calculate ∫_C ((9(z^2 +2))/(z(z+1)^3 (z−2)))dz with C is the circle C ={z∈C/ ∣z∣ =3}


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Question Number 37299 by math khazana by abdo last updated on 11/Jun/18
$calculate ∫_C ((9(z^2 +2))/(z(z+1)^3 (z−2)))dz with C is the circle C ={z∈C/ ∣z∣ =3}$
$c a l c u l a t e \int_{C} \frac{9 (z^{2} + 2)}{z {(z + 1)}^{3} (z - 2)} d z w i t h C i s t h e$ $c i r c l e C = {z \in C / ∣ z ∣ = 3}$
Commented by math khazana by abdo last updated on 17/Jun/18
$let consider the complex function ϕ(z)= ((9(z^2 +2))/(z(z+1)^3 (z−2))) the poles of ϕ are 0, −1,2 (all are at interior of circle C) ∫_C ϕ−z)dz =2iπ { Res(ϕ,0)+Res(ϕ,−1)+Res(ϕ,2) Res(ϕ,0)=lim_(z→0) zϕ(z)= ((18)/(−2)) =−9 Res(ϕ,2) =lim_(z→2) (z−2)ϕ(z) = ((36)/(2.27)) =((18)/(27))= ((2.9)/(3.9)) =(2/3) Res(ϕ,−1) =lim_(z→−1) (1/((3−1)!)){(z+1)^3 ϕ(z)}^((2)) =lim_(z→−1) (9/2){ ((z^2 +2)/(z^2 −2z))}^((2)) =lim_(z→−1) (9/2){((2z(z^2 −2z) −(2z−2)(z^2 +2))/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ ((2z^3 −4z^2 −2z^3 −4z +2z^2 +4)/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ ((−2z^2 −4z +4)/((z^2 −2z)^2 ))}^((1)) =lim_(z→−1) (9/2){ (((−4z−4)(z^2 −2z)^2 −2(2z−2)(z^2 −2z)(−2z^2 −4z +4))/((z^2 −2z)^4 ))} =lim_(z→−1) (9/2){ (((−4z−4)(z^2 −2z) −4(z−1)(−2z^2 −4z +4))/((z^2 −2z)^3 ))} =(9/2) ((8(6))/(27)) = ((8.6)/(2.3)) = 4.2=8 ∫_C ϕ(z)dz=2iπ{−9 +(2/3) +8} =2iπ(−1+(2/3))=2iπ(−(1/3))=−((2iπ)/3)$
$l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{9 (z^{2} + 2)}{z {(z + 1)}^{3} (z - 2)} t h e p o l e s o f φ a r e$ $0, - 1, 2 (a l l a r e a t i n t e r i o r o f c i r c l e C)$ $\int_{C} φ - z) d z = 2 i π {R e s (φ, 0) + R e s (φ, - 1) + R e s (φ, 2)$ $R e s (φ, 0) = {l i m}_{z \to 0} z φ (z) = \frac{18}{- 2} = - 9$ $R e s (φ, 2) = {l i m}_{z \to 2} (z - 2) φ (z)$ $= \frac{36}{2.27} = \frac{18}{27} = \frac{2.9}{3.9} = \frac{2}{3}$ $R e s (φ, - 1) = {l i m}_{z \to - 1} \frac{1}{(3 - 1)!} {{(z + 1)}^{3} φ (z)}^{(2)}$ $= {l i m}_{z \to - 1} \frac{9}{2} {\frac{z^{2} + 2}{z^{2} - 2 z}}^{(2)}$ $= {l i m}_{z \to - 1} \frac{9}{2} {\frac{2 z (z^{2} - 2 z) - (2 z - 2) (z^{2} + 2)}{{(z^{2} - 2 z)}^{2}}}^{(1)}$ $= {l i m}_{z \to - 1} \frac{9}{2} {\frac{2 z^{3} - 4 z^{2} - 2 z^{3} - 4 z + 2 z^{2} + 4}{{(z^{2} - 2 z)}^{2}}}^{(1)}$ $= {l i m}_{z \to - 1} \frac{9}{2} {\frac{- 2 z^{2} - 4 z + 4}{{(z^{2} - 2 z)}^{2}}}^{(1)}$ $= {l i m}_{z \to - 1} \frac{9}{2} {\frac{(- 4 z - 4) {(z^{2} - 2 z)}^{2} - 2 (2 z - 2) (z^{2} - 2 z) (- 2 z^{2} - 4 z + 4)}{{(z^{2} - 2 z)}^{4}}}$ $= {l i m}_{z \to - 1} \frac{9}{2} {\frac{(- 4 z - 4) (z^{2} - 2 z) - 4 (z - 1) (- 2 z^{2} - 4 z + 4)}{{(z^{2} - 2 z)}^{3}}}$ $= \frac{9}{2} \frac{8 (6)}{27} = \frac{8.6}{2.3} = 4.2 = 8$ $\int_{C} φ (z) d z = 2 i π {- 9 + \frac{2}{3} + 8}$ $= 2 i π (- 1 + \frac{2}{3}) = 2 i π (- \frac{1}{3}) = - \frac{2 i π}{3}$
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