Question Number 30522 by abdo imad last updated on 22/Feb/18
$${let}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{ix}\right)^{{n}} \:−\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{and}\:{factorize}\:{p}\left({x}\right) \\ $$$$\left.\right)\:{give}\:{p}\left({x}\right)\:{at}\:{form}\:{of}\:{arcs}. \\ $$$$ \\ $$
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).](Q30699.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{p}\left({x}\right)=\mathrm{2}{i}\:{Im}\left(\mathrm{1}+{ix}\right)^{{n}} \:{but}\:\mid\mathrm{1}+{ix}\mid=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}+{ix}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\left(\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{i}\:\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${and}\:{cos}\theta\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{and}\:{sin}\theta\:=\:\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\Rightarrow{tan}\theta={x}\: \\ $$$$\Rightarrow\theta=\:{artanx}\:\Rightarrow\left(\mathrm{1}+{ix}\right)^{{n}} =\left(\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}\:^{{n}} \:{e}^{{inarctanx}} \:\Rightarrow\right. \\ $$$${p}\left({x}\right)=\:\mathrm{2}{i}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} \:{sin}\left({narctanx}\right) \\ $$$${p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{narctanx}={k}\pi\:\Leftrightarrow{arctanx}=\frac{{k}\pi}{{n}}\:\Leftrightarrow\:{x}_{{k}} ={tan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${and}\:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{and}\:{p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right){let} \\ $$$${find}\:\lambda\:{we}\:{have}\:{p}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({ix}\right)^{{k}} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−{ix}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(\left({i}\right)^{{k}} \:−\left(−{i}\right)^{{k}} \right){x}^{{k}} \:\:=\mathrm{2}{i}\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{x}^{\mathrm{2}{p}+\mathrm{1}} \\ $$$$\Rightarrow\lambda=\:\mathrm{2}{i}\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\Rightarrow \\ $$$${p}\left({x}\right)=\:\mathrm{2}{i}\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right)\:. \\ $$$$\left.\mathrm{2}\right)\:{p}\left({x}\right)=\mathrm{2}{i}\:\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{{n}} \:{sin}\left({narctanx}\right). \\ $$
Answered by sma3l2996 last updated on 23/Feb/18
$${p}\left({x}\right)=\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\left({e}^{{in}\theta} −{e}^{−{in}\theta} \right)\:\:\backslash\theta={tan}^{−\mathrm{1}} {x} \\ $$$$=\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\left(\mathrm{2}{isin}\left({n}\theta\right)\right) \\ $$$${p}\left({x}\right)=\mathrm{2}{i}\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{sin}\left({ntan}^{−\mathrm{1}} \left({x}\right)\right) \\ $$$${tan}^{−\mathrm{1}} \left({x}_{{k}} \right)=\frac{{k}\pi}{{n}}\:\:\Rightarrow{x}_{{k}} ={tan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${so}\:\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:\alpha_{{k}} ={tan}\left(\frac{{k}\pi}{{n}}\right)\:\:\backslash{k}=\left(\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},...,{n}−\mathrm{1}\right) \\ $$$${p}\left({x}\right)=\mathrm{2}{i}\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$ \\ $$
Commented by abdo imad last updated on 24/Feb/18
$${you}\:{have}\:{commited}\:{a}\:{error}\:{in}\:{leading}\:{coefficient}\:{of}\:{p}\left({x}\right) \\ $$$${sir}. \\ $$