The Laurent series of $ g(z)=\frac{z^n+z^{-n}}{z^2-(a+\frac{1}{a})z+1}$
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calculus
complex-analysis
power-series
taylor-expansion
1 Answers
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Assuming your work is right, the key idea you seem to be overlooking is
$ \begin{align} g(z) &= \sum_{p=o}^{\infty} \frac{(1-a^{2p+2})}{a^p(1-a^2)}z^{p+n} + \sum_{p=o}^{\infty} \frac{(1-a^{2p+2})}{a^p(1-a^2)}z^{p-n} \\ &= \sum_{(p-n)=o}^{\infty} \frac{(1-a^{2(p-n)+2})}{a^{p-n}(1-a^2)}z^{(p-n)+n} + \sum_{(p+n)=o}^{\infty} \frac{(1-a^{2(p+n)+2})}{a^{p+n}(1-a^2)}z^{(p+n)-n} \end{align} $