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 .](https://www.tinkutara.com/question/Q30593.png)
$${factorize}\:{inside}\:{C}\left[{x}\right]\:{p}\left({x}\right)=\left(\mathrm{1}+{i}\frac{{x}}{{n}}\right)^{{n}} \:−\left(\mathrm{1}−{i}\frac{{x}}{{n}}\right)^{{n}} . \\ $$
Answered by sma3l2996 last updated on 23/Feb/18
$${p}\left({x}\right)=\left(\mathrm{1}+{ix}/{n}\right)^{{n}} −\left(\mathrm{1}−{ix}/{n}\right)^{{n}} \\ $$$$={n}^{−{n}} \left(\left({n}+{ix}\right)^{{n}} −\left({n}−{ix}\right)^{{n}} \right) \\ $$$$=\frac{\mathrm{1}}{{n}}\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }\left(\left({e}^{{i}\theta} \right)^{{n}} −\left({e}^{−{i}\theta} \right)^{{n}} \right)\:\:\backslash\theta={tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right) \\ $$$$=\frac{\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{n}}\left({e}^{{in}\theta} −{e}^{−{in}\theta} \right)=\frac{\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{n}}\left(\mathrm{2}{isin}\left({n}\theta\right)\right) \\ $$$${p}\left({x}\right)=\mathrm{2}{i}\frac{\sqrt{{n}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{{n}}{sin}\left({ntan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right)\right) \\ $$$${tan}^{−\mathrm{1}} \left({x}_{{k}} /{n}\right)=\frac{{k}\pi}{{n}}\:\Rightarrow{x}_{{k}} ={ntan}\left(\frac{{k}\pi}{{n}}\right)\:\:\:\backslash{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},…,{n}−\mathrm{1}\right) \\ $$$${so}\:{roots}\:{of}\:\:{p}\left({x}\right)\:{are}\:\:\alpha_{{k}} ={ntan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${so}\:\:{p}\left({x}\right)=\mathrm{2}{i}\sqrt{\mathrm{1}+\left(\frac{{x}}{{n}}\right)^{\mathrm{2}} }\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{ntan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$