QJEEM19Macf66: JEE Main 2019 Mathematics - Class 11 JEEM19Macf66 Solution: If and be the roots of the equation x 2 – 2x + 2 = 0, then the…

Question

Verified Solution • 2025–26 Session moderate

If

\alpha

and

\beta

be the roots of the equation x² – 2x + 2 = 0, then the least value of n for which

{\left( {{\alpha \over \beta }} \right)^n} = 1

is :

Accepted Answer

x² – 2x + 2 = 0

\therefore

x =

{{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i

Now,

{\alpha \over \beta } = {{1 + i} \over {1 - i}} = {{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}} = i

or

{\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( {1 - i} \right)}^2}} \over {1 + {i^2}}} = - i

\therefore

{\left( { i} \right)^n}

= 1

and

{\left( { - i} \right)^n}

= 1

So n must be a multiple of 4

\therefore

Minimum value of n = 4

Step-by-Step Verified Solution