Question Number 145810 by tabata last updated on 08/Jul/21
$${find}\:{residue}\:{f}\left({z}\right)={z}^{−\mathrm{3}} {csc}\left({z}^{\mathrm{2}} \right) \\ $$
Commented by tabata last updated on 08/Jul/21
$${how}\:{can}\:{it}\:{solve}\:{this} \\ $$
Commented by tabata last updated on 08/Jul/21
$$????? \\ $$
Answered by Olaf_Thorendsen last updated on 09/Jul/21
$${f}\left({z}\right)\:=\:{z}^{−\mathrm{3}} \mathrm{csc}\left({z}^{\mathrm{2}} \right)\:=\:\frac{\mathrm{1}}{{z}^{\mathrm{3}} \mathrm{sin}\left({z}^{\mathrm{2}} \right)} \\ $$$$\mathrm{The}\:\mathrm{value}\:{z}=\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{pole}\:\mathrm{of}\:{f}. \\ $$$${f}\left({z}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{1}}{{z}^{\mathrm{3}} ×\left({z}^{\mathrm{2}} −\frac{{z}^{\mathrm{6}} }{\mathrm{3}!}+\frac{{z}^{\mathrm{10}} }{\mathrm{5}!}…\right)} \\ $$$${f}\left({z}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{1}}{{z}^{\mathrm{5}} \left(\mathrm{1}−\frac{{z}^{\mathrm{4}} }{\mathrm{3}!}+\frac{{z}^{\mathrm{8}} }{\mathrm{5}!}…\right)}\: \\ $$$${f}\left({z}\right)\:\underset{\mathrm{0}} {\sim}\:\frac{\mathrm{1}}{{z}^{\mathrm{5}} }\left(\mathrm{1}+\frac{{z}^{\mathrm{4}} }{\mathrm{3}!}−\frac{{z}^{\mathrm{8}} }{\mathrm{5}!}+…\right)\: \\ $$$$\mathrm{The}\:\mathrm{coef}\:\mathrm{of}\:\frac{\mathrm{1}}{{z}}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{3}!}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\:\mathrm{Res}_{\mathrm{0}} {f}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$