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Question Number 25822 by abdo imad last updated on 15/Dec/17

answer to the question of p(X)= (1+iX)^n −(1−iX)^n   key of solutionafter resolving p(X)=0  the roots of p(X)  are  x_k =tan(kπ/n) with k  in [[0.n−1]] so  p(X)= ∝Π_(k=0) ^(k=n−1) (X−x_k^  ) let searsh ∝ by using binome formula  p(X)= 2iΣ_(p=0) ^ (−1)^(p ) C_n ^(2p+1) X^(2p+1) so  ∝= 2i(−1)^([n−1/2])  C_n ^   case1 n=2N   p(X)=∝ Π_(k=0) ^(k=2N−1) (X−tan(kπ/2N))  and ∝=4in(−1)^(N−1)   case2  n=2N+1    p(X)=∝Π_(k=0) ^(k=2N) (X−tan(kπ/2N+1)  and ∝=2i(−1)^N

$${answer}\:{to}\:{the}\:{question}\:{of}\:{p}\left({X}\right)=\:\left(\mathrm{1}+{iX}\right)^{{n}} −\left(\mathrm{1}−{iX}\right)^{{n}} \\ $$$${key}\:{of}\:{solutionafter}\:{resolving}\:{p}\left({X}\right)=\mathrm{0}\:\:{the}\:{roots}\:{of}\:{p}\left({X}\right) \\ $$$${are}\:\:{x}_{{k}} ={tan}\left({k}\pi/{n}\right)\:{with}\:{k}\:\:{in}\:\left[\left[\mathrm{0}.{n}−\mathrm{1}\right]\right]\:{so} \\ $$$${p}\left({X}\right)=\:\propto\prod_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \left({X}−{x}_{{k}^{} } \right)\:{let}\:{searsh}\:\propto\:{by}\:{using}\:{binome}\:{formula} \\ $$$${p}\left({X}\right)=\:\mathrm{2}{i}\sum_{{p}=\mathrm{0}} ^{} \left(−\mathrm{1}\right)^{{p}\:} {C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} {X}^{\mathrm{2}{p}+\mathrm{1}} {so} \\ $$$$\propto=\:\mathrm{2}{i}\left(−\mathrm{1}\right)^{\left[{n}−\mathrm{1}/\mathrm{2}\right]} \:{C}_{{n}} ^{} \\ $$$${case}\mathrm{1}\:{n}=\mathrm{2}{N}\:\:\:{p}\left({X}\right)=\propto\:\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}−\mathrm{1}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}\right)\right) \\ $$$${and}\:\propto=\mathrm{4}{in}\left(−\mathrm{1}\right)^{{N}−\mathrm{1}} \\ $$$${case}\mathrm{2}\:\:{n}=\mathrm{2}{N}+\mathrm{1}\:\:\:\:{p}\left({X}\right)=\propto\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}+\mathrm{1}\right)\right. \\ $$$${and}\:\propto=\mathrm{2}{i}\left(−\mathrm{1}\right)^{{N}} \\ $$

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