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Question Number 119136 by Hassen_Timol last updated on 22/Oct/20

Commented by MJS_new last updated on 22/Oct/20

$to understand what happens draw it!$ $use p = 2 \Rightarrow n = 5$ $w^{n} = 1 \Rightarrow draw x - axis (real) and y - axis$ $(imaginary) and a unit circle . then scetch$ $a regular pentagon beginning at (1 / 0)$ $the 2 vertices above the x - axis sum up to$ $S, the ones below sum up to T$ ${it}^{'} s really easy to see; everything else follows$
Commented by Hassen_Timol last updated on 22/Oct/20
Okay thank you very much
	Answered by mathmax by abdo last updated on 22/Oct/20
	$S =Σ_(k=1) ^p w^k and T=Σ_(k=1) ^p w^(k+p) ⇒ S =Σ_(k=1) ^p (e^((2iπ)/(2p+1)) )^k =Σ_(k=0) ^p (e^((2iπ)/(2p+1)) )^k −1 =((1−(e^((2iπ)/(2p+1)) )^(p+1) )/(1−e^((2iπ)/(2p+1)) )) −1 =((1−cos(((2π(p+1))/(2p+1)))−isin(((2π(p+1))/(2p+1))))/(1−cos(((2π)/(2p+1)))−isin(((2π)/(2p+1)))))−1 =((2sin^2 ((((p+1)π)/(2p+1)))−2isin((((p+1)π)/(2p+1)))cos((((p+1)π)/(2p+1))))/(2sin^2 ((π/(2p+1)))−2isin((π/(2p+1)))cos((π/(2p+1)))))−1 =((−isin((((p+1)π)/(2p+1)))e^(i(((p+1)π)/(2p+1))) )/(−isin((π/(2p+1)))e^((iπ)/(2p+1)) ))−1 =((sin((((p+1)π)/(2p+1))))/(sin((π/(2p+1)))))e^((ipπ)/(2p+1)) −1 T =Σ_(k=1) ^p w^(k+p) =w^p (Σ_(k=0) ^p w^k −1) =(e^((2iπ)/(2p+1)) )^p ×{((sin((((p+1)π)/(2p+1))))/(sin((π/(2p+1))))) e^((ipπ)/(2p+1)) −1} Im(S) =((sin((((p+1)π)/(2p+1))))/(sin((π/(2p+1)))))×sin(((pπ)/(2p+1)))>0 Re(S) =((sin((((p+1)π)/(2p+1))))/(sin((π/(2p+1)))))×cos(((pπ)/(2p+1)))−1$
	$S = \sum_{k = 1}^{p} w^{k} and T = \sum_{k = 1}^{p} w^{k + p} \Rightarrow$ $S = \sum_{k = 1}^{p} {(e^{\frac{2 i π}{2 p + 1}})}^{k} = \sum_{k = 0}^{p} {(e^{\frac{2 i π}{2 p + 1}})}^{k} - 1$ $= \frac{1 - {(e^{\frac{2 i π}{2 p + 1}})}^{p + 1}}{1 - e^{\frac{2 i π}{2 p + 1}}} - 1 = \frac{1 - \cos (\frac{2 π (p + 1)}{2 p + 1}) - isin (\frac{2 π (p + 1)}{2 p + 1})}{1 - \cos (\frac{2 π}{2 p + 1}) - isin (\frac{2 π}{2 p + 1})} - 1$ $= \frac{{2 \sin}^{2} (\frac{(p + 1) π}{2 p + 1}) - 2 isin (\frac{(p + 1) π}{2 p + 1}) \cos (\frac{(p + 1) π}{2 p + 1})}{{2 \sin}^{2} (\frac{π}{2 p + 1}) - 2 isin (\frac{π}{2 p + 1}) \cos (\frac{π}{2 p + 1})} - 1$ $= \frac{- isin (\frac{(p + 1) π}{2 p + 1}) e^{i \frac{(p + 1) π}{2 p + 1}}}{- isin (\frac{π}{2 p + 1}) e^{\frac{i π}{2 p + 1}}} - 1 = \frac{\sin (\frac{(p + 1) π}{2 p + 1})}{\sin (\frac{π}{2 p + 1})} e^{\frac{ip π}{2 p + 1}} - 1$ $T = \sum_{k = 1}^{p} w^{k + p} = w^{p} (\sum_{k = 0}^{p} w^{k} - 1)$ $= {(e^{\frac{2 i π}{2 p + 1}})}^{p} \times {\frac{\sin (\frac{(p + 1) π}{2 p + 1})}{\sin (\frac{π}{2 p + 1})} e^{\frac{ip π}{2 p + 1}} - 1}$ $Im (S) = \frac{\sin (\frac{(p + 1) π}{2 p + 1})}{\sin (\frac{π}{2 p + 1})} \times \sin (\frac{p π}{2 p + 1}) > 0$ $Re (S) = \frac{\sin (\frac{(p + 1) π}{2 p + 1})}{\sin (\frac{π}{2 p + 1})} \times \cos (\frac{p π}{2 p + 1}) - 1$
	Commented by Hassen_Timol last updated on 26/Oct/20
	Thank you so much you save my entire life
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