let C={z/∣z∣=2} find ∫_C ((z^2 sinz cosz)/((z^2 +1)^2 ))dz


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Question Number 125506 by mathmax by abdo last updated on 11/Dec/20
$let C={z/∣z∣=2} find ∫_C ((z^2 sinz cosz)/((z^2 +1)^2 ))dz$
$let C = {z / ∣ z ∣= 2} find \int_{C} \frac{z^{2} sinz cosz}{{(z^{2} + 1)}^{2}} dz$
Commented by ZiYangLee last updated on 11/Dec/20
$First, we let the complex function f(z)=((z^2 sin(z)cos(z))/((1+z^2 )^2 )) Noted that f(z) has two second polar poles inside the circle ∣z∣=2, By Residue Theorom, we have ∮_(∣z∣=2) ((z^2 sin(z)cos(z))/((1+z^2 )^2 )) =2πi{Re_(z=i) sf(z)+Re_(z=−i) s f(z)} =2πi{lim_(z→i) [(z−i)^2 f(z)]+lim_(z→i) (d/dz)[(z+i)^2 f(z)]} =2πi(((3e^4 +1)/(16e^2 ))+((3e^4 +1)/(16e^2 ))) =((3e^4 +1)/4)πi$
$First, we let the complex function$ $f (z) = \frac{z^{2} \sin (z) \cos (z)}{{(1 + z^{2})}^{2}}$ $Noted that f (z) has two second polar$ $poles inside the circle ∣ z ∣= 2,$ $By Residue Theorom, we have$ $\oint_{∣ z ∣= 2} \frac{z^{2} \sin (z) \cos (z)}{{(1 + z^{2})}^{2}}$ $= 2 π i {R \underset{z = i}{e s} f (z) + R \underset{z = - i}{e s} f (z)}$ $= 2 π i {\lim_{z \to i} [{(z - i)}^{2} f (z)] + \lim_{z \to i} \frac{d}{d z} [{(z + i)}^{2} f (z)]}$ $= 2 π i (\frac{3 e^{4} + 1}{16 e^{2}} + \frac{3 e^{4} + 1}{16 e^{2}})$ $= \frac{3 e^{4} + 1}{4} π i$
	Answered by mathmax by abdo last updated on 11/Dec/20
	$let ϕ(z)=((z^2 sinz cosz)/((z^2 +1)^2 )) ⇒ϕ(z)=(1/2)((z^2 sin(2z))/((z−i)^2 (z+i)^2 )) residus theorem give ∫_C ϕ(z)dz=2iπ{Res(ϕ,i)+Res(ϕ,−i)} (i snd −i are double poles) ⇒ Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {((z^2 sin(2z))/(2(z+i)^2 ))}^((1)) =(1/2)lim_(z→i) (((2zsin(2z)+2z^2 cos(2z))(z+i)^2 −2(z+i)z^2 sin(2z))/((z+i)^4 )) =lim_(z→i) (((zsin(2z)+z^2 cos(2z))(z+i)−z^2 sin2z))/((z+i)^3 )) =(((2i)(isin(2i)−cos(2i))+sin(2i))/((2i)^3 )) =((−2sin(2i)−2icos(2i)+sin(2i))/(−8i)) =((sin(2i)+2icos(2i))/(8i)) Res(ϕ,−i)=lim_(z→−i) (1/((2−1)!)){(z+i)^2 ϕ(z)}^((1)) =(1/2)lim_(z→−i) { ((z^2 sin(2z))/((z−i)^2 ))}^((1)) =(1/2)lim_(z→−i) (((2zsin(2z)+2z^2 cos(2z))(z−i)^2 −2(z−i)z^2 sin(2z))/((z−i)^4 )) =lim_(z→−i) (((zsin(2z)+z^2 cos(2z))(z−i)−z^2 sin(2z))/((z−i)^3 )) =(((−2i)(isin(2i)−cos(2i))−sin(2i))/((−2i)^3 )) =(((−2i)(isin(2i)−cos(2i))−sin(2i))/(8i))=((2sin(2i)+2icos(2i)−sin(2i))/(8i)) =((sin(2i)+2icos(2i))/(8i)) ⇒ ∫_C ϕ(z)dz =2iπ{((sin(2i)+2icos(2i)+sin(2i)+2icos(2i))/(8i))} =(π/4)(2sin(2i)+4icos(2i)) =(π/2)sin(2i)+iπ cos(2i) sinz =((e^(iz) −e^(−iz) )/(2i)) ⇒sin(2i)=((e^(i(2i)) −e^(−i(2i)) )/(2i))=((e^(−2) −e^2 )/(2i)) =ish(2) cosz =((e^(iz) +e^(−iz) )/2)⇒cos(2i)=((e^(−2) +e^2 )/2)=ch(2) ⇒ ⇒∫_C ϕ(z)dz =((iπ)/2)sh(2)+iπch(2)$
	$let φ (z) = \frac{z^{2} sinz cosz}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{1}{2} \frac{z^{2} \sin (2 z)}{{(z - i)}^{2} {(z + i)}^{2}}$ $residus theorem give \int_{C} φ (z) dz = 2 i π {Res (φ, i) + Res (φ, - i)}$ $(i snd - i are double poles) \Rightarrow$ $Res (φ, i) = \lim_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= \lim_{z \to i} {\frac{z^{2} \sin (2 z)}{2 {(z + i)}^{2}}}^{(1)} = \frac{1}{2} \lim_{z \to i} \frac{(2 zsin (2 z) + {2 z}^{2} \cos (2 z)) {(z + i)}^{2} - 2 (z + i) z^{2} \sin (2 z)}{{(z + i)}^{4}}$ $= \lim_{z \to i} \frac{(zsin (2 z) + z^{2} \cos (2 z)) (z + i) - z^{2} \sin 2 z)}{{(z + i)}^{3}}$ $= \frac{(2 i) (isin (2 i) - \cos (2 i)) + \sin (2 i)}{{(2 i)}^{3}} = \frac{- 2 \sin (2 i) - 2 icos (2 i) + \sin (2 i)}{- 8 i}$ $= \frac{\sin (2 i) + 2 icos (2 i)}{8 i}$ $Res (φ, - i) = \lim_{z \to - i} \frac{1}{(2 - 1)!} {{(z + i)}^{2} φ (z)}^{(1)}$ $= \frac{1}{2} \lim_{z \to - i} {\frac{z^{2} \sin (2 z)}{{(z - i)}^{2}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to - i} \frac{(2 zsin (2 z) + {2 z}^{2} \cos (2 z)) {(z - i)}^{2} - 2 (z - i) z^{2} \sin (2 z)}{{(z - i)}^{4}}$ $= \lim_{z \to - i} \frac{(zsin (2 z) + z^{2} \cos (2 z)) (z - i) - z^{2} \sin (2 z)}{{(z - i)}^{3}}$ $= \frac{(- 2 i) (isin (2 i) - \cos (2 i)) - \sin (2 i)}{{(- 2 i)}^{3}}$ $= \frac{(- 2 i) (isin (2 i) - \cos (2 i)) - \sin (2 i)}{8 i} = \frac{2 \sin (2 i) + 2 icos (2 i) - \sin (2 i)}{8 i}$ $= \frac{\sin (2 i) + 2 icos (2 i)}{8 i} \Rightarrow$ $\int_{C} φ (z) dz = 2 i π {\frac{\sin (2 i) + 2 icos (2 i) + \sin (2 i) + 2 icos (2 i)}{8 i}}$ $= \frac{π}{4} (2 \sin (2 i) + 4 icos (2 i)) = \frac{π}{2} \sin (2 i) + i π \cos (2 i)$ $sinz = \frac{e^{iz} - e^{- iz}}{2 i} \Rightarrow \sin (2 i) = \frac{e^{i (2 i)} - e^{- i (2 i)}}{2 i} = \frac{e^{- 2} - e^{2}}{2 i}$ $= ish (2)$ $cosz = \frac{e^{iz} + e^{- iz}}{2} \Rightarrow \cos (2 i) = \frac{e^{- 2} + e^{2}}{2} = ch (2) \Rightarrow$ $\Rightarrow \int_{C} φ (z) dz = \frac{i π}{2} sh (2) + i π ch (2)$
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