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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
$$\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:\mathrm{C}} \:{f}\left({z}\right)\mathrm{d}{z} \\ $$$$\mathrm{Res}_{{z}=\mathrm{1}} \left\{\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{3}} }\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:{C}} \:\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{3}} }\mathrm{d}{z} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:{C}} \:\:\frac{\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{2}} }}{{z}−\mathrm{1}}\mathrm{d}{z}=\underset{{z}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{L}'\mathrm{hosiptal}\::\right) \\ $$$$\underset{{z}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{21}{z}^{\mathrm{20}} +\mathrm{2}{z}+\mathrm{1}}{\mathrm{2}\left({z}−\mathrm{1}\right)}\:\:\mathrm{and}…\:\mathrm{Twice}!! \\ $$$$\underset{{z}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{420}{z}^{\mathrm{19}} +\mathrm{2}}{\mathrm{2}}=\mathrm{211} \\ $$$$\therefore\mathrm{Res}_{{z}=\mathrm{1}} \left\{{f}\left({z}\right)\right\}=\mathrm{211} \\ $$$$\bigstar\mathrm{Caution}\bigstar \\ $$$${f}\left(\alpha\right)''=''\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:{C}} \:\:\frac{{f}\left({z}\right)}{{z}−\alpha}\:\mathrm{d}{z}\: \\ $$$$\mathrm{Why}\:\mathrm{did}\:\mathrm{I}\:\mathrm{use}\:\mathrm{big}\:\mathrm{quotes}\:\mathrm{for}\:\mathrm{this}\: \\ $$$$\mathrm{equation}?? \\ $$$$\mathrm{because}\:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for}\:\mathrm{establshing} \\ $$$$\mathrm{this}\:\mathrm{equation}\:\mathrm{are}\:\mathrm{that}\:\mathrm{path}\:{C}\: \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{curve} \\ $$$$\mathrm{and}\:\mathrm{there}\:\mathrm{must}\:\mathrm{be}\:\mathrm{no}\:\mathrm{singularity} \\ $$$$\mathrm{in}\:\mathrm{path}\:\mathrm{C} \\ $$
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
$$\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\underset{{z}\rightarrow{c}} {\mathrm{lim}}\left(\left({z}+{c}\right){f}\left({z}\right)\right) \\ $$$$\oint_{{C}} {f}\left({z}\right){dz}=\mathrm{2}\pi{i}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{Res}_{{z}={c}_{{k}} } \left\{{f}\left({z}\right)\right\} \\ $$$$\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\oint_{{C}} {f}\left({z}\right){dz} \\ $$$$\oint_{{C}} {f}\left({z}\right){dz}=\mathrm{2}\pi{i}\centerdot\mathrm{211} \\ $$$$\therefore\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\mathrm{211} \\ $$

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