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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
(x_m +iy_m )^(2(n+(1/2))) =1⇔[(x_m +iy_m )^(n+(1/2)) −1]×[(x_m +iy_m )^(n+(1/n)) +1]=0 { ((x_m +iy_m =1⇒1+x_m +iy_m =2)),((1−x_m +iy_m =1−(x_m +iy_m )+2iy_m =2iy_m )) :} ⇒((1−x_m +iy_m )/(1+x_m +iy_m ))=((2iy_m )/2)=iy_m Σ_(k=1) ^(k=2020) ((1−x_k +iy_k )/(1+x_k +iy_k ))=iΣ_(k=1) ^(k=2020) y_k ⇒P=[((2020(2021))/2)]i (P/(43))=((1010×2021)/(43))=47470i (P/(43))=4747i
$$\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{2}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)} =\mathrm{1}\Leftrightarrow\left[\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}\right]×\left[\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)^{\mathrm{n}+\frac{\mathrm{1}}{\mathrm{n}}} +\mathrm{1}\right]=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} =\mathrm{1}\Rightarrow\mathrm{1}+\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} =\mathrm{2}}\\{\mathrm{1}−\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} =\mathrm{1}−\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)+\mathrm{2iy}_{\mathrm{m}} =\mathrm{2iy}_{\mathrm{m}} }\end{cases} \\ $$$$\Rightarrow\frac{\mathrm{1}−\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} }{\mathrm{1}+\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} }=\frac{\mathrm{2iy}_{\mathrm{m}} }{\mathrm{2}}=\mathrm{iy}_{\mathrm{m}} \\ $$$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{k}=\mathrm{2020}} \frac{\mathrm{1}−\mathrm{x}_{\mathrm{k}} +\mathrm{iy}_{\mathrm{k}} }{\mathrm{1}+\mathrm{x}_{\mathrm{k}} +\mathrm{iy}_{\mathrm{k}} }=\mathrm{i}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{k}=\mathrm{2020}} \mathrm{y}_{\mathrm{k}} \\ $$$$\Rightarrow\mathrm{P}=\left[\frac{\mathrm{2020}\left(\mathrm{2021}\right)}{\mathrm{2}}\right]\mathrm{i} \\ $$$$\frac{\mathrm{P}}{\mathrm{43}}=\frac{\mathrm{1010}×\mathrm{2021}}{\mathrm{43}}=\mathrm{47470i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\mathrm{P}}}{\mathrm{43}}=\mathrm{4747}\boldsymbol{\mathrm{i}} \\ $$
Commented by York12 last updated on 24/May/23
I guess you have commited some mistake right here
$${I}\:{guess}\:{you}\:{have}\:{commited}\:{some}\:{mistake}\:{right} \\ $$$${here} \\ $$
Answered by witcher3 last updated on 24/May/23
(x_m +iy_m )=e^((2imπ)/(2n+1)) ,m∈{0,......2n} Z_m =e^(2i((mπ)/(2n+1))) ,ia_(k,n) =((2ikπ)/(2n+1)) for all the reste n=1010 Σ_(k=1) ^(2020) ((1−x_k +iy_k )/(1+x_k +iy_k ))=Σ_(k=0) ^(2020) ((1−(x_k −iy_k ))/(1+(x_k +iy_k )))=Σ((1−e^(−ia_k ) )/(1+e^(ia_k ) )) =Σ((e^(ia_k ) −1)/(e^(ia_k ) (1+e^(ia_k ) )))=Σ_k (2/(1+e^(ia_k ) ))−(1/e^(ia_k ) ) ler p(x)=x^(2n+1) −1 ((p′(x))/(p(x)))=Σ_(k=0) ^(2n) (1/(X−e^(ia_k ) ))⇒((p′(−1))/(p(−1)))=−Σ(1/(1+e^(ia_k ) )) ⇒Σ_(k=0) ^(2020) (1/(1+e^(ia_k ) ))=−(((2021))/(−2))=((2021)/2) Σ_(k=0) ^(2020) e^(−ia_k ) =((1−(e^(−i((2π)/(2021))) )^(2021) )/(1−e^(−((i2π)/(2021))) ))=0 P=2.((2021)/2)=2021=43.47 (p/(43))=47
$$\left(\mathrm{x}_{\mathrm{m}} +\mathrm{iy}_{\mathrm{m}} \right)=\mathrm{e}^{\frac{\mathrm{2im}\pi}{\mathrm{2n}+\mathrm{1}}} ,\mathrm{m}\in\left\{\mathrm{0},……\mathrm{2n}\right\} \\ $$$$\mathrm{Z}_{\mathrm{m}} =\mathrm{e}^{\mathrm{2i}\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}} ,\mathrm{ia}_{\mathrm{k},\mathrm{n}} =\frac{\mathrm{2ik}\pi}{\mathrm{2n}+\mathrm{1}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{the}\:\mathrm{reste}\:\mathrm{n}=\mathrm{1010} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\sum}}\frac{\mathrm{1}−\mathrm{x}_{\mathrm{k}} +\mathrm{iy}_{\mathrm{k}} }{\mathrm{1}+\mathrm{x}_{\mathrm{k}} +\mathrm{iy}_{\mathrm{k}} }=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\frac{\mathrm{1}−\left(\mathrm{x}_{\mathrm{k}} −\mathrm{iy}_{\mathrm{k}} \right)}{\mathrm{1}+\left(\mathrm{x}_{\mathrm{k}} +\mathrm{iy}_{\mathrm{k}} \right)}=\Sigma\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{ia}_{\mathrm{k}} } }{\mathrm{1}+\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } } \\ $$$$=\Sigma\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{ia}}_{\boldsymbol{\mathrm{k}}} } −\mathrm{1}}{\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } \left(\mathrm{1}+\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } \right)}=\underset{\mathrm{k}} {\sum}\frac{\mathrm{2}}{\mathrm{1}+\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } }−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } } \\ $$$$\mathrm{ler}\:\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2n}+\mathrm{1}} −\mathrm{1} \\ $$$$\frac{\mathrm{p}'\left(\mathrm{x}\right)}{\mathrm{p}\left(\mathrm{x}\right)}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\frac{\mathrm{1}}{\mathrm{X}−\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } }\Rightarrow\frac{\mathrm{p}'\left(−\mathrm{1}\right)}{\mathrm{p}\left(−\mathrm{1}\right)}=−\Sigma\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } } \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{ia}_{\mathrm{k}} } }=−\frac{\left(\mathrm{2021}\right)}{−\mathrm{2}}=\frac{\mathrm{2021}}{\mathrm{2}} \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\mathrm{e}^{−\mathrm{ia}_{\mathrm{k}} } =\frac{\mathrm{1}−\left(\mathrm{e}^{−\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{2021}}} \right)^{\mathrm{2021}} }{\mathrm{1}−\mathrm{e}^{−\frac{\mathrm{i2}\pi}{\mathrm{2021}}} }=\mathrm{0} \\ $$$$\mathrm{P}=\mathrm{2}.\frac{\mathrm{2021}}{\mathrm{2}}=\mathrm{2021}=\mathrm{43}.\mathrm{47} \\ $$$$\frac{\mathrm{p}}{\mathrm{43}}=\mathrm{47} \\ $$$$ \\ $$

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