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Question Number 30522 by abdo imad last updated on 22/Feb/18
let p(x)= (1+ix)^n  −(1−ix)^n   1) find the roots of p(x) and factorize p(x)  ) give p(x) at form of arcs.
$${let}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{ix}\right)^{{n}} \:−\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{and}\:{factorize}\:{p}\left({x}\right) \\ $$$$\left.\right)\:{give}\:{p}\left({x}\right)\:{at}\:{form}\:{of}\:{arcs}. \\ $$$$ \\ $$
Commented by abdo imad last updated on 24/Feb/18
1) we have p(x)=2i Im(1+ix)^n  but ∣1+ix∣=(√(1+x^2 ))  ⇒1+ix =(√(1+x^2 )) ( (1/( (√(1+x^2 )))) +i (x/( (√(1+x^2 )))))=r e^(iθ)  ⇒r=(√(1+x^2 ))  and cosθ =(1/( (√(1+x^2 )))) and sinθ = (x/( (√(1+x^2 )))) ⇒tanθ=x   ⇒θ= artanx ⇒(1+ix)^n =((√(1+x^2 )))^n  e^(inarctanx)  ⇒  p(x)= 2i((√(1+x^2 )) )^n  sin(narctanx)  p(x)=0 ⇔ narctanx=kπ ⇔arctanx=((kπ)/n) ⇔ x_k =tan(((kπ)/n))  and  k∈[[0,n−1]] and p(x)=λ Π_(k=0) ^(n−1)  (x−tan(((kπ)/n)))let  find λ we have 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   =2i Σ_(p=0) ^([((n−1)/2)])   C_n ^(2p+1)  x^(2p+1)   ⇒λ= 2i C_n ^(2[((n−1)/2)]+1)  ⇒  p(x)= 2i C_n ^(2[((n−1)/2)]+1)   Π_(k=0) ^(n−1)  (x−tan(((kπ)/n))) .  2) p(x)=2i ((√(1+x^2 )) )^n  sin(narctanx).
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{p}\left({x}\right)=\mathrm{2}{i}\:{Im}\left(\mathrm{1}+{ix}\right)^{{n}} \:{but}\:\mid\mathrm{1}+{ix}\mid=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}+{ix}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{i}\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${and}\:{cos}\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{and}\:{sin}\theta\:=\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\Rightarrow{tan}\theta={x}\: \\ $$$$\Rightarrow\theta=\:{artanx}\:\Rightarrow\left(\mathrm{1}+{ix}\right)^{{n}} =\left(\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\:^{{n}} \:{e}^{{inarctanx}} \:\Rightarrow\right. \\ $$$${p}\left({x}\right)=\:\mathrm{2}{i}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} \:{sin}\left({narctanx}\right) \\ $$$${p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{narctanx}={k}\pi\:\Leftrightarrow{arctanx}=\frac{{k}\pi}{{n}}\:\Leftrightarrow\:{x}_{{k}} ={tan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${and}\:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{and}\:{p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right){let} \\ $$$${find}\:\lambda\:{we}\:{have}\:{p}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({ix}\right)^{{k}} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−{ix}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(\left({i}\right)^{{k}} \:−\left(−{i}\right)^{{k}} \right){x}^{{k}} \:\:=\mathrm{2}{i}\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{x}^{\mathrm{2}{p}+\mathrm{1}} \\ $$$$\Rightarrow\lambda=\:\mathrm{2}{i}\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\Rightarrow \\ $$$${p}\left({x}\right)=\:\mathrm{2}{i}\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right)\:. \\ $$$$\left.\mathrm{2}\right)\:{p}\left({x}\right)=\mathrm{2}{i}\:\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} \:{sin}\left({narctanx}\right). \\ $$
Answered by sma3l2996 last updated on 23/Feb/18
p(x)=(√((1+x^2 )^n ))(e^(inθ) −e^(−inθ) )  \θ=tan^(−1) x  =(√((1+x^2 )^n ))(2isin(nθ))  p(x)=2i(√((1+x^2 )^n ))sin(ntan^(−1) (x))  tan^(−1) (x_k )=((kπ)/n)  ⇒x_k =tan(((kπ)/n))  so  roots of p(x) are α_k =tan(((kπ)/n))  \k=(0,1,2,3,...,n−1)  p(x)=2i(√((1+x^2 )^n ))Π_(k=0) ^(n−1) (x−tan(((kπ)/n)))
$${p}\left({x}\right)=\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\left({e}^{{in}\theta} −{e}^{−{in}\theta} \right)\:\:\backslash\theta={tan}^{−\mathrm{1}} {x} \\ $$$$=\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\left(\mathrm{2}{isin}\left({n}\theta\right)\right) \\ $$$${p}\left({x}\right)=\mathrm{2}{i}\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{sin}\left({ntan}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$${tan}^{−\mathrm{1}} \left({x}_{{k}} \right)=\frac{{k}\pi}{{n}}\:\:\Rightarrow{x}_{{k}} ={tan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${so}\:\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:\alpha_{{k}} ={tan}\left(\frac{{k}\pi}{{n}}\right)\:\:\backslash{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},…,{n}−\mathrm{1}\right) \\ $$$${p}\left({x}\right)=\mathrm{2}{i}\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$ \\ $$
Commented by abdo imad last updated on 24/Feb/18
you have commited a error in leading coefficient of p(x)  sir.
$${you}\:{have}\:{commited}\:{a}\:{error}\:{in}\:{leading}\:{coefficient}\:{of}\:{p}\left({x}\right) \\ $$$${sir}. \\ $$

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