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Question Number 69873 by Masumsiddiqui399@gmail.com last updated on 28/Sep/19

Commented by mathmax by abdo last updated on 28/Sep/19

P_(n−1) =Π_(k=1) ^(n−1) (1−z_k ) =Π_(k=1) ^(n−1) (1−e^((i2kπ)/n) ) =Π_(k=1) ^(n−1) (1−cos(((2kπ)/n))−isin(((2kπ)/n))) =Π_(k=1) ^(n−1) (2sin^2 (((kπ)/n))−2isin(((kπ)/n))cos(((kπ)/n))) =Π_(k=1) ^(n−1) (−2isin(((kπ)/n)))e^((ikπ)/n) =(−2i)^(n−1) Π_(k=1) ^(n−1) sin(((kπ)/n))e^(((iπ)/n)Σ_(k=1) ^(n−1) k) =(−2i)^(n−1) Π_(k=1) ^(n−1) sin(((kπ)/n)) e^(((iπ)/n)(((n−1)n)/2)) =(−2i)^(n−1) Π_(k=1) ^(n−1) sin(((kπ)/n))(i)^(n−1) =2^(n−1) Π_(k=1) ^(n−1) sin(((kπ)/n))

$${P}_{{n}−\mathrm{1}} =\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{1}−{z}_{{k}} \right)\:=\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{1}−{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \right) \\ $$$$=\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)\right) \\ $$$$=\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$=\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(−\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right)\right){e}^{\frac{{ik}\pi}{{n}}} \:=\left(−\mathrm{2}{i}\right)^{{n}−\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {sin}\left(\frac{{k}\pi}{{n}}\right){e}^{\frac{{i}\pi}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {k}} \\ $$$$=\left(−\mathrm{2}{i}\right)^{{n}−\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right)\:{e}^{\frac{{i}\pi}{{n}}\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}} \\ $$$$=\left(−\mathrm{2}{i}\right)^{{n}−\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {sin}\left(\frac{{k}\pi}{{n}}\right)\left({i}\right)^{{n}−\mathrm{1}} \\ $$$$=\mathrm{2}^{{n}−\mathrm{1}} \:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{sin}\left(\frac{{k}\pi}{{n}}\right) \\ $$

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