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Question-201873




Question Number 201873 by sonukgindia last updated on 14/Dec/23
Answered by witcher3 last updated on 14/Dec/23
Re((1/(1+e^(iθ) )))=((1+e^(−iθ) )/((1+e^(iθ) )(1+e^(−iθ) )))=((1+cos(θ))/(2+2cos(θ)))=(1/2);∀θ∈R−{(1+2k)π}  I==((ϕ^3 −1)/2)  ϕ^2 =1+ϕ  ϕ^3 =ϕ^2 +ϕ=1+2ϕ⇒ϕ^3 −1=2ϕ  I=ϕ=((1+(√5))/2)
$$\mathrm{Re}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }\right)=\frac{\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} \right)\left(\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} \right)}=\frac{\mathrm{1}+\mathrm{cos}\left(\theta\right)}{\mathrm{2}+\mathrm{2cos}\left(\theta\right)}=\frac{\mathrm{1}}{\mathrm{2}};\forall\theta\in\mathbb{R}−\left\{\left(\mathrm{1}+\mathrm{2k}\right)\pi\right\} \\ $$$$\mathrm{I}==\frac{\varphi^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\ $$$$\varphi^{\mathrm{2}} =\mathrm{1}+\varphi \\ $$$$\varphi^{\mathrm{3}} =\varphi^{\mathrm{2}} +\varphi=\mathrm{1}+\mathrm{2}\varphi\Rightarrow\varphi^{\mathrm{3}} −\mathrm{1}=\mathrm{2}\varphi \\ $$$$\mathrm{I}=\varphi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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