Res_(z=c) {f(z)}=(1/(2πi)) ∮_( C) f(z)dz Res_(z=1) {((z^(21) +z^2 +z+1)/((z−1)^3 ))}=(1/(2πi)) ∮_( C) ((z^(21) +z^2 +z+1)/((z−1)^3 ))dz (1/(2πi)) ∮_( C) (((z^(21) +z^2 +z+1)/((z−1)^2 ))/(z−1))dz=lim_(z→1) ((z^(21) +z^2 +z+1)/((z−1)^2 )) L′hosiptal :) lim_(z→1) ((21z^(20) +2z+1)/(2(z−1))) and... Twice!! lim_(z→1) ((420z^(19) +2)/2)=211 ∴Res_(z=1) {f(z)}=211 ★Caution★ f(α)′′=′′(1/(2πi)) ∮_( C) ((f(z))/(z−α)) dz Why did I use big quotes for this equation?? because the conditions for establshing this equation are that path C must be a simple closed curve and there must be no singularity in path C


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Question Number 216493 by issac last updated on 09/Feb/25
$Res_(z=c) {f(z)}=(1/(2πi)) ∮_( C) f(z)dz Res_(z=1) {((z^(21) +z^2 +z+1)/((z−1)^3 ))}=(1/(2πi)) ∮_( C) ((z^(21) +z^2 +z+1)/((z−1)^3 ))dz (1/(2πi)) ∮_( C) (((z^(21) +z^2 +z+1)/((z−1)^2 ))/(z−1))dz=lim_(z→1) ((z^(21) +z^2 +z+1)/((z−1)^2 )) L′hosiptal :) lim_(z→1) ((21z^(20) +2z+1)/(2(z−1))) and... Twice!! lim_(z→1) ((420z^(19) +2)/2)=211 ∴Res_(z=1) {f(z)}=211 ★Caution★ f(α)′′=′′(1/(2πi)) ∮_( C) ((f(z))/(z−α)) dz Why did I use big quotes for this equation?? because the conditions for establshing this equation are that path C must be a simple closed curve and there must be no singularity in path C$
${Res}_{z = c} {f (z)} = \frac{1}{2 π i} \oint_{C} f (z) d z$ ${Res}_{z = 1} {\frac{z^{21} + z^{2} + z + 1}{{(z - 1)}^{3}}} = \frac{1}{2 π i} \oint_{C} \frac{z^{21} + z^{2} + z + 1}{{(z - 1)}^{3}} d z$ $\frac{1}{2 π i} \oint_{C} \frac{\frac{z^{21} + z^{2} + z + 1}{{(z - 1)}^{2}}}{z - 1} d z = \lim_{z \to 1} \frac{z^{21} + z^{2} + z + 1}{{(z - 1)}^{2}}$ $L^{'} hosiptal :)$ $\lim_{z \to 1} \frac{21 z^{20} + 2 z + 1}{2 (z - 1)} and . . . Twice!!$ $\lim_{z \to 1} \frac{420 z^{19} + 2}{2} = 211$ $∴ {Res}_{z = 1} {f (z)} = 211$ $★ Caution ★$ $f {(α)}^{″} =^{″} \frac{1}{2 π i} \oint_{C} \frac{f (z)}{z - α} d z$ $Why did I use big quotes for this$ $equation ? ?$ $because the conditions for establshing$ $this equation are that path C$ $must be a simple closed curve$ $and there must be no singularity$ $in path C$
	Answered by MrGaster last updated on 09/Feb/25
	$Res_(z=c) {f(z)}=lim_(z→c) ((z+c)f(z)) ∮_C f(z)dz=2πiΣ_(k=1) ^n Res_(z=c_k ) {f(z)} Res_(z=c) {f(z)}=(1/(2πi))∮_C f(z)dz ∮_C f(z)dz=2πi∙211 ∴Res_(z=c) {f(z)}=211$
	${Res}_{z = c} {f (z)} = \lim_{z \to c} ((z + c) f (z))$ $\oint_{C} f (z) d z = 2 π i \underset{k = 1}{\sum^{n}} {Res}_{z = c_{k}} {f (z)}$ ${Res}_{z = c} {f (z)} = \frac{1}{2 π i} \oint_{C} f (z) d z$ $\oint_{C} f (z) d z = 2 π i \cdot 211$ $∴ {Res}_{z = c} {f (z)} = 211$
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