Menu Close

Question-164553

Tinku Tara 0 Comments
Pin




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
$$\Psi=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{d}\theta}{\mathrm{1}+\mathrm{tcos}\theta}\:\:\mathrm{changement}\:\mathrm{e}^{\mathrm{i}\theta} =\mathrm{z}\:\mathrm{give} \\ $$$$\Psi=\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{dz}}{\mathrm{iz}\left(\mathrm{1}+\mathrm{t}\frac{\mathrm{z}+\mathrm{z}^{−\mathrm{1}} }{\mathrm{2}}\right)}=\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2idz}}{\mathrm{z}\left(\mathrm{2}+\mathrm{tz}+\mathrm{tz}^{−\mathrm{1}} \right)} \\ $$$$=\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2idz}}{\mathrm{2z}+\mathrm{tz}^{\mathrm{2}} \:+\mathrm{t}}=\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2idz}}{\mathrm{tz}^{\mathrm{2}} +\mathrm{2z}+\mathrm{t}} \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{−\mathrm{2i}}{\mathrm{tz}^{\mathrm{2}} \:+\mathrm{2z}+\mathrm{t}} \\ $$$$\mathrm{tz}^{\mathrm{2}\:} +\mathrm{2z}+\mathrm{t}=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{1}−\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{case1}\:\:\Delta^{'} <\mathrm{0}\:\Rightarrow\:\mathrm{2complex}\:\mathrm{roots}\:\:\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{t}}\:\:\:\left(\mathrm{t}\neq\mathrm{0}\right) \\ $$$$\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{t}} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \mid−\mathrm{1}=\frac{\mathrm{1}}{\mid\mathrm{t}\mid}\sqrt{\mathrm{1}+\mathrm{t}^{\mathrm{2}} −\mathrm{1}}=\mathrm{1}−\mathrm{1}=\mathrm{0}\:\Rightarrow\mid\mathrm{z}_{\mathrm{1}} \mid=\mathrm{1}\:\:\mathrm{also} \\ $$$$\mid\mathrm{z}_{\mathrm{2}} \mid−\mathrm{1}=\mathrm{0}\:\Rightarrow\mid\mathrm{z}_{\mathrm{2}} \mid=\mathrm{1} \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{1}} \varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)+\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{2}} \right)\right\}=\mathrm{0} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{2}} \right)=−\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right) \\ $$$$\mathrm{case2}\:\:\Delta^{'} >\mathrm{0}\Rightarrow\mathrm{1}−\mathrm{t}^{\mathrm{2}} >\mathrm{0}\:\Rightarrow\mid\mathrm{t}\mid<\mathrm{1}\:\Rightarrow\mathrm{2racines} \\ $$$$\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}} \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \mid−\mathrm{1}=\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mid\mathrm{t}\mid}−\mathrm{1}=\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }−\mid\mathrm{t}\mid}{\mid\mathrm{g}\mid}=\frac{\mathrm{1}−\mid\mathrm{t}\mid−\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mid\mathrm{t}\mid} \\ $$$$\left(\mathrm{1}−\mid\mathrm{t}\mid\right)^{\mathrm{2}} −\mathrm{1}+\mathrm{t}^{\mathrm{2}} =\mathrm{t}^{\mathrm{2}} −\mathrm{2}\mid\mathrm{t}\mid+\mathrm{1}−\mathrm{1}+\mathrm{t}^{\mathrm{2}} =\mathrm{2t}^{\mathrm{2}} −\mathrm{2}\mid\mathrm{t}\mid=\mathrm{2}\mid\mathrm{t}\mid\left(\mathrm{t}−\mathrm{1}\right)<\mathrm{0}\:\Rightarrow \\ $$$$\mid\mathrm{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid\mathrm{z}_{\mathrm{2}} \mid−\mathrm{1}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mid\mathrm{t}\mid}−\mathrm{1}=\frac{\mathrm{1}−\mid\mathrm{t}\mid+\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mid\mathrm{t}\mid}>\mathrm{0}\:\Rightarrow\mid\mathrm{z}_{\mathrm{2}} \mid>\mathrm{1}\left(\mathrm{out}\:\mathrm{of}\:\mathrm{circle}\right) \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)\:\mathrm{wehave}\:\varphi\left(\mathrm{z}\right)=\frac{−\mathrm{2i}}{\mathrm{t}\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)} \\ $$$$\Rightarrow\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)=\frac{−\mathrm{2i}}{\mathrm{t}\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)}=\frac{−\mathrm{2i}}{\mathrm{t}×\frac{\mathrm{2}\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}}}=\frac{−\mathrm{i}}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi.\frac{−\mathrm{i}}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}=\Psi \\ $$
Answered by mkam last updated on 19/Jan/22
Answered by mkam last updated on 19/Jan/22

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *