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Question Number 201873 by sonukgindia last updated on 14/Dec/23

Answered by witcher3 last updated on 14/Dec/23

$$\mathrm{Re}\left(\frac{1}{1+{\mathrm{e}}^{\mathrm{i}\theta }}\right)=\frac{1+{\mathrm{e}}^{-\mathrm{i}\theta }}{(1+{\mathrm{e}}^{\mathrm{i}\theta })(1+{\mathrm{e}}^{-\mathrm{i}\theta })}=\frac{1+\cos \left(\theta \right)}{2+2\cos \left(\theta \right)}=\frac{1}{2};\mathrm{\forall }\theta \in \mathbb{R}-\left\{(1+2\mathrm{k})\pi \right\}$$$$\mathrm{I}==\frac{{\phi }^3-1}{2}$$$${\phi }^2=1+\phi$$$${\phi }^3={\phi }^2+\phi =1+2\phi \Rightarrow {\phi }^3-1=2\phi$$$$\mathrm{I}=\phi =\frac{1+\sqrt{5}}{2}$$

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