Official solution
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