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Question Number 30593 by abdo imad last updated on 23/Feb/18

factorize inside C[x] p(x)=(1+i(x/n))^n  −(1−i(x/n))^n .

$${factorize}\:{inside}\:{C}\left[{x}\right]\:{p}\left({x}\right)=\left(\mathrm{1}+{i}\frac{{x}}{{n}}\right)^{{n}} \:−\left(\mathrm{1}−{i}\frac{{x}}{{n}}\right)^{{n}} . \\ $$

Answered by sma3l2996 last updated on 23/Feb/18

p(x)=(1+ix/n)^n −(1−ix/n)^n   =n^(−n) ((n+ix)^n −(n−ix)^n )  =(1/n)(√(n^2 +x^2 ))((e^(iθ) )^n −(e^(−iθ) )^n )  \θ=tan^(−1) ((x/n))  =((√(n^2 +x^2 ))/n)(e^(inθ) −e^(−inθ) )=((√(n^2 +x^2 ))/n)(2isin(nθ))  p(x)=2i((√(n^2 +x^2 ))/n)sin(ntan^(−1) ((x/n)))  tan^(−1) (x_k /n)=((kπ)/n) ⇒x_k =ntan(((kπ)/n))   \k=(0,1,2,...,n−1)  so roots of  p(x) are  α_k =ntan(((kπ)/n))  so  p(x)=2i(√(1+((x/n))^2 ))Π_(k=0) ^(n−1) (x−ntan(((kπ)/n)))

$${p}\left({x}\right)=\left(\mathrm{1}+{ix}/{n}\right)^{{n}} −\left(\mathrm{1}−{ix}/{n}\right)^{{n}} \\ $$$$={n}^{−{n}} \left(\left({n}+{ix}\right)^{{n}} −\left({n}−{ix}\right)^{{n}} \right) \\ $$$$=\frac{\mathrm{1}}{{n}}\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }\left(\left({e}^{{i}\theta} \right)^{{n}} −\left({e}^{−{i}\theta} \right)^{{n}} \right)\:\:\backslash\theta={tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right) \\ $$$$=\frac{\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{n}}\left({e}^{{in}\theta} −{e}^{−{in}\theta} \right)=\frac{\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{n}}\left(\mathrm{2}{isin}\left({n}\theta\right)\right) \\ $$$${p}\left({x}\right)=\mathrm{2}{i}\frac{\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{n}}{sin}\left({ntan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right)\right) \\ $$$${tan}^{−\mathrm{1}} \left({x}_{{k}} /{n}\right)=\frac{{k}\pi}{{n}}\:\Rightarrow{x}_{{k}} ={ntan}\left(\frac{{k}\pi}{{n}}\right)\:\:\:\backslash{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},...,{n}−\mathrm{1}\right) \\ $$$${so}\:{roots}\:{of}\:\:{p}\left({x}\right)\:{are}\:\:\alpha_{{k}} ={ntan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${so}\:\:{p}\left({x}\right)=\mathrm{2}{i}\sqrt{\mathrm{1}+\left(\frac{{x}}{{n}}\right)^{\mathrm{2}} }\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{ntan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$

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