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Question Number 216493 Answers: 1 Comments: 0

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

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

Question Number 216491 Answers: 2 Comments: 0

Question Number 216489 Answers: 1 Comments: 1

find residuo ((x^(21) +x^2 +x+1)/((x−1)^3 ))

$${find}\:\:{residuo} \\ $$$$\:\:\:\:\:\:\:\:\frac{{x}^{\mathrm{21}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$

Question Number 216487 Answers: 1 Comments: 2

Find the value of ω^7 + ω^8 + ω^(12) where ω is omega function.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\omega^{\mathrm{7}} \:\:+\:\:\omega^{\mathrm{8}} \:\:+\:\:\omega^{\mathrm{12}} \:\:\mathrm{where} \\ $$$$\omega\:\:\mathrm{is}\:\mathrm{omega}\:\mathrm{function}. \\ $$

Question Number 216486 Answers: 1 Comments: 0

∫_( 0) ^( 1) x(√(x ((x ((x ((x ...))^(1/5) ))^(1/4) ))^(1/3) )) dx

$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}\sqrt{\mathrm{x}\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:\:\sqrt[{\mathrm{4}}]{\mathrm{x}\:\:\sqrt[{\mathrm{5}}]{\mathrm{x}\:...}}}}\:\:\mathrm{dx} \\ $$

Question Number 216485 Answers: 2 Comments: 0

Solve for x in: i^x = 2

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:\:\mathrm{in}:\:\:\:\mathrm{i}^{\mathrm{x}} \:\:=\:\:\mathrm{2} \\ $$

Question Number 216478 Answers: 1 Comments: 1

Question Number 216477 Answers: 2 Comments: 0

Question Number 216471 Answers: 1 Comments: 3

Question Number 216445 Answers: 2 Comments: 2

Prove that Γ((1/2)) = (√π)

$$\mathrm{Prove}\:\mathrm{that}\:\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:=\:\:\sqrt{\pi} \\ $$

Question Number 216437 Answers: 3 Comments: 0

Question Number 216454 Answers: 0 Comments: 7

Reponse a l exercice N8: Reponses par ordre:(1,2,3,4,5,6) imsge 1 imsge 2 image 3 imsge 5 imsge 4 imsge 6

$$\mathrm{Reponse}\:\mathrm{a}\:\:\mathrm{l}\:\mathrm{exercice}\:\:\mathrm{N8}: \\ $$$$\mathrm{Reponses}\:\mathrm{par}\:\mathrm{ordre}:\left(\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right) \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{1}\: \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{2} \\ $$$$\boldsymbol{\mathrm{image}}\:\mathrm{3} \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{5} \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{4} \\ $$$$\boldsymbol{\mathrm{imsge}}\:\mathrm{6} \\ $$

Question Number 216425 Answers: 2 Comments: 1

Question Number 216421 Answers: 1 Comments: 0

If asinθ + bcosθ = acosecθ + bsecθ then prove that each term is equal to (a^(2/3) − b^(2/3) )(√(a^(2/3) + b^(2/3) )).

$$\mathrm{If}\:{a}\mathrm{sin}\theta\:+\:{b}\mathrm{cos}\theta\:=\:{a}\mathrm{cosec}\theta\:+\:{b}\mathrm{sec}\theta\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{each}\:\mathrm{term}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left({a}^{\frac{\mathrm{2}}{\mathrm{3}}} \:−\:{b}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)\sqrt{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:{b}^{\frac{\mathrm{2}}{\mathrm{3}}} }. \\ $$

Question Number 216416 Answers: 1 Comments: 4

Question Number 216411 Answers: 1 Comments: 0

(dx/dx)

$$\frac{{dx}}{{dx}} \\ $$

Question Number 216408 Answers: 0 Comments: 1

∫_(−1) ^1 (1/x)(√((1+x)/(1−x)))ln(((2x^2 +2x+1)/(2x^2 −2x+1)))dx

$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\mathrm{ln}\left(\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\right){dx} \\ $$

Question Number 216390 Answers: 0 Comments: 3

f(x)=ax

$${f}\left({x}\right)={ax}\: \\ $$

Question Number 216381 Answers: 1 Comments: 0

∫(lnx)^2 dx

$$\int\left({lnx}\right)^{\mathrm{2}} {dx} \\ $$

Question Number 216388 Answers: 1 Comments: 1

Question Number 216387 Answers: 1 Comments: 0

Question Number 216372 Answers: 2 Comments: 0

∫((xe^x )/((x+1)^2 ))dx

$$\int\frac{{xe}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 216369 Answers: 2 Comments: 0

Question Number 216355 Answers: 1 Comments: 1

given that ϕ,β are the roots of the equation 3x2−x−5=0 from the equation whose roots are 2ϕ−1/β,2β−1/ϕ

$${given}\:{that}\:\varphi,\beta\:{are}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:\mathrm{3}{x}\mathrm{2}−{x}−\mathrm{5}=\mathrm{0}\:{from}\:{the}\:{equation}\:{whose}\:{roots}\:{are}\:\mathrm{2}\varphi−\mathrm{1}/\beta,\mathrm{2}\beta−\mathrm{1}/\varphi \\ $$

Question Number 216352 Answers: 1 Comments: 1

Question Number 216351 Answers: 1 Comments: 0

Vector field F^→ ;R^3 →R^3 F^→ (x,y,z)=xye_1 ^→ −5ye_2 ^→ −3yze_3 ^→ ∫∫_(S;x^2 +y^2 +z^2 =r^2 ) F^→ ∙dS^→ = ?

$$\mathrm{Vector}\:\mathrm{field}\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}};\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}^{\mathrm{3}} \\ $$$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y},{z}\right)={xy}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\mathrm{5}{y}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} −\mathrm{3}{yz}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\underset{\mathcal{S};{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} } {\int\int}\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{S}}}=\:? \\ $$

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