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Question Number 192596 by York12 last updated on 22/May/23

Answered by a.lgnaoui last updated on 24/May/23

$${({\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}})}^{2(\mathrm{n}+\frac{1}{2})}=1\iff [{({\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}})}^{\mathrm{n}+\frac{1}{2}}-1]\times [{({\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}})}^{\mathrm{n}+\frac{1}{\mathrm{n}}}+1]=0$$$$\{\begin{array}{l}{\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}}=1\Rightarrow 1+{\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}}=2\\ 1-{\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}}=1-({\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}})+{2\mathrm{iy}}_{\mathrm{m}}={2\mathrm{iy}}_{\mathrm{m}}\end{array}$$$$\Rightarrow \frac{1-{\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}}}{1+{\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}}}=\frac{{2\mathrm{iy}}_{\mathrm{m}}}{2}={\mathrm{iy}}_{\mathrm{m}}$$$$\sum _{\mathrm{k}=1}^{\mathrm{k}=2020}\frac{1-{\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}}}{1+{\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}}}=\mathrm{i}\sum _{\mathrm{k}=1}^{\mathrm{k}=2020}{\mathrm{y}}_{\mathrm{k}}$$$$\Rightarrow \mathrm{P}=\left[\frac{2020\left(2021\right)}{2}\right]\mathrm{i}$$$$\frac{\mathrm{P}}{43}=\frac{1010\times 2021}{43}=47470\mathrm{i}$$$$\frac{\mathbf{P}}{43}=4747\mathbf{i}$$

Commented by York12 last updated on 24/May/23

$$Iguessyouhavecommitedsomemistakeright$$$$here$$

Answered by witcher3 last updated on 24/May/23

$$({\mathrm{x}}_{\mathrm{m}}+{\mathrm{iy}}_{\mathrm{m}})={\mathrm{e}}^{\frac{2\mathrm{im}\pi }{2\mathrm{n}+1}},\mathrm{m}\in \{0,......2\mathrm{n}\}$$$${\mathrm{Z}}_{\mathrm{m}}={\mathrm{e}}^{2\mathrm{i}\frac{\mathrm{m}\pi }{2\mathrm{n}+1}},{\mathrm{ia}}_{\mathrm{k},\mathrm{n}}=\frac{2\mathrm{ik}\pi }{2\mathrm{n}+1}$$$$\mathrm{for}\mathrm{all}\mathrm{the}\mathrm{reste}\mathrm{n}=1010$$$$\underset{\mathrm{k}=1}{\sum ^{2020}}\frac{1-{\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}}}{1+{\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}}}=\underset{\mathrm{k}=0}{\sum ^{2020}}\frac{1-({\mathrm{x}}_{\mathrm{k}}-{\mathrm{iy}}_{\mathrm{k}})}{1+({\mathrm{x}}_{\mathrm{k}}+{\mathrm{iy}}_{\mathrm{k}})}=\mathrm{\Sigma }\frac{1-{\mathrm{e}}^{-{\mathrm{ia}}_{\mathrm{k}}}}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}$$$$=\mathrm{\Sigma }\frac{{\mathbf{e}}^{{\mathbf{ia}}_{\mathbf{k}}}-1}{{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}(1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}})}=\sum _{\mathrm{k}}\frac{2}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}-\frac{1}{{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}$$$$\mathrm{ler}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{2\mathrm{n}+1}-1$$$$\frac{{\mathrm{p}}^{\prime }\left(\mathrm{x}\right)}{\mathrm{p}\left(\mathrm{x}\right)}=\underset{\mathrm{k}=0}{\sum ^{2\mathrm{n}}}\frac{1}{\mathrm{X}-{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}\Rightarrow \frac{{\mathrm{p}}^{\prime }(-1)}{\mathrm{p}(-1)}=-\mathrm{\Sigma }\frac{1}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}$$$$\Rightarrow \underset{\mathrm{k}=0}{\sum ^{2020}}\frac{1}{1+{\mathrm{e}}^{{\mathrm{ia}}_{\mathrm{k}}}}=-\frac{\left(2021\right)}{-2}=\frac{2021}{2}$$$$\underset{\mathrm{k}=0}{\sum ^{2020}}{\mathrm{e}}^{-{\mathrm{ia}}_{\mathrm{k}}}=\frac{1-{\left({\mathrm{e}}^{-\mathrm{i}\frac{2\pi }{2021}}\right)}^{2021}}{1-{\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{2021}}}=0$$$$\mathrm{P}=2.\frac{2021}{2}=2021=43.47$$$$\frac{\mathrm{p}}{43}=47$$$$$$

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