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

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

$$\underset{{C}:\mid{z}\mid=\mathrm{1}} {\int}\frac{−\mathrm{2}{i}}{{Az}^{\mathrm{2}} +\mathrm{2}{z}+{A}}{dz}\:{where}\:{C}\:{is}\:{a}\:{unit}\:{circle}\:{with}\:{radius}\:\mathrm{1} \\ $$

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

$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{−\mathrm{2i}}{\mathrm{az}^{\mathrm{2}} \:+\mathrm{2z}+\mathrm{a}}\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\Delta^{'} \:=\mathrm{1}−\mathrm{a}^{\mathrm{2}} \:\:\mathrm{case}\:\mathrm{1}\:\:\mid\mathrm{a}\mid<\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} \:=\frac{−\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}} \\ $$$$\left(\mathrm{we}\:\mathrm{suppose}\:\mathrm{a}>\mathrm{0}\right)\:\:\mid\mathrm{z}_{\mathrm{1}} \mid−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{a}}\left(\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }\right)−\mathrm{1}\:=\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }−\mathrm{a}}{\mathrm{a}} \\ $$$$=\frac{\mathrm{1}−\mathrm{a}−\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}}<\mathrm{0}\:\:\mathrm{because} \\ $$$$\left(\mathrm{1}−\mathrm{a}\right)^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)\:=\mathrm{a}^{\mathrm{2}} −\mathrm{2a}+\mathrm{1}−\mathrm{1}+\mathrm{a}^{\mathrm{2}} \:=\mathrm{2a}^{\mathrm{2}} −\mathrm{2a}\:=\mathrm{2a}\left(\mathrm{a}−\mathrm{1}\right)<\mathrm{0} \\ $$$$\mid\mathrm{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}}−\mathrm{1}\:=\frac{\mathrm{1}−\mathrm{a}+\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}}>\mathrm{0}\left(\mathrm{out}\:\mathrm{of}\:\mathrm{circle}\right)\:\Rightarrow \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)\:=\mathrm{2i}\pi×\frac{−\mathrm{2i}}{\mathrm{a}\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{4}\pi}{\mathrm{a}.\frac{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\mathrm{a}}}\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }} \\ $$$$\mathrm{case}\:\mathrm{2}\:\:\mid\mathrm{a}\mid>\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{a}}\:\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{a}} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \mid\:=\frac{\mathrm{1}}{\mathrm{a}}\sqrt{\mathrm{1}+\mathrm{a}^{\mathrm{2}} −\mathrm{1}}=\mathrm{1}\:\:\mathrm{and}\:\mid\mathrm{z}_{\mathrm{2}} \mid=\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)+\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{2}} \right)\right\} \\ $$$$=\mathrm{2i}\pi\left\{\frac{−\mathrm{2i}}{\mathrm{a}\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)}+\frac{−\mathrm{2i}}{\mathrm{a}\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)}\right\}=\mathrm{0} \\ $$

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