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Question Number 145166 by tabata last updated on 02/Jul/21

Commented by tabata last updated on 02/Jul/21

$$helpmesir$$

Answered by mathmax by abdo last updated on 03/Jul/21

$$\mathrm{f}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2}{(\mathrm{z}-2)({\mathrm{z}}^2+1)}\Rightarrow$$$$\mathrm{Res}(\mathrm{f},2)={\lim }_{\mathrm{z}\to 2}(\mathrm{z}-2)\mathrm{f}\left(\mathrm{z}\right)={\lim }_{\mathrm{z}\to 2}\frac{{\mathrm{z}}^2}{{\mathrm{z}}^2+1}=\frac{4}{2^2+1}=\frac{4}{5}$$$$\mathrm{Res}(\mathrm{f},\mathrm{i})={\lim }_{\mathrm{z}\to \mathrm{i}}(\mathrm{z}-\mathrm{i})\mathrm{f}\left(\mathrm{z}\right)={\lim }_{\mathrm{z}\to \mathrm{i}}(\mathrm{z}-\mathrm{i}).\frac{{\mathrm{z}}^2}{(\mathrm{z}-2)(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}$$$$={\lim }_{\mathrm{z}\to \mathrm{i}}\frac{{\mathrm{z}}^2}{(\mathrm{z}-2)(\mathrm{z}+\mathrm{i})}=\frac{-1}{(\mathrm{i}-2)\left(2\mathrm{i}\right)}=\frac{-1}{-2-4\mathrm{i}}=\frac{1}{2+4\mathrm{i}}$$$$=\frac{2-4\mathrm{i}}{2^2-{\left(4\mathrm{i}\right)}^2}=\frac{2-4\mathrm{i}}{4+16}=\frac{2-4\mathrm{i}}{20}=\frac{1-2\mathrm{i}}{10}=\frac{1}{10}-\frac{1}{5}\mathrm{i}$$

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