Question

If ${\left[ {\dfrac{3}{2} + i\dfrac{{\sqrt 3 }}{2}} \right] ^{50}} = {3^{25}}(x + iy)$ where x and y are real then the ordered pair $\left( {x{\text{ }},{\text{ }}y} \right)$ is
$\left( 1 \right)$ $\left( { - 3{\text{ }},{\text{ }}0} \right)$
$\left( 2 \right)$ $\left( {0{\text{ }},{\text{ }}3} \right)$
$\left( 3 \right)$ $\left( {0{\text{ }},{\text{ }} - 3} \right)$
$\left( 4 \right)$ $\left( {\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}} \right)$

Answer 1

Hint : We have to find an ordered pair of $\left( {x,y} \right)$ . We solve this question using the concept of the cube root of unity . We should also have the knowledge of the identities of complex numbers . Firstly we have to make the equation in terms of one of the roots of unity and then comparing both the sides and then evaluating the value of $x$ and $y$ .

Complete step-by-step answer :
Given : ${\left[ {\dfrac{3}{2} + i\dfrac{{\sqrt 3 }}{2}} \right] ^{50}} = {3^{25}}(x + iy)$
Taking , $\sqrt 3$ common from the L.H.S. , we get
${\left[ {\sqrt 3 \left( {\dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}} \right)} \right] ^{50}} = {3^{25}}(x + iy)$
Taking $\sqrt 3$ out of the bracket , we get
${3^{25}}{\left[ {\dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}} \right] ^{50}} = {3^{25}}(x + iy)$
Cancelling the terms
${\left[ {\dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}} \right] ^{50}} = (x + iy)$
We know , ${i^2} = - 1$
${\left[ {\dfrac{{ - {i^2}\sqrt 3 }}{2} + \dfrac{i}{2}} \right] ^{50}} = (x + iy)$
Taking in common from the L.H.S.
${i^{50}}{\left[ {\dfrac{{ - i\sqrt 3 }}{2} + \dfrac{1}{2}} \right] ^{50}} = (x + iy)$
We also know , $i = \sqrt { - 1}$ and the values of i repeats in multiples of $4$
So, simplifying the equation
${i^{4 \times 12 + 2}}{\left[ {\dfrac{1}{2} - i \times \dfrac{{\sqrt 3 }}{2}} \right] ^{50}} = (x + iy)$
We know , ${i^{(4n + 2)}} = {i^2}$ and ${i^2} = - 1$
$- {\left[ {\dfrac{1}{2} - i \times \dfrac{{\sqrt 3 }}{2}} \right] ^{50}} = (x + iy)$
Also ,
$- {\left[ { - \left( {\dfrac{{ - 1}}{2} + i \times \dfrac{{\sqrt 3 }}{2}} \right)} \right] ^{50}} = (x + iy)$
As ,
$\left[ {\dfrac{{ - 1}}{2} + i \times \dfrac{{\sqrt 3 }}{2}} \right]$ is one root of unity
So ,
Let $\omega = - \dfrac{1}{2} + i \times \dfrac{{\sqrt 3 }}{2}$
Then , the equation becomes ${\omega ^{50}} = x + iy$
Similarly , roots of unity also follows the rule of iota( $\;i$ ) i.e. the values of ω repeats in multiples of 4
So,
$- {\omega ^{4 \times 12 + 2}} = x + iy$
We know , ${i^{(4n + 2)}} = {i^2}$ and ${i^2} = - 1$
Similarly , ${\omega ^{(4n + 2)}} = {\omega ^2}$
$- {\omega ^2} = x + iy$
As we assumed that $\left( { - \dfrac{1}{2} + i \times \dfrac{{\sqrt 3 }}{2}} \right)$ is one of the complex root of unity
So , the other complex root of unity is $\left( { - \dfrac{1}{2} - i \times \dfrac{{\sqrt 3 }}{2}} \right)$
i.e.
${\omega ^2} = \left( { - \dfrac{1}{2} - i \times \dfrac{{\sqrt 3 }}{2}} \right)$
Putting ${\omega ^2}$ in equation , then
$- \left( {\dfrac{-1}{2} - i \times \dfrac{{\sqrt 3 }}{2}} \right) = \left( {x + iy} \right)$
$\left( {\dfrac{{ 1}}{2} + i \times \dfrac{{\sqrt 3 }}{2}} \right) = \left( {x + iy} \right)$
Comparing the real part with the real part and complex part with complex part , we get
$x{\text{ }} = {\text{ }}\dfrac{1}{2}$ and $y = \dfrac{{\sqrt 3 }}{2}$
Hence , The value of ordered pair $\left( {x,y} \right) = \left( {\dfrac{{ 1}}{2},\dfrac{{\sqrt 3 }}{2}} \right)$ .
Thus , the correct option is $\left( 4 \right)$ .
So, the correct answer is “Option 4”.

Note : The equation of the cube root of the unit is given as : ${\omega ^2} + \omega + 1 = 0$ . We can calculate the value of $\omega$ by using the quadratic formula . The quadratic formula is $\dfrac{{\sqrt { - b \pm [{b^2} - 4ac] } }}{{2a}}$ . Where $a$ is the coefficient of ${\omega ^2}$ , $b$ is the coefficient of $\omega$ and $c$ is the coefficient of the constant term in the quadratic equation . Two roots of unity are complex numbers and one is a real number .