Question Number 19434 by Tinkutara last updated on 11/Aug/17
$$\mathrm{If}\:\left(\frac{\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}}{\mathrm{1}\:−\:{i}\sqrt{\mathrm{3}}}\right)^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{integer},\:\mathrm{then}\:{n}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 11/Aug/17
![(((1+i(√3))/(1−i(√3))))^n =((e^(iπ/6) /e^(−iπ/6) ))^n =[x] ⇒ e^(inπ/3) =±1 =e^(ikπ) ⇒ n=3k where k∈Z .](https://www.tinkutara.com/question/Q19443.png)
$$\left(\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{1}−{i}\sqrt{\mathrm{3}}}\right)^{\mathrm{n}} =\left(\frac{\mathrm{e}^{{i}\pi/\mathrm{6}} }{\mathrm{e}^{−{i}\pi/\mathrm{6}} }\right)^{\mathrm{n}} =\left[\mathrm{x}\right] \\ $$$$\Rightarrow\:\mathrm{e}^{\mathrm{in}\pi/\mathrm{3}} =\pm\mathrm{1}\:=\mathrm{e}^{\mathrm{ik}\pi} \\ $$$$\Rightarrow\:\:\boldsymbol{\mathrm{n}}=\mathrm{3}\boldsymbol{\mathrm{k}}\:\mathrm{where}\:\mathrm{k}\in{Z}\:. \\ $$
Commented by Tinkutara last updated on 11/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$