Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$$\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\underset{{z}\rightarrow{c}} {\mathrm{lim}}\left(\left({z}+{c}\right){f}\left({z}\right)\right) \\ $$$$\oint_{{C}} {f}\left({z}\right){dz}=\mathrm{2}\pi{i}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{Res}_{{z}={c}_{{k}} } \left\{{f}\left({z}\right)\right\} \\ $$$$\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\oint_{{C}} {f}\left({z}\right){dz} \\ $$$$\oint_{{C}} {f}\left({z}\right){dz}=\mathrm{2}\pi{i}\centerdot\mathrm{211} \\ $$$$\therefore\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\mathrm{211} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com