Say we are trying to find the Laurent series (at zero) of $\frac{cos(z)}{z^{3}}$ why is it we can factor out $z^{-3}$ and multiply the Taylor series expansion of $cos(z)$ at zero when we are trying to find the expansion of the whole of the function not just cos. Surely we also miss out all the terms that are summed to negative infinity also by only using the Taylor expansion.
2026-03-27 01:43:59.1774575839
Why when finding a Laurent series can we factor the function
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Well, those terms are zero... The point is that since $$ \frac{\cos z}{z^3} = \frac{1}{z^3}\left( 1 -\frac{z^2}{2} + \frac{z^4}{24} - \cdots \right) = \frac{1}{z^3}-\frac{1/2}{z} + \frac{z}{24}-\cdots $$
the non-zero terms of the Laurent series begin at index -3. Not all singularities are essential singularities!