Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 164553 by mkam last updated on 18/Jan/22

Commented by tabata last updated on 18/Jan/22

$? ? ? ?$
	Answered by mathmax by abdo last updated on 19/Jan/22
	$Ψ=∫_0 ^(2π) (dθ/(1+tcosθ)) changement e^(iθ) =z give Ψ=∫_(∣z∣=1) (dz/(iz(1+t((z+z^(−1) )/2))))=∫_(∣z∣=1) ((−2idz)/(z(2+tz+tz^(−1) ))) =∫_(∣z∣=1) ((−2idz)/(2z+tz^2 +t))=∫_(∣z∣=1) ((−2idz)/(tz^2 +2z+t)) ϕ(z)=((−2i)/(tz^2 +2z+t)) tz^(2 ) +2z+t=0→Δ^′ =1−t^2 case1 Δ^′ <0 ⇒ 2complex roots z_1 =((−1+i(√(t^2 −1)))/t) (t≠0) z_2 =((−1−i(√(t^2 −1)))/t) ∣z_1 ∣−1=(1/(∣t∣))(√(1+t^2 −1))=1−1=0 ⇒∣z_1 ∣=1 also ∣z_2 ∣−1=0 ⇒∣z_2 ∣=1 ∫_(∣z∣=1) ϕ(z)dz=2iπ{Res(ϕ,z_1 )+Res(ϕ,z_2 )}=0 due to Res(ϕ,z_2 )=−Res(ϕ,z_1 ) case2 Δ^′ >0⇒1−t^2 >0 ⇒∣t∣<1 ⇒2racines z_1 =((−1+(√(1−t^2 )))/t) and z_2 =((−1−(√(1−t^2 )))/t) ∣z_1 ∣−1=((1−(√(1−t^2 )))/(∣t∣))−1=((1−(√(1−t^2 ))−∣t∣)/(∣g∣))=((1−∣t∣−(√(1−t^2 )))/(∣t∣)) (1−∣t∣)^2 −1+t^2 =t^2 −2∣t∣+1−1+t^2 =2t^2 −2∣t∣=2∣t∣(t−1)<0 ⇒ ∣z_1 ∣<1 ∣z_2 ∣−1=((1+(√(1−t^2 )))/(∣t∣))−1=((1−∣t∣+(√(1−t^2 )))/(∣t∣))>0 ⇒∣z_2 ∣>1(out of circle) ∫_(∣z∣=1) ϕ(z)dz=2iπ Res(ϕ,z_1 ) wehave ϕ(z)=((−2i)/(t(z−z_1 )(z−z_2 ))) ⇒Res(ϕ,z_1 )=((−2i)/(t(z_1 −z_2 )))=((−2i)/(t×((2(√(1−t^2 )))/t)))=((−i)/( (√(1−t^2 )))) ⇒ ∫_(∣z∣=1) ϕ(z)dz=2iπ.((−i)/( (√(1−t^2 ))))=((2π)/( (√(1−t^2 ))))=Ψ$
	$Ψ = \int_{0}^{2 π} \frac{d θ}{1 + tcos θ} changement e^{i θ} = z give$ $Ψ = \int_{∣ z ∣= 1} \frac{dz}{iz (1 + t \frac{z + z^{- 1}}{2})} = \int_{∣ z ∣= 1} \frac{- 2 idz}{z (2 + tz + {tz}^{- 1})}$ $= \int_{∣ z ∣= 1} \frac{- 2 idz}{2 z + {tz}^{2} + t} = \int_{∣ z ∣= 1} \frac{- 2 idz}{{tz}^{2} + 2 z + t}$ $φ (z) = \frac{- 2 i}{{tz}^{2} + 2 z + t}$ ${tz}^{2} + 2 z + t = 0 \to Δ^{^{'}} = 1 - t^{2}$ $case 1 Δ^{^{'}} < 0 \Rightarrow 2 complex roots z_{1} = \frac{- 1 + i \sqrt{t^{2} - 1}}{t} (t \neq 0)$ $z_{2} = \frac{- 1 - i \sqrt{t^{2} - 1}}{t}$ $∣ z_{1} ∣ - 1 = \frac{1}{∣ t ∣} \sqrt{1 + t^{2} - 1} = 1 - 1 = 0 \Rightarrow∣ z_{1} ∣= 1 also$ $∣ z_{2} ∣ - 1 = 0 \Rightarrow∣ z_{2} ∣= 1$ $\int_{∣ z ∣= 1} φ (z) dz = 2 i π {Res (φ, z_{1}) + Res (φ, z_{2})} = 0$ $due to Res (φ, z_{2}) = - Res (φ, z_{1})$ $case 2 Δ^{^{'}} > 0 \Rightarrow 1 - t^{2} > 0 \Rightarrow∣ t ∣< 1 \Rightarrow 2 racines$ $z_{1} = \frac{- 1 + \sqrt{1 - t^{2}}}{t} and z_{2} = \frac{- 1 - \sqrt{1 - t^{2}}}{t}$ $∣ z_{1} ∣ - 1 = \frac{1 - \sqrt{1 - t^{2}}}{∣ t ∣} - 1 = \frac{1 - \sqrt{1 - t^{2}} - ∣ t ∣}{∣ g ∣} = \frac{1 - ∣ t ∣ - \sqrt{1 - t^{2}}}{∣ t ∣}$ ${(1 - ∣ t ∣)}^{2} - 1 + t^{2} = t^{2} - 2 ∣ t ∣ + 1 - 1 + t^{2} = {2 t}^{2} - 2 ∣ t ∣= 2 ∣ t ∣ (t - 1) < 0 \Rightarrow$ $∣ z_{1} ∣< 1$ $∣ z_{2} ∣ - 1 = \frac{1 + \sqrt{1 - t^{2}}}{∣ t ∣} - 1 = \frac{1 - ∣ t ∣ + \sqrt{1 - t^{2}}}{∣ t ∣} > 0 \Rightarrow∣ z_{2} ∣> 1 (out of circle)$ $\int_{∣ z ∣= 1} φ (z) dz = 2 i π Res (φ, z_{1}) wehave φ (z) = \frac{- 2 i}{t (z - z_{1}) (z - z_{2})}$ $\Rightarrow Res (φ, z_{1}) = \frac{- 2 i}{t (z_{1} - z_{2})} = \frac{- 2 i}{t \times \frac{2 \sqrt{1 - t^{2}}}{t}} = \frac{- i}{\sqrt{1 - t^{2}}} \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) dz = 2 i π . \frac{- i}{\sqrt{1 - t^{2}}} = \frac{2 π}{\sqrt{1 - t^{2}}} = Ψ$
	Answered by mkam last updated on 19/Jan/22

	Answered by mkam last updated on 19/Jan/22

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum