Question Number 56145 by gunawan last updated on 11/Mar/19
$$\mathrm{find}\:\mathrm{residu}\:\mathrm{of}\:\mathrm{function} \\ $$$${f}\left({z}\right)=\frac{{e}^{\frac{\mathrm{1}}{{z}}} }{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{in}\:{z}=\mathrm{0} \\ $$
Answered by einsteindrmaths@hotmail.fr last updated on 05/Apr/19
$${if}\:{f}\left({z}\right)=\Sigma{a}_{{n}} \left({z}−{a}\right)^{{n}} \:{res}\left({f}.{a}\right)={a}_{−\mathrm{1}} \\ $$$${e}\left(\mathrm{1}/{z}\right)=\Sigma\frac{\mathrm{1}}{{n}!{z}^{{n}} } \\ $$$$\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}=\Sigma\left(−{z}^{\mathrm{2}} \right)^{{n}} \\ $$$${f}\left({z}\right)=\Sigma\:\:\:\:\left(−\mathrm{1}\right)^{{m}} \left({z}\right)^{\mathrm{2}{m}} .\Sigma\frac{\mathrm{1}}{{n}!{z}^{{n}} } \\ $$$${a}_{−\mathrm{1}} =\underset{\mathrm{2}{m}−{n}=−\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{m}} \frac{\mathrm{1}}{{n}!}=\underset{{m}} {\sum}\left(−\mathrm{1}\right)^{{m}} .\frac{\mathrm{1}}{\left(\mathrm{2}{m}+\mathrm{1}\right)!}=\mathrm{sin}\:\left(\mathrm{1}\right) \\ $$