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