Question

If we have the rots as $\alpha \text{ and }\beta $ of a quadratic equation ${{x}^{2}}-2x+2=0$ then the least value of n for which ${{\left( \dfrac{\alpha }{\beta } \right)}^{n}}=1$ is:
\[\begin{align}
  & \text{A}.\text{ 2} \\
 & \text{B}.\text{ 3} \\
 & \text{C}.\text{ 4} \\
 & \text{D}.\text{ 5} \\
\end{align}\]

Answer 1

Hint: The above question uses the concept of quadratic equations and complex numbers. As we know that, if quadratic equation $a{{x}^{2}}+bx+c=0$ then \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] where expression ${{b}^{2}}-4ac=D$ (Discriminant). We have three cases for any quadratic equation i.e.
$D\text{ }>\text{ }0$: roots are real and distinct (unequal).
$D\text{ }=\text{ }0$: roots are real and coincident (equal).
$D\text{ }<\text{ }0$: roots are imaginary.
In the about question, we will find the third case of $D\text{ } <\text{ }0$ and then further we apply the complex number basics (like $i=\sqrt{-1}$)

Complete step-by-step solution:
Now, we have been given equation as:
\[{{x}^{2}}-2x+2=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
As we know that for a quadratic equation $a{{x}^{2}}+bx+c=0$, the root is given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
Hence, from equation (i) we have: a = 1, b = -2, c = 2. Now, substituting them in above equation, and simplifying using $i=\sqrt{-1}$, we have
\[\begin{align}
  & x=\dfrac{-\left( -2 \right)\pm \sqrt{4-4\times 1\times 2}}{2\times 1} \\
 &\Rightarrow x=\dfrac{2\pm \left( \sqrt{4} \right)\left( \sqrt{-1} \right)}{2} \\
 &\Rightarrow x=\dfrac{2\pm i\times 2}{2} \\
 &\Rightarrow x=\dfrac{1\pm i}{1} \\
 &\Rightarrow x=\left( 1\pm i \right) \\
\end{align}\]
Hence, we have got the roots as \[\alpha =1+i,\beta =1-i\]
Now, from given condition we have:
\[\begin{align}
  & {{\left( \dfrac{\alpha }{\beta } \right)}^{n}}=1 \\
 &\Rightarrow {{\left( \dfrac{1+i}{1-i} \right)}^{n}}=1 \\
 &\Rightarrow {{\left( \dfrac{\left( 1+i \right)\left( 1+i \right)}{{{1}^{2}}-{{i}^{2}}} \right)}^{n}}=1\text{ (rationalise)} \\
 &\Rightarrow {{\left( \dfrac{{{1}^{2}}+{{i}^{2}}+2i}{1-\left( -1 \right)} \right)}^{n}}=1\text{ (}\because {{\text{i}}^{2}}=-1) \\
 &\Rightarrow {{\left( \dfrac{1-1+2i}{2} \right)}^{n}}=1 \\
 &\Rightarrow {{i}^{n}}=1 \\
\end{align}\]
Now for values of n as
\[\begin{align}
  & n=2,{{i}^{2}}=-1 \\
 &\Rightarrow n=3,{{i}^{3}}={{i}^{2}}\left( i \right)=\left( -i \right) \\
 &\Rightarrow n=4,{{i}^{4}}=\left( {{i}^{2}} \right)\left( {{i}^{2}} \right)=\left( -1 \right)\left( -1 \right)=\left( 1 \right) \\
 &\Rightarrow n=5,{{i}^{5}}=\left( {{i}^{2}} \right)\left( {{i}^{3}} \right)=\left( -1 \right)\left( -i \right)=\left( i \right) \\
\end{align}\]
Hence, option (C) 4 is the correct answer.


Note: This is one of the basic part of quadratic equation where we have to see the nature of roots and solve accordingly. From exam point of view, there are two other most important categories (topics) present in quadratic equation which are asked many times in several competitive examinations. These topics are: common roots of quadratic equation, location of roots and maximum and minimum value of quadratic equation.
Now, come to the complex number part in that given question i.e. $i=\sqrt{-1}$. If anywhere, we see this term i in denominator then immediately try to convert this to the form or state where it should not be present. And the most important step for converting such condition is concept of rationalization.