using the Laurent expansion i got the answer to be $$-(z+1)\sum_{n=0}^\infty \frac{z^{n-1}}{2^{n+1}}$$ however, I've got a feeling I've made a mistake somewhere?
2026-03-30 14:09:32.1774879772
laurent series expansion about $z=0$
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By partial fractions, $g(z) = - \frac{1}{2} \frac{1}{z} - \frac{3}{4} \frac{1}{1-\frac{z}{2}} = -\frac{1}{2} \frac{1}{z} - \frac{3}{4} \sum_{i=0}^\infty \frac{z^i}{2^i}$.