Question Number 155899 by Tawa11 last updated on 05/Oct/21
$$\mathrm{Find}\:\mathrm{the}\:\:\:\mathrm{nth}\:\:\:\mathrm{root}\:\mathrm{of}\:\:\:\mathrm{1}.\:\:\mathrm{Find}\:\mathrm{these}\:\mathrm{values}\:\mathrm{when}\:\:\:\mathrm{n}\:\:=\:\:\mathrm{6}. \\ $$
Answered by Ar Brandon last updated on 05/Oct/21
![1=e^(2kπi) , k∈Z z_n =(1)^(1/n) =e^(((2kπ)/n)i) , k∈[0, n−1] z_6 =e^(((kπ)/3)i) , k∈[0, 5] z_6 =1, (1/2)(1+i(√3)), −(1/2)(1−i(√3)) −1, −(1/2)(1+i(√3)), (1/2)(1−i(√3))](https://www.tinkutara.com/question/Q155904.png)
$$\mathrm{1}={e}^{\mathrm{2}{k}\pi{i}} ,\:{k}\in\mathbb{Z} \\ $$$${z}_{{n}} =\sqrt[{{n}}]{\mathrm{1}}={e}^{\frac{\mathrm{2}{k}\pi}{{n}}{i}} ,\:{k}\in\left[\mathrm{0},\:{n}−\mathrm{1}\right] \\ $$$${z}_{\mathrm{6}} ={e}^{\frac{{k}\pi}{\mathrm{3}}{i}} ,\:{k}\in\left[\mathrm{0},\:\mathrm{5}\right] \\ $$$${z}_{\mathrm{6}} =\mathrm{1},\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right),\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{i}\sqrt{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:−\mathrm{1},\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{i}\sqrt{\mathrm{3}}\right),\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{i}\sqrt{\mathrm{3}}\right) \\ $$