Question

Find the Least positive integer n For which [ (1+i√3)/1-i√3) ]^n=1

Answer 1

it is given that, [(1 + i√3)/(1 - i√3)]ⁿ = 1

first of all, resolve (1 + i√3)/(1 - i√3)

(1 + i√3)(1 + i√3)/(1 - i√3)(1 + i√3)

= (1 + i√3)²/(1² - i²√3²)

= (1 + i²√3² + 2i√3)/(1 + 3)

= (1 - 3 + 2i√3)/(1 + 3)

= (-1 + i√3)/2

= (-1/2 + i√3/2)

= sin(-π/6) + icos(-π/6)

now, [(1 + i√3)/(1 - i√3)]ⁿ = {sin(-π/6) + icos(-π/6)}ⁿ
= sin(-nπ/6) + i cos(-nπ/6)

now, if choose n in such a way that, sin(-nπ/6) + i cos(-nπ/6) becomes 1

so, n = 9 is answer .

Answer 2

Answer:

3

Step-by-step explanation:

Concept:

We know that cube roots of unity are 1, ω, ω²

Also ω³=1

$\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\\\\=\frac{1+i\sqrt{3}}{1-i\sqrt{3}}.\frac{1+i\sqrt{3}}{1+i\sqrt{3}}\\\\=\frac{(1+i\sqrt{3})^2}{1+3}\\\\=\frac{1-3+i2\sqrt3}{4}\\\\=\frac{-2+i2\sqrt3}{4}\\\\=\frac{-1}{2}+i\frac{\sqrt3}{2}\\\\=cos\frac{2\pi}{3}+i\:sin\frac{2\pi}{3}\\\\=\omega\\\\Given:\:(\frac{1+i\sqrt{3}}{1-i\sqrt{3}})^{n}=1$

$This\:implies\:{\omega}^{n}=1$

since ω is a cube root of of unity, the least value of n satisfying the above condition is 3.