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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-kpi-n-with-k-in-0-n-1-so-p-X-k-0-k-n-1-X-x-k-let-searsh-




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
answertothequestionofp(X)=(1+iX)n(1iX)nkeyofsolutionafterresolvingp(X)=0therootsofp(X)arexk=tan(kπ/n)withkin[[0.n1]]sop(X)=k=0k=n1(Xxk)letsearshbyusingbinomeformulap(X)=2ip=0(1)pCn2p+1X2p+1so∝=2i(1)[n1/2]Cncase1n=2Np(X)=∝k=0k=2N1(Xtan(kπ/2N))and∝=4in(1)N1case2n=2N+1p(X)=∝k=0k=2N(Xtan(kπ/2N+1)and∝=2i(1)N

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