Question Number 36352 by prof Abdo imad last updated on 01/Jun/18
$${let}\:{p}\left({x}\right)\:={x}^{{n}} \:−{e}^{{in}\alpha} \:\:\:\:{with}\:{n}\:{integr}\:{and}\:\alpha\:{fromR} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right]\:. \\ $$
Commented by math khazana by abdo last updated on 10/Jul/18
$${changement}\:{x}\:={e}^{{i}\alpha\:} {z}\:{give} \\ $$$${p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{e}^{{in}\alpha} \:{z}^{{n}} \:−{e}^{{in}\alpha} =\mathrm{0}\:\Leftrightarrow{z}^{{n}} −\mathrm{1}\:=\mathrm{0}\:\Leftrightarrow \\ $$$${z}_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \:{and}\:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{so}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$${are}\:{x}_{{k}} ={e}^{{i}\alpha} \:{e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:={e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}}\:+\alpha\right)} \:\:{with}\:\mathrm{0}\leqslant{k}\leqslant{n}−\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{wehave}\:\:{p}\left({x}\right)\:=\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left\{{x}−\:{e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}}\:+\alpha\right)} \right\} \\ $$