# Find the residue of $\frac{1 - \cos z}{z^{3} (z-3)}$

Is my solution correct? Also, are there removable singularities? Im having trouble classifying singularities

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Yes......you can also see that around $0$ the series expansion looks like $$-1/6z -1/18 - z/216 \ - .... etc$$ So at zero you have a simple pole of order 1. This clearly follows as for $z \rightarrow 0$ you have $$z(1-\cos(z))/z^3(z-3) = 2\sin^2(z/2)/z^2(z-3)\sim 1/2(z-3) \rightarrow -1/6$$ Hence $z^k(1-\cos(z))/z^3(z-3) \sim z^{k-1}/2(z-3) \rightarrow 0$ for all $k \geq 2$. Thus the Laurent series starts from $1/z$.