{1}{1+z^2}+\frac{1}{3-z}$ in the region $|z|>3$. Laurent expansion of $f(z)=\frac{1}{1+z^2}+\frac{1}{3-z}$ in the region $|z|>3$.
According to the solution the expansion is
$$\sum_{n=0}^{\infty}(-1)^nz^{-2n}+\sum_{n=0}^{\infty}\frac{3^n}{z^{n+1}}.$$
But I don't understand why. My work is as follows:
\begin{align*}
f(z)&=\frac{1}{z^2}\frac{1}{1-(-z^{-2})}-\frac{1}{z}\frac{1}{1-\frac{3}{z}}\\
&=\frac{1}{z^2}\sum_{n=0}^{\infty}(-1)^nz^{-2n}-\frac{1}{z}\sum_{n=0}^{\infty}\left(\frac{3}{z}\right)^n\\
&=\sum_{n=0}^{\infty}(-1)^nz^{-2n-2}-\sum_{n=0}^{\infty}\frac{3^n}{z^{n+1}}
\end{align*}
But in the correct answer, what happens to the $\frac{1}{z^2}$ and the negative from the $-\frac{1}{z}$ in the second term?

I think the given answer, the first summatoy BEGINS in $n=1$.
$$\frac{1}{z^2}\frac{1}{1-(-z^2)} = - \frac{-z^{-2}}{1-(-z^2)}= -\sum_{n=1}^\infty (-z^2)^n=\sum_{n=1}^\infty (-1)^{n+1} z^{2n}
$$
