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Question Number 190324 by Shrinava last updated on 31/Mar/23

Commented by senestro last updated on 01/Apr/23

$$howcanwesolveit?$$

Answered by witcher3 last updated on 01/Apr/23

$$\mathrm{S}=\underset{\mathrm{k}=1}{\prod ^{4\mathrm{n}+2}}\frac{1}{2}({\mathrm{e}}^{4\mathrm{i}\frac{\pi \mathrm{k}}{4\mathrm{n}+3}}+1).{\mathrm{e}}^{-\frac{2\pi }{4\mathrm{n}+3}\left(\frac{4\mathrm{n}+3}{2}(4\mathrm{n}+1)\right)}$$$$=\frac{1}{2^{4\mathrm{n}+2}}\prod _{\mathrm{t}\in {\mathrm{S}}_{\mathrm{k}}}(1+{\mathrm{t}}^2),{\mathrm{s}}_{\mathrm{k}}=\{{\mathrm{e}}^{\frac{2\mathrm{i}\pi \mathrm{k}}{4\mathrm{n}+3}},\mathrm{k}\in \{0,4\mathrm{n}+2\}\}$$$${\mathrm{Z}}^{4\mathrm{n}+3}=1\iff \mathrm{z}\in {\mathrm{S}}_{\mathrm{k}}$$$${\mathrm{Z}}^{8\mathrm{n}+6}-1=\frac{1}{{16}^{\mathrm{n}}.4}\prod _{\mathrm{t}\in {\mathrm{S}}_{\mathrm{k}}-\left\{0\right\}}({\mathrm{Z}}^2-{\mathrm{t}}^2)\iff \frac{{\mathrm{Z}}^{8\mathrm{n}+6}-1}{{\mathrm{z}}^2-1}=\mathrm{P}\left(\mathrm{z}\right)$$$$\underset{\mathrm{k}=1}{\prod ^{4\mathrm{n}+2}}({\mathrm{e}}^{\frac{4\mathrm{ik}\pi }{4\mathrm{n}+3}}+1)=\mathrm{P}\left(\mathrm{i}\right)=1$$$$\mathrm{S}=\frac{1}{{16}^{\mathrm{n}}.4}$$$$\Rightarrow \mathrm{\Omega }=\frac{1}{4}$$$$$$

Commented by Shrinava last updated on 03/Apr/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

Commented by witcher3 last updated on 05/Apr/23

$$\mathrm{withe}\mathrm{Pleasur}\mathrm{God}\mathrm{bless}\mathrm{You}$$

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