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

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1 Answer 1

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 $.

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