Question Number 36198 by prof Abdo imad last updated on 30/May/18
$${let}\:{f}\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{1}} \\ $$$${find}\:\left({a}_{\left.{k}\right)} {the}\:{poles}\:{of}\:{f}\:{and}\:{calculate}\:\right. \\ $$$${Res}\left({f},{a}_{{k}} \right) \\ $$
Commented by prof Abdo imad last updated on 01/Jun/18
$${we}\:{have}\:{f}\left({z}\right)\:=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}−\mathrm{1}\right)\left({z}+\mathrm{1}\right)\left(\:{z}−{i}\right)\left({z}+{i}\right)}\:{so}\:{the}\:{poles}\:{of}\:{f} \\ $$$${are}\:\mathrm{1},−\mathrm{1},{i},−{i}\:\:{and} \\ $$$$\:{Res}\left({f},\mathrm{1}\right)\:=\:{lim}_{{z}\rightarrow\mathrm{1}} \left({z}−\mathrm{1}\right)\varphi\left({z}\right)=\:\frac{\mathrm{2}}{\mathrm{2}\left(\mathrm{2}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Res}\left({f},−\mathrm{1}\right)\:={lim}_{{z}\rightarrow−\mathrm{1}} \left({z}+\mathrm{1}\right)\varphi\left({z}\right) \\ $$$$=\:\frac{\mathrm{2}}{\left(−\mathrm{2}\right).\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Res}\left({f},\:{i}\right)={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)=\:\mathrm{0} \\ $$$${Res}\left(\varphi,−{i}\right)={lim}_{{z}\rightarrow−{i}} \left({z}+{i}\right)\varphi\left({z}\right)\:=\:\mathrm{0} \\ $$