Calculate 1/ I = ∮_c^+ ((zdz)/((z−1)^2 (z^2 −2z+1−2i))) ,C={z/∣z∣=2} 2/ J =∮_c^+ ((ch(z)dz)/(z(e^z −1))) , C={z/∣z−3i∣=4} 3/ K=∮_c^+ ((sin(z)dz)/(z^3 (z+1)^2 )) , C={z/∣z∣=2}


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Question Number 131094 by Chhing last updated on 01/Feb/21
$Calculate 1/ I = ∮_c^+ ((zdz)/((z−1)^2 (z^2 −2z+1−2i))) ,C={z/∣z∣=2} 2/ J =∮_c^+ ((ch(z)dz)/(z(e^z −1))) , C={z/∣z−3i∣=4} 3/ K=∮_c^+ ((sin(z)dz)/(z^3 (z+1)^2 )) , C={z/∣z∣=2}$
$Calculate$ $1 / I = \oint_{c^{+}} \frac{zdz}{{(z - 1)}^{2} (z^{2} - 2 z + 1 - 2 i)}, C = {z / ∣ z ∣= 2}$ $2 / J = \oint_{c^{+}} \frac{ch (z) dz}{z (e^{z} - 1)}, C = {z / ∣ z - 3 i ∣= 4}$ $3 / K = \oint_{c^{+}} \frac{\sin (z) dz}{z^{3} {(z + 1)}^{2}}, C = {z / ∣ z ∣= 2}$
Commented by mathmax by abdo last updated on 02/Feb/21

$sir what do you mean by C^{+} ? define it . . .$
	Answered by mathmax by abdo last updated on 01/Feb/21
	$1) I =∫_C^+ ((zdz)/((z−1)^2 (z^2 −2z+1−2i))) let ϕ(z)=(z/((z−1)^2 (z^2 −2z+1−2i))) z^2 −2z+1−2i=0 ⇒Δ^′ =1−(1−2i)=2i ⇒z_1 =1+(√(2i))=1+(√2)e^((iπ)/4) z_2 =1−(√2)e^((iπ)/4) we have z_1 =1+(√2){(1/( (√2)))+(i/( (√2)))}=1+1+i=2+i ⇒ ∣z_1 ∣=(√(4+1))=(√5)>(√2) and z_2 =1−(√2){(1/( (√2)))−(i/( (√2)))} =1−1+i=i ⇒∣z_1 ∣<2 z_2 ∈C^+ also z=1 is double pole for ϕ ⇒ I =∫_C^+ ϕ(z)dz =2iπ{Res(ϕ,1) +Res(ϕ,i)} Res(ϕ,1) =lim_(z→1 ) (1/((2−1)!)){(z−1)^2 ϕ(z)}^((1)) =lim_(z→1) ((z/(z^2 −2z+1−2i)))^((1)) =lim_(z→1) (((z^2 −2z+1−2i−z(2z−2))/((z^2 −2z+1−2i)^2 ))) =((1−2+1−2i−o)/((1−2+1−2i)^2 ))=((−2i)/((−2i)^2 ))=((−2i)/(−4)) =(i/2) Res(ϕ,i)=lim_(z→i) (z−i)ϕ(z) =lim_(z→i) (z−i)(z/((z−1)^2 (z−i)(z−2−i))) =lim_(z→i) (z/((z−1)^2 (z−2−i)))=(i/((i−1)^2 (−2))) =((−i)/(2(−2i)))=(1/4) ⇒ I =2iπ{(i/2)+(1/4)} =−π +((iπ)/2)$
	$1) I = \int_{C^{+}} \frac{zdz}{{(z - 1)}^{2} (z^{2} - 2 z + 1 - 2 i)} let φ (z) = \frac{z}{{(z - 1)}^{2} (z^{2} - 2 z + 1 - 2 i)}$ $z^{2} - 2 z + 1 - 2 i = 0 \Rightarrow Δ^{^{'}} = 1 - (1 - 2 i) = 2 i \Rightarrow z_{1} = 1 + \sqrt{2 i} = 1 + \sqrt{2} e^{\frac{i π}{4}}$ $z_{2} = 1 - \sqrt{2} e^{\frac{i π}{4}} we have z_{1} = 1 + \sqrt{2} {\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}} = 1 + 1 + i = 2 + i \Rightarrow$ $∣ z_{1} ∣= \sqrt{4 + 1} = \sqrt{5} > \sqrt{2} and z_{2} = 1 - \sqrt{2} {\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}} = 1 - 1 + i = i \Rightarrow∣ z_{1} ∣< 2$ $z_{2} \in C^{+} also z = 1 is double pole for φ \Rightarrow$ $I = \int_{C^{+}} φ (z) dz = 2 i π {Res (φ, 1) + Res (φ, i)}$ $Res (φ, 1) = \lim_{z \to 1} \frac{1}{(2 - 1)!} {{(z - 1)}^{2} φ (z)}^{(1)}$ $= \lim_{z \to 1} {(\frac{z}{z^{2} - 2 z + 1 - 2 i})}^{(1)}$ $= \lim_{z \to 1} (\frac{z^{2} - 2 z + 1 - 2 i - z (2 z - 2)}{{(z^{2} - 2 z + 1 - 2 i)}^{2}})$ $= \frac{1 - 2 + 1 - 2 i - o}{{(1 - 2 + 1 - 2 i)}^{2}} = \frac{- 2 i}{{(- 2 i)}^{2}} = \frac{- 2 i}{- 4} = \frac{i}{2}$ $Res (φ, i) = \lim_{z \to i} (z - i) φ (z) = \lim_{z \to i} (z - i) \frac{z}{{(z - 1)}^{2} (z - i) (z - 2 - i)}$ $= \lim_{z \to i} \frac{z}{{(z - 1)}^{2} (z - 2 - i)} = \frac{i}{{(i - 1)}^{2} (- 2)} = \frac{- i}{2 (- 2 i)} = \frac{1}{4} \Rightarrow$ $I = 2 i π {\frac{i}{2} + \frac{1}{4}} = - π + \frac{i π}{2}$
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