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 .