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Question-145817




Question Number 145817 by tabata last updated on 08/Jul/21
Commented by tabata last updated on 08/Jul/21
solve by complex number
$${solve}\:{by}\:{complex}\:{number} \\ $$
Answered by mathmax by abdo last updated on 09/Jul/21
Ψ=_(e^(iθ)  =z)     ∫_C^+      (dz/(iz(5+3((z+z^(−1) )/2)))) =∫_C^+    ((2dz)/(iz(10+3z+3z^(−1) )))  =∫_C^+    ((−2idz)/(10z+3z^2  +3))=∫_C^+    ((−2idz)/(3z^2  +10z +3)) →ϕ(z)=((−2i)/(3(z+(1/3))(z+3)))  Δ^′  =5^2 −9=16 ⇒z_1 =((−5+4)/3)=−(1/3)  z_2 =((−5−4)/3)=−3  ∫_C^+    ϕ(z)dz=iπRes(ϕ,−(1/3))=iπ×((−2i)/(3(3−(1/3))))  =((2π)/(3((8/3))))=((2π)/8)=(π/4)   (C^+  is half circle positif)
$$\Psi=_{\mathrm{e}^{\mathrm{i}\theta} \:=\mathrm{z}} \:\:\:\:\int_{\mathrm{C}^{+} } \:\:\:\:\frac{\mathrm{dz}}{\mathrm{iz}\left(\mathrm{5}+\mathrm{3}\frac{\mathrm{z}+\mathrm{z}^{−\mathrm{1}} }{\mathrm{2}}\right)}\:=\int_{\mathrm{C}^{+} } \:\:\frac{\mathrm{2dz}}{\mathrm{iz}\left(\mathrm{10}+\mathrm{3z}+\mathrm{3z}^{−\mathrm{1}} \right)} \\ $$$$=\int_{\mathrm{C}^{+} } \:\:\frac{−\mathrm{2idz}}{\mathrm{10z}+\mathrm{3z}^{\mathrm{2}} \:+\mathrm{3}}=\int_{\mathrm{C}^{+} } \:\:\frac{−\mathrm{2idz}}{\mathrm{3z}^{\mathrm{2}} \:+\mathrm{10z}\:+\mathrm{3}}\:\rightarrow\varphi\left(\mathrm{z}\right)=\frac{−\mathrm{2i}}{\mathrm{3}\left(\mathrm{z}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{z}+\mathrm{3}\right)} \\ $$$$\Delta^{'} \:=\mathrm{5}^{\mathrm{2}} −\mathrm{9}=\mathrm{16}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{5}+\mathrm{4}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{5}−\mathrm{4}}{\mathrm{3}}=−\mathrm{3} \\ $$$$\int_{\mathrm{C}^{+} } \:\:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{i}\pi\mathrm{Res}\left(\varphi,−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{i}\pi×\frac{−\mathrm{2i}}{\mathrm{3}\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{3}\left(\frac{\mathrm{8}}{\mathrm{3}}\right)}=\frac{\mathrm{2}\pi}{\mathrm{8}}=\frac{\pi}{\mathrm{4}}\:\:\:\left(\mathrm{C}^{+} \:\mathrm{is}\:\mathrm{half}\:\mathrm{circle}\:\mathrm{positif}\right) \\ $$

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