Question Number 116964 by saorey0202 last updated on 08/Oct/20
Answered by Bird last updated on 08/Oct/20
![we have 1+ix =(√(1+x^2 )) e^(iarctan(x)) 1−ix =(√(1+x^2 )) e^(−i arctan(x)) ⇒ ((1+ix)/(1−ix)) =e^(2i arctan(x)) also ((1+ia)/(1−ia)) =(((√(1+a^2 ))e^(iarctsn(a)) )/( (√(1+a^2 ))e^(−isrctana) )) =e^(2iarctan(a)) (e) ⇒ e^(2in arctanx ) =e^(2i arctana) ⇒ 2narctanx =2arctana +2kπ ⇒ arctanx =((arctan(a))/n) +((kπ)/n) ⇒x_k =tan(((arctan(a))/n) +((kπ)/n)) with k∈[[0,n−1]] all x_k are reals and distincts](https://www.tinkutara.com/question/Q116979.png)
$${we}\:{have}\:\mathrm{1}+{ix}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{{iarctan}\left({x}\right)} \\ $$$$\mathrm{1}−{ix}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{−{i}\:{arctan}\left({x}\right)} \:\Rightarrow \\ $$$$\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\:={e}^{\mathrm{2}{i}\:{arctan}\left({x}\right)} \\ $$$${also}\:\frac{\mathrm{1}+{ia}}{\mathrm{1}−{ia}}\:=\frac{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }{e}^{{iarctsn}\left({a}\right)} }{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }{e}^{−{isrctana}} } \\ $$$$={e}^{\mathrm{2}{iarctan}\left({a}\right)} \\ $$$$\left({e}\right)\:\Rightarrow\:{e}^{\mathrm{2}{in}\:{arctanx}\:} \:={e}^{\mathrm{2}{i}\:{arctana}} \:\Rightarrow \\ $$$$\mathrm{2}{narctanx}\:=\mathrm{2}{arctana}\:+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$${arctanx}\:=\frac{{arctan}\left({a}\right)}{{n}}\:+\frac{{k}\pi}{{n}} \\ $$$$\Rightarrow{x}_{{k}} ={tan}\left(\frac{{arctan}\left({a}\right)}{{n}}\:+\frac{{k}\pi}{{n}}\right) \\ $$$${with}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${all}\:{x}_{{k}} {are}\:{reals}\:{and}\:{distincts} \\ $$