Question

If \[\alpha \] and \[\beta \] are the roots of \[{x^2} - 2x + 2 = 0\], then find the minimum value of \[n\] such that \[{\left( {\dfrac{\alpha }{\beta }} \right)^n} = 1\].
A. 4
B. 3
C. 2
D. 5

Answer 1

Hint: In the given question, one quadratic equation is given. First, we will factorize the quadratic equation and find the roots of the equation. Then substitute values of \[\alpha \] and \[\beta \] in the equation \[{\left( {\dfrac{\alpha }{\beta }} \right)^n} = 1\] and find the minimum value of \[n\].


Formula Used: The conjugate of a complex number \[z = a + ib\] is \[\overline z = a - ib\].
\[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\]

Complete step by step solution:
The given quadratic equation is \[{x^2} - 2x + 2 = 0\]. \[\alpha \] and \[\beta \] are the roots of the equation.
Let’s factorize the quadratic equation.
\[{x^2} - 2x + 2 = 0\]
\[ \Rightarrow \]\[{x^2} - 2x + 1 + 1 = 0\]
\[ \Rightarrow \]\[{x^2} - 2x + 1 = - 1\]
\[ \Rightarrow \]\[{\left( {x - 1} \right)^2} = - 1\]
Take square root of both sides
\[ \Rightarrow \]\[\left( {x - 1} \right) = \pm i\]
\[ \Rightarrow \] \[x = 1 + i\] and \[x = 1 - i\]
Let’s consider \[\alpha = 1 + i\] and \[\beta = 1 - i\].

Now,
\[\dfrac{\alpha }{\beta } = \dfrac{{1 + i}}{{1 - i}}\]
multiply the numerator and denominator of the equation by the conjugate of the denominator.
The conjugate of the denominator \[\left( {1 - i} \right)\] is \[\left( {1 + i} \right)\].
Multiply the numerator and denominator by \[\left( {1 + i} \right)\].
\[\dfrac{\alpha }{\beta } = \dfrac{{\left( {1 + i} \right)\left( {1 + i} \right)}}{{\left( {1 - i} \right)\left( {1 + i} \right)}}\]
Simplify the above equation.
Apply the formula \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\] for the denominator
\[\dfrac{\alpha }{\beta } = \dfrac{{1 + i + i + {i^2}}}{{\left( {{1^2} + {1^2}} \right)}}\]
\[ \Rightarrow \]\[\dfrac{\alpha }{\beta } = \dfrac{{1 + 2i - 1}}{2}\] [since, \[{i^2} = - 1\]]
\[ \Rightarrow \]\[\dfrac{\alpha }{\beta } = \dfrac{{2i}}{2}\]
\[ \Rightarrow \]\[\dfrac{\alpha }{\beta } = i\]
Therefore,
\[{\left( {\dfrac{\alpha }{\beta }} \right)^n} = {i^n}\]
\[ \Rightarrow \]\[1 = {i^n}\]
We know that, \[{i^{4a}} = 1\] where \[a = 1,2,3,....\]
Hence, the minimum value of \[n\] is 4.

Option ‘A’ is correct

Note: Students are often confused with the formulas \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\] and \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} - {b^2}} \right)\] . The correct formula of the product of a complex number and its conjugate is \[\left( {a + ib} \right)\left( {a - ib} \right) = \left( {{a^2} + {b^2}} \right)\]. Because the square of the imaginary number \[i\] is \[ - 1\], which is multiplied by \[- {b^2}\].