Question

If α and β be the roots of the equation x² + 2x + 2 = 0, then the least value of n for which (α/β)ⁿ = 1
is:
(A) 4 (B) 2
(C) 5 (D) 3

Answer 1

$\text{Given equation is}$

$x^2+2x+2=0$

$x^2+2x+1=-1$

$(x+1)^2=i^2$

$x+1=\pm\,i$

$\implies\,x=-1\pm\,i$

$\text{Then,}$

$\alpha=-1+i\;\text{and}\;\beta=-1-i$

$\frac{\alpha}{\beta}$

$=\frac{-1+i}{-1-i}$

$=\frac{1-i}{1+i}{\times}\frac{1-i}{1-i}$

$=\frac{1-1-2i}{1+1}$

$=\frac{-2i}{2}$

$=-i$

$\text{Condition given:}$

$(\frac{\alpha}{\beta})^n=1$

$(-i)^n=1$

$\implies\text{n=4, 8, 12,.......}$

$\therefore\textbf{The least value of n satisfying the given condition is 4}$

$\implies\textbf{Option (A) is correct}$

Answer 2

for n = 4 , (α/β)ⁿ = 1

•Given,

x² + 2x + 2 = 0

Now, a= 1

b=2

c=2

•b²-4ac = (2)²-4(1)(2)

= 4 - 8

= -4

•Using quadratic formula ,

x = [-b ± √(b²-4ac)]/2a

x = [-2 ± √-4]/2

x = [-2±2i]/2

x = -1 ± i

=> α = -1 + i and β = -1 - i

α/β = (-1 + i )/( -1 - i )

•Realising the denominator,

α/β = (-1 + i )²/( -1 - i )(-1 + i )

α/β = (1 + i² - 2i )/(-1)( i +1 )( i - 1 )

α/β = ( 1 - 1 -2i)/(-1)(i²-1)

α/β = (-2i)/(2)

α/β = ( - i )

•we need,

(α/β)ⁿ = 1

( -i )ⁿ = 1

•clearly for n = 4

(-i)⁴ = i⁴ = 1

f•or n = 4 , (α/β)ⁿ = 1