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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?

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    $\begingroup$ I think your answer is correct and the given answer is wrong. $\endgroup$ Apr 6, 2020 at 9:38

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I think the given answer, the first summatoy BEGINS in $n=1$.

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

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