Question Number 27384 by abdo imad last updated on 05/Jan/18
![let give p(x)= (((1+ix)/(1−ix)))^n − ((1+itanα )/(1−itanα)) factorize p(x) inside C[x].](https://www.tinkutara.com/question/Q27384.png)
$${let}\:{give}\:\:{p}\left({x}\right)=\:\left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\right)^{{n}} −\:\frac{\mathrm{1}+{itan}\alpha\:}{\mathrm{1}−{itan}\alpha}\:\:{factorize}\:{p}\left({x}\right)\:{inside} \\ $$$${C}\left[{x}\right]. \\ $$
Commented by abdo imad last updated on 07/Jan/18
![roots of p(x) p(x)=0⇔ (((1+ix)/(1−ix)) )^n = ((1+itanα)/(1−itanα))=((cosα +isinα)/(cosα −isinα)) =e^(i(2α)) let put ((1+ix)/(1−ix))= t so p(x)=0⇔ t^n = e^(i(2α)) wich have for solutions t_k =e^(i2(α+kπ)(1/n)) and k∈[[0,n−1]] but ((1+ix)/(1−ix))=t_k ⇔ 1+ix=(1−ix)t_k ⇔ ix(1+t_k )=t_k −1 ix =((t_k −1)/(1+t_k )) ⇔x =−i((t_k −1)/(1+tk)) =i ((1−t_k )/(1+t_k )) so the complex wich verify p(x)=0 are x_k = i ((1−t_k )/(1+t_k )) and k∈[[0,n−1]] but x_k = i ((1−e^(i2((α+kπ)/n)) )/(1+ e^(i2((α+kπ)/n)) )) =i((1−cos(2((α+kπ)/n)) −i sin(2((α+kπ)/n)))/(1+cos(2((α+kπ)/n))+isin(2((α+kπ)/n)))) x_(k ) =tan(((α+kπ)/n)) with k from[[0,n−1]] so p(x)= λ Π_(k=0) ^(n−1) (x−tan(((α+kπ)/n))) .](https://www.tinkutara.com/question/Q27505.png)
$${roots}\:{of}\:{p}\left({x}\right)\:\: \\ $$$${p}\left({x}\right)=\mathrm{0}\Leftrightarrow\:\:\left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\:\right)^{{n}} =\:\frac{\mathrm{1}+{itan}\alpha}{\mathrm{1}−{itan}\alpha}=\frac{{cos}\alpha\:+{isin}\alpha}{{cos}\alpha\:−{isin}\alpha}\:={e}^{{i}\left(\mathrm{2}\alpha\right)} \\ $$$${let}\:{put}\:\:\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}=\:{t}\:{so}\:{p}\left({x}\right)=\mathrm{0}\Leftrightarrow\:{t}^{{n}} =\:{e}^{{i}\left(\mathrm{2}\alpha\right)} \:{wich}\:{have}\:{for}\:{solutions} \\ $$$${t}_{{k}} ={e}^{{i}\mathrm{2}\left(\alpha+{k}\pi\right)\frac{\mathrm{1}}{{n}}} \:\:\:\:{and}\:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{but}\:\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}={t}_{{k}} \\ $$$$\Leftrightarrow\:\mathrm{1}+{ix}=\left(\mathrm{1}−{ix}\right){t}_{{k}} \:\Leftrightarrow\:{ix}\left(\mathrm{1}+{t}_{{k}} \right)={t}_{{k}} \:−\mathrm{1} \\ $$$${ix}\:=\frac{{t}_{{k}} −\mathrm{1}}{\mathrm{1}+{t}_{{k}} }\:\:\:\:\Leftrightarrow{x}\:=−{i}\frac{{t}_{{k}} −\mathrm{1}}{\mathrm{1}+{tk}}\:={i}\:\frac{\mathrm{1}−{t}_{{k}} }{\mathrm{1}+{t}_{{k}} }\:{so}\:{the}\:{complex}\:{wich}\: \\ $$$${verify}\:{p}\left({x}\right)=\mathrm{0}\:{are}\:{x}_{{k}} =\:{i}\:\frac{\mathrm{1}−{t}_{{k}} }{\mathrm{1}+{t}_{{k}} }\:\:\:\:{and}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${but}\:{x}_{{k}} =\:{i}\:\frac{\mathrm{1}−{e}^{{i}\mathrm{2}\frac{\alpha+{k}\pi}{{n}}} }{\mathrm{1}+\:{e}^{{i}\mathrm{2}\frac{\alpha+{k}\pi}{{n}}} }\:={i}\frac{\mathrm{1}−{cos}\left(\mathrm{2}\frac{\alpha+{k}\pi}{{n}}\right)\:−{i}\:{sin}\left(\mathrm{2}\frac{\alpha+{k}\pi}{{n}}\right)}{\mathrm{1}+{cos}\left(\mathrm{2}\frac{\alpha+{k}\pi}{{n}}\right)+{isin}\left(\mathrm{2}\frac{\alpha+{k}\pi}{{n}}\right)} \\ $$$${x}_{{k}\:} ={tan}\left(\frac{\alpha+{k}\pi}{{n}}\right)\:\:\:\:{with}\:{k}\:{from}\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${so}\:{p}\left({x}\right)=\:\lambda\:\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{tan}\left(\frac{\alpha+{k}\pi}{{n}}\right)\right)\:\:. \\ $$
Answered by sma3l2996 last updated on 06/Jan/18
$${p}\left({x}\right)=\left(\frac{{i}\left(−{i}+{x}\right)}{−{i}\left({i}+{x}\right)}\right)^{{n}} −\frac{{cos}\alpha+{isin}\alpha}{{cos}\alpha−{isin}\alpha}=\left(\frac{−\left(−\mathrm{2}{i}+{i}+{x}\right)}{{i}+{x}}\right)^{{n}} −\frac{{e}^{{i}\alpha} }{{e}^{−{i}\alpha} } \\ $$$${p}\left({x}\right)=\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{1}−\frac{\mathrm{2}}{{i}+{x}}\right)^{{n}} −{e}^{\mathrm{2}{i}\alpha} \\ $$$${p}\left({x}\right)=\left(−\mathrm{1}\right)^{{n}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{2}}{{i}+{x}}\right)^{{k}} \:^{{n}} {C}_{{k}} −{e}^{\mathrm{2}{i}\alpha} \\ $$