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

Commented by tabata last updated on 02/Jul/21

help me sir

$${help}\:{me}\:{sir} \\ $$

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

f(z)=(z^2 /((z−2)(z^2 +1))) ⇒ Res(f,2)=lim_(z→2) (z−2)f(z)=lim_(z→2) (z^2 /(z^(2 ) +1))=(4/(2^2 +1))=(4/5) Res(f,i)=lim_(z→i) (z−i)f(z) =lim_(z→i) (z−i).(z^2 /((z−2)(z−i)(z+i))) =lim_(z→i) (z^2 /((z−2)(z+i)))=((−1)/((i−2)(2i)))=((−1)/(−2−4i))=(1/(2+4i)) =((2−4i)/(2^2 −(4i)^2 ))=((2−4i)/(4+16))=((2−4i)/(20))=((1−2i)/(10))=(1/(10))−(1/5)i

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

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