∫_(C:∣z∣=1) ((−2i)/(Az^2 +2z+A))dz where C is a unit circle with radius 1


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Question Number 131743 by frc2crc last updated on 08/Feb/21

$\int_{C :∣ z ∣= 1} \frac{- 2 i}{{A z}^{2} + 2 z + A} d z w h e r e C i s a u n i t c i r c l e w i t h r a d i u s 1$
	Answered by mathmax by abdo last updated on 08/Feb/21
	$let ϕ(z)=((−2i)/(az^2 +2z+a)) poles of ϕ? Δ^′ =1−a^2 case 1 ∣a∣<1 ⇒z_1 =((−1+(√(1−a^2 )))/a) and z_2 =((−1−(√(1−a^2 )))/a) (we suppose a>0) ∣z_1 ∣−1=(1/a)(1−(√(1−a^2 )))−1 =((1−(√(1−a^2 ))−a)/a) =((1−a−(√(1−a^2 )))/a)<0 because (1−a)^2 −(1−a^2 ) =a^2 −2a+1−1+a^2 =2a^2 −2a =2a(a−1)<0 ∣z_2 ∣−1 =((1+(√(1−a^2 )))/a)−1 =((1−a+(√(1−a^2 )))/a)>0(out of circle) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπRes(ϕ,z_1 ) =2iπ×((−2i)/(a(z_1 −z_2 ))) =((4π)/(a.((2(√(1−a^2 )))/a))) =((2π)/( (√(1−a^2 )))) case 2 ∣a∣>1 ⇒z_1 =((−1+i(√(a^2 −1)))/a) and z_2 =((−1−i(√(a^2 −1)))/a) ∣z_1 ∣ =(1/a)(√(1+a^2 −1))=1 and ∣z_2 ∣=1 ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ{ Res(ϕ,z_1 )+Res(ϕ,z_2 )} =2iπ{((−2i)/(a(z_1 −z_2 )))+((−2i)/(a(z_2 −z_1 )))}=0$
	$let φ (z) = \frac{- 2 i}{{az}^{2} + 2 z + a} poles of φ ?$ $Δ^{^{'}} = 1 - a^{2} case 1 ∣ a ∣< 1 \Rightarrow z_{1} = \frac{- 1 + \sqrt{1 - a^{2}}}{a} and z_{2} = \frac{- 1 - \sqrt{1 - a^{2}}}{a}$ $(we suppose a > 0) ∣ z_{1} ∣ - 1 = \frac{1}{a} (1 - \sqrt{1 - a^{2}}) - 1 = \frac{1 - \sqrt{1 - a^{2}} - a}{a}$ $= \frac{1 - a - \sqrt{1 - a^{2}}}{a} < 0 because$ ${(1 - a)}^{2} - (1 - a^{2}) = a^{2} - 2 a + 1 - 1 + a^{2} = {2 a}^{2} - 2 a = 2 a (a - 1) < 0$ $∣ z_{2} ∣ - 1 = \frac{1 + \sqrt{1 - a^{2}}}{a} - 1 = \frac{1 - a + \sqrt{1 - a^{2}}}{a} > 0 (out of circle) \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) dz = 2 i π Res (φ, z_{1}) = 2 i π \times \frac{- 2 i}{a (z_{1} - z_{2})}$ $= \frac{4 π}{a . \frac{2 \sqrt{1 - a^{2}}}{a}} = \frac{2 π}{\sqrt{1 - a^{2}}}$ $case 2 ∣ a ∣> 1 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{a^{2} - 1}}{a} and z_{2} = \frac{- 1 - i \sqrt{a^{2} - 1}}{a}$ $∣ z_{1} ∣ = \frac{1}{a} \sqrt{1 + a^{2} - 1} = 1 and ∣ z_{2} ∣= 1 \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) dz = 2 i π {Res (φ, z_{1}) + Res (φ, z_{2})}$ $= 2 i π {\frac{- 2 i}{a (z_{1} - z_{2})} + \frac{- 2 i}{a (z_{2} - z_{1})}} = 0$
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