detirmine the residues f(z)=((cosz)/(z^2 (z−π)^3 ))


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Question Number 147405 by tabata last updated on 20/Jul/21

$d e t i r m i n e t h e r e s i d u e s f (z) = \frac{c o s z}{z^{2} {(z - π)}^{3}}$
Commented by Mrsof last updated on 20/Jul/21

$? ? ?$
	Answered by mathmax by abdo last updated on 20/Jul/21
	$f(z)=((cosz)/(z^2 (z−π)^3 )) the poles are o and π Res(f,0)=lim_(z→0) (1/((2−1)!)){z^2 f(z)}^((1)) =lim_(z→o) {((cosz)/((z−π)^3 ))}^((1)) =lim_(z→0) ((−sinz(z−π)^3 −3(z−π)^2 cosz)/((z−π)^6 )) =lim_(z→0) ((−sinz(z−π)−3cosz)/((z−π)^4 ))=((−3)/π^4 ) Res(f,π)=lim_(z→π) (1/((3−1)!)){(z−π)^3 f(z)}^((2)) =(1/2)lim_(z→π) {((cosz)/z^2 )}^((2)) =(1/2)lim_(z→π) {((−z^2 sinz−2zcosz)/z^4 )}^((1)) =−(1/2)lim_(z→π) {((zsinz+2cosz)/z^3 )}^((1)) =−(1/2)lim_(z→π) (((sinz+zcosz−2sinz)z^3 −3z^2 (zsinz+2cosz))/z^6 ) =−(1/2)lim_(z→π) (((zcosz−sinz)z−3(zsinz+2cosz))/z^4 ) =−(1/2)×(((−π)π−3(−2))/π^4 )=−(1/2)×((−π^2 +6)/π^4 ) =((π^2 −6)/(2π^4 ))$
	$f (z) = \frac{cosz}{z^{2} {(z - π)}^{3}} the poles are o and π$ $Res (f, 0) = \lim_{z \to 0} \frac{1}{(2 - 1)!} {z^{2} f (z)}^{(1)}$ $= \lim_{z \to o} {\frac{cosz}{{(z - π)}^{3}}}^{(1)} = \lim_{z \to 0} \frac{- sinz {(z - π)}^{3} - 3 {(z - π)}^{2} cosz}{{(z - π)}^{6}}$ $= \lim_{z \to 0} \frac{- sinz (z - π) - 3 cosz}{{(z - π)}^{4}} = \frac{- 3}{π^{4}}$ $Res (f, π) = \lim_{z \to π} \frac{1}{(3 - 1)!} {{(z - π)}^{3} f (z)}^{(2)}$ $= \frac{1}{2} \lim_{z \to π} {\frac{cosz}{z^{2}}}^{(2)} = \frac{1}{2} \lim_{z \to π} {\frac{- z^{2} sinz - 2 zcosz}{z^{4}}}^{(1)}$ $= - \frac{1}{2} \lim_{z \to π} {\frac{zsinz + 2 cosz}{z^{3}}}^{(1)}$ $= - \frac{1}{2} \lim_{z \to π} \frac{(sinz + zcosz - 2 sinz) z^{3} - {3 z}^{2} (zsinz + 2 cosz)}{z^{6}}$ $= - \frac{1}{2} \lim_{z \to π} \frac{(zcosz - sinz) z - 3 (zsinz + 2 cosz)}{z^{4}}$ $= - \frac{1}{2} \times \frac{(- π) π - 3 (- 2)}{π^{4}} = - \frac{1}{2} \times \frac{- π^{2} + 6}{π^{4}}$ $= \frac{π^{2} - 6}{2 π^{4}}$
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