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Res-z-c-f-z-1-2pii-C-f-z-dz-Res-z-1-z-21-z-2-z-1-z-1-3-1-2pii-C-z-21-z-2-z-1-z-1-3-dz-1-2pii-C-z-21-z-2-z-1-z-1-2-z-1-dz-l

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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
Resz=c{f(z)}=12πi∮Cf(z)dzResz=1{z21+z2+z+1(z−1)3}=12πi∮Cz21+z2+z+1(z−1)3dz12πi∮Cz21+z2+z+1(z−1)2z−1dz=limz→1z21+z2+z+1(z−1)2L′hosiptal:)limz→121z20+2z+12(z−1)and…Twice!!limz→1420z19+22=211∴Resz=1{f(z)}=211★Caution★f(α)″=″12πi∮Cf(z)z−αdzWhydidIusebigquotesforthisequation??becausetheconditionsforestablshingthisequationarethatpathCmustbeasimpleclosedcurveandtheremustbenosingularityinpathC
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
Resz=c{f(z)}=limz→c((z+c)f(z))∮Cf(z)dz=2πi∑nk=1Resz=ck{f(z)}Resz=c{f(z)}=12πi∮Cf(z)dz∮Cf(z)dz=2πi⋅211∴Resz=c{f(z)}=211

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