Question Number 28267 by abdo imad last updated on 22/Jan/18
$${let}\:{give}\:{the}\:{polynomial} \\ $$$${P}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\left(\mathrm{1}+{ix}\right)^{{n}} \:−\left(\mathrm{1}−{ix}\right)^{{n}} \right)\:.{find}\:{the}\:{roots}\:{of}\:{P}\left({x}\right) \\ $$$${and}\:{factorize}\:{P}\left({x}\right). \\ $$
Commented by abdo imad last updated on 23/Jan/18
![P(x)=0 (1+ix)^n =(1 −ix)^n ⇔ ( ((1−ix)/(1+ix)) )^n =1 let put ((1−ix)/(1+ix))=z (e)⇔ z^n =1 wich have for roots z_k = e^(i((2kπ)/n)) and k∈[[0,n−1]] so the roots of P(x) are the complexes x_(k ) / ((1−x_k )/(1+x_k ))=z_k ⇔ 1−x_k =z_k +z_k x_k ⇔(1+z_k )x_k = 1−z_k ⇔ x_k =((1−z_k )/(1+z_k ))=((1−cos(((2kπ)/n))− isin(((2kπ)/n)))/(1+cos(((2kπ)/n))+isin(((2kπ)/n)))) = ((2sin^2 (((kπ)/n))−2isin(((kπ)/n))cos(((kπ)/n)))/(2cos^2 (((kπ)/n))+2isin(((kπ)/n))cos(((kπ)/n)))) =((−isin(((kπ)/n))e^(i(π/n)) )/(cos(((kπ)/n))e^(i(π/n)) ))=−itan (((kπ)/n)) so the roots of P(x) are x_k = −itan(((kπ)/n)) and k∈[[0,n−1]] P(x)=λ Π_(k=0) ^(n−1) (x +itan(((kπ)/n))) let find λ? we have 2i P(x)= Σ_(k=0) ^n C_n ^k (ix)^k −Σ_(k=0) ^n C_n ^k (−ix)^k =Σ_(k=0) ^n C_n ^k (i^k −(−i)^k )x^k =2iΣ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (−1)^p x^(2p+1) ⇒ P(x)= Σ_(p=0) ^([((n−1)/2)]) (−1)^p C_n ^(2p+1) x^(2p+1) and λ= (−1)^([((n−1)/2)]) C_n ^(2[((n−1)/2)]+1) finally P(x)=(−1)^([((n−1)/2)]) C_n ^(2[((n−1)/2)]+1) Π_(k=0) ^(k=n−1) ( x+itan(((kπ)/n))) .](Q28277.png)
$${P}\left({x}\right)=\mathrm{0}\:\:\left(\mathrm{1}+{ix}\right)^{{n}} \:=\left(\mathrm{1}\:−{ix}\right)^{{n}} \Leftrightarrow\:\:\left(\:\frac{\mathrm{1}−{ix}}{\mathrm{1}+{ix}}\:\right)^{{n}} =\mathrm{1}\:{let}\:{put} \\ $$$$\frac{\mathrm{1}−{ix}}{\mathrm{1}+{ix}}={z}\:\:\left({e}\right)\Leftrightarrow\:\:{z}^{{n}} \:=\mathrm{1}\:\:{wich}\:{have}\:{for}\:{roots} \\ $$$$\:{z}_{{k}} =\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:\:\:\:\:{and}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{so}\:{the}\:{roots}\:{of}\:{P}\left({x}\right)\:{are}\:{the} \\ $$$${complexes}\:{x}_{{k}\:} /\:\:\frac{\mathrm{1}−{x}_{{k}} }{\mathrm{1}+{x}_{{k}} }={z}_{{k}} \Leftrightarrow\:\mathrm{1}−{x}_{{k}} ={z}_{{k}} \:+{z}_{{k}} {x}_{{k}} \\ $$$$\Leftrightarrow\left(\mathrm{1}+{z}_{{k}} \right){x}_{{k}} =\:\mathrm{1}−{z}_{{k}} \:\Leftrightarrow\:{x}_{{k}} =\frac{\mathrm{1}−{z}_{{k}} }{\mathrm{1}+{z}_{{k}} }=\frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−\:{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{1}+{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)+{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)} \\ $$$$=\:\:\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)+\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)} \\ $$$$=\frac{−{isin}\left(\frac{{k}\pi}{{n}}\right){e}^{{i}\frac{\pi}{{n}}} }{{cos}\left(\frac{{k}\pi}{{n}}\right){e}^{{i}\frac{\pi}{{n}}} }=−{itan}\:\left(\frac{{k}\pi}{{n}}\right)\:\:{so}\:{the}\:{roots}\:{of}\:{P}\left({x}\right)\:{are} \\ $$$${x}_{{k}} =\:−{itan}\left(\frac{{k}\pi}{{n}}\right)\:\:{and}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${P}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}\:+{itan}\left(\frac{{k}\pi}{{n}}\right)\right)\:\:{let}\:{find}\:\lambda? \\ $$$${we}\:{have}\:\:\mathrm{2}{i}\:{P}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:{C}_{{n}} ^{{k}} \:\left({ix}\right)^{{k}} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−{ix}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\left({i}^{{k}} \:\:−\left(−{i}\right)^{{k}} \right){x}^{{k}} \:=\mathrm{2}{i}\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \left(−\mathrm{1}\right)^{{p}} \:{x}^{\mathrm{2}{p}+\mathrm{1}} \\ $$$$\Rightarrow\:\:{P}\left({x}\right)=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{x}^{\mathrm{2}{p}+\mathrm{1}} \:\:\:{and} \\ $$$$\lambda=\:\left(−\mathrm{1}\right)^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:{finally} \\ $$$${P}\left({x}\right)=\left(−\mathrm{1}\right)^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \prod_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \left(\:{x}+{itan}\left(\frac{{k}\pi}{{n}}\right)\right)\:\:. \\ $$$$ \\ $$
Answered by sma3l2996 last updated on 23/Jan/18
$${P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\left(\mathrm{1}+{ix}\right)^{{n}} −\left(\mathrm{1}−{ix}\right)^{{n}} \right) \\ $$$${let}'{s}\:{put}\:\:{z}=\mathrm{1}+{ix}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{e}^{{i}\theta} \:{with}\:\:\theta={tan}^{−\mathrm{1}} {x}\:\: \\ $$$${so}\:{P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{e}^{{i}\theta} \right)^{{n}} −\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{e}^{−{i}\theta} \right)^{{n}} \right) \\ $$$$=\frac{\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}} }{\mathrm{2}{i}}\left({e}^{{in}\theta} −{e}^{−{in}\theta} \right)=\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{{n}} {sin}\left({n}\theta\right) \\ $$$${P}\left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}/\mathrm{2}} {sin}\left({ntan}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$$ \\ $$