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Question Number 144085 by mohammad17 last updated on 21/Jun/21

Commented by mohammad17 last updated on 21/Jun/21

$h e l p m e s i r$
	Answered by mathmax by abdo last updated on 21/Jun/21
	$Φ=∫_C ((3z^2 +2)/((z−1)^2 (z−π)))dz with C→∣z−2∣=2 ϕ(z)=((3z^2 +2)/((z−1)^2 (z−π))) the poles are 1 and π( interior on the circle) ⇒Φ=2iπ {Res(ϕ,1)+Res(ϕ,π)} Res(ϕ,1)=lim_(z→1) (1/((2−1)!)){(z−1)^2 ϕ(z)}^((1)) =lim_(z→1) (((3z^2 +2)/((z−π))))^((1)) =lim_(z→1) ((6z(z−π)−(3z^2 +2))/((z−π)^2 )) =((6(1−π)−5)/((1−π)^2 )) Res(ϕ,π)=lim_(z→π) (z−π)ϕ(z) =lim_(z→π) ((3z^2 +2)/((z−1)^2 ))=((3π^2 +2)/((π−1)^2 )) ⇒ Φ=2iπ{((1−6π)/((1−π)^2 ))+((3π^2 +2)/((π−1)^2 ))} =((2iπ)/((π−1)^2 ))(3π^2 −6π+3)$
	$Φ = \int_{C} \frac{{3 z}^{2} + 2}{{(z - 1)}^{2} (z - π)} dz with C \to∣ z - 2 ∣= 2$ $φ (z) = \frac{{3 z}^{2} + 2}{{(z - 1)}^{2} (z - π)} the poles are 1 and π (interior on the circle)$ $\Rightarrow Φ = 2 i π {Res (φ, 1) + Res (φ, π)}$ $Res (φ, 1) = \lim_{z \to 1} \frac{1}{(2 - 1)!} {{(z - 1)}^{2} φ (z)}^{(1)}$ $= \lim_{z \to 1} {(\frac{{3 z}^{2} + 2}{(z - π)})}^{(1)} = \lim_{z \to 1} \frac{6 z (z - π) - ({3 z}^{2} + 2)}{{(z - π)}^{2}}$ $= \frac{6 (1 - π) - 5}{{(1 - π)}^{2}}$ $Res (φ, π) = \lim_{z \to π} (z - π) φ (z)$ $= \lim_{z \to π} \frac{{3 z}^{2} + 2}{{(z - 1)}^{2}} = \frac{3 π^{2} + 2}{{(π - 1)}^{2}} \Rightarrow$ $Φ = 2 i π {\frac{1 - 6 π}{{(1 - π)}^{2}} + \frac{3 π^{2} + 2}{{(π - 1)}^{2}}} = \frac{2 i π}{{(π - 1)}^{2}} (3 π^{2} - 6 π + 3)$
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