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Question Number 35056 by math khazana by abdo last updated on 14/May/18

let p(x)=(1+jx)^n −(1−jx)^n 1) find the roots of p(x) 2)factorize p(x) inside C[x] j =e^(i((2π)/3)) .

$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{jx}\right)^{{n}} \:−\left(\mathrm{1}−{jx}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$${j}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:. \\ $$

Commented by math khazana by abdo last updated on 10/Jun/18

p(x)=0 ⇔ (((1−jx)^n )/((1+jx)^n )) =1⇔(((1−jx)/(1+jx)))^n =1 the roots of z^n =1 are the complex z_k =e^(i((2kπ)/n)) k∈[[0,n−1]] so the roots of p(x) are the complex Z_k / ((1−jZ_k )/(1+jZ_k )) =z_k ⇔1−jZ_k =z_k +jz_k Z_k ⇔ j(1+z_k )Z_k =1−z_k ⇔Z_k =(1/j) ((1−z_k )/(1+z_k )) ⇒ j Z_k = ((1 −cos(((2kπ)/n)) −isin(((2kπ)/n)))/(1+cos(((2kπ)/n)) +i sin(((2kπ)/n)))) =((2sin^2 (((kπ)/n)) −2i sin(((kπ)/n))cos(((kπ)/n)))/(2cos^2 (((kπ)/n)) +2isin(((kπ)/n))cos(((kπ)/n)))) =((−isin(((kπ)/n))e^(i((kπ)/n)) )/(cos(((kπ)/n))e^(i((kπ)/n)) )) =−i tan(((kπ)/n)) ⇒ Z_(k ) = ((−i)/j)tan(((kπ)/n)) =e^(−i(π/2)) e^(−i((2π)/3)) tan(((kπ)/n)) = e^(−i( ((7π)/6))) tan(((kπ)/n)) with n ∈[[1,n−1]]

$${p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:\frac{\left(\mathrm{1}−{jx}\right)^{{n}} }{\left(\mathrm{1}+{jx}\right)^{{n}} }\:=\mathrm{1}\Leftrightarrow\left(\frac{\mathrm{1}−{jx}}{\mathrm{1}+{jx}}\right)^{{n}} \:=\mathrm{1}\:{the}\:{roots} \\ $$$${of}\:{z}^{{n}} =\mathrm{1}\:{are}\:{the}\:{complex}\:{z}_{{k}} \:={e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${so}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:{the}\:{complex}\:{Z}_{{k}} \:/ \\ $$$$\frac{\mathrm{1}−{jZ}_{{k}} }{\mathrm{1}+{jZ}_{{k}} }\:={z}_{{k}} \:\Leftrightarrow\mathrm{1}−{jZ}_{{k}} \:={z}_{{k}} \:+{jz}_{{k}} {Z}_{{k}} \:\:\Leftrightarrow \\ $$$${j}\left(\mathrm{1}+{z}_{{k}} \right){Z}_{{k}} =\mathrm{1}−{z}_{{k}} \:\Leftrightarrow{Z}_{{k}} \:=\frac{\mathrm{1}}{{j}}\:\frac{\mathrm{1}−{z}_{{k}} }{\mathrm{1}+{z}_{{k}} }\:\Rightarrow \\ $$$${j}\:{Z}_{{k}} =\:\frac{\mathrm{1}\:−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)\:−{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{1}+{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)\:+{i}\:{sin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)\:−\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)\:+\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)} \\ $$$$=\frac{−{isin}\left(\frac{{k}\pi}{{n}}\right){e}^{{i}\frac{{k}\pi}{{n}}} }{{cos}\left(\frac{{k}\pi}{{n}}\right){e}^{{i}\frac{{k}\pi}{{n}}} }\:=−{i}\:{tan}\left(\frac{{k}\pi}{{n}}\right)\:\Rightarrow \\ $$$${Z}_{{k}\:} =\:\frac{−{i}}{{j}}{tan}\left(\frac{{k}\pi}{{n}}\right)\:={e}^{−{i}\frac{\pi}{\mathrm{2}}} \:{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:{tan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$$=\:{e}^{−{i}\left(\:\frac{\mathrm{7}\pi}{\mathrm{6}}\right)} {tan}\left(\frac{{k}\pi}{{n}}\right)\:\:\:{with}\:{n}\:\in\left[\left[\mathrm{1},{n}−\mathrm{1}\right]\right] \\ $$

Commented by math khazana by abdo last updated on 10/Jun/18

p(x) =λ Π_(k=1) ^(n−1 ) ( x−Z_k ) =λ Π_(k=1) ^(n−1) (x−e^(−i((7π)/6)) tan(((kπ)/n))) λ is the dominentcoefficient let find it we have p(x) =(1+jx)^n −(1−jx)^n = Σ_(k=0) ^n C_n ^k j^k x^k −Σ_(k=0) ^n C_n ^k (−j)^k x^k = Σ_(k=0) ^n C_n ^k (j^k −(−j)^k )x^k ⇒ λ = C_n ^n (j^n −(−j)^n ) =e^(i((2nπ)/3)) −(−1)^n e^(i((2nπ)/3)) =(1−(−1)^n ) e^(i((2nπ)/3)) ⇒ p(x) ={1−(−1)^n }e^(i((2nπ)/3)) Π_(k=1) ^n (x−e^(−i((7π)/6)) tan(((kπ)/n))).

$${p}\left({x}\right)\:=\lambda\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}\:} \:\left(\:{x}−{Z}_{{k}} \right)\:=\lambda\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left({x}−{e}^{−{i}\frac{\mathrm{7}\pi}{\mathrm{6}}} {tan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$\lambda\:{is}\:{the}\:{dominentcoefficient}\:\:{let}\:{find}\:{it} \\ $$$${we}\:{have}\:{p}\left({x}\right)\:=\left(\mathrm{1}+{jx}\right)^{{n}} \:−\left(\mathrm{1}−{jx}\right)^{{n}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:{j}^{{k}} {x}^{{k}} \:\:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−{j}\right)^{{k}} \:{x}^{{k}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \left({j}^{{k}} \:−\left(−{j}\right)^{{k}} \right){x}^{{k}} \:\Rightarrow \\ $$$$\lambda\:=\:{C}_{{n}} ^{{n}} \:\left({j}^{{n}} \:−\left(−{j}\right)^{{n}} \right)\:={e}^{{i}\frac{\mathrm{2}{n}\pi}{\mathrm{3}}} \:−\left(−\mathrm{1}\right)^{{n}} \:{e}^{{i}\frac{\mathrm{2}{n}\pi}{\mathrm{3}}} \\ $$$$=\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right)\:{e}^{{i}\frac{\mathrm{2}{n}\pi}{\mathrm{3}}} \:\Rightarrow \\ $$$${p}\left({x}\right)\:=\left\{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right\}{e}^{{i}\frac{\mathrm{2}{n}\pi}{\mathrm{3}}} \:\prod_{{k}=\mathrm{1}} ^{{n}} \left({x}−{e}^{−{i}\frac{\mathrm{7}\pi}{\mathrm{6}}} \:{tan}\left(\frac{{k}\pi}{{n}}\right)\right). \\ $$

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