Question

If \[{\left( {\dfrac{3}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\], where \[x\] and \[y\] are real. Then find the ordered pair \[\left( {x,y} \right)\].
A. \[\left( { - 3,0} \right)\]
B. \[\left( {0,3} \right)\]
C. \[\left( {0, - 3} \right)\]
D. \[\left( {\dfrac{1}{2},\sqrt {\dfrac{3}{2}} } \right)\]

Answer 1

Hint We know that, the cube root of 1 are 1, \[\omega = \dfrac{{ - 1 + i\sqrt 3 }}{2}\] and \[{\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}\]. First, we will take common \[\sqrt 3 \] from the left side of the equation and put \[\omega = \dfrac{{ - 1 + i\sqrt 3 }}{2}\]. Then simplify the equation and compare the real part and imaginary part to get the value of \[x\] and \[y\].

Formula used:
The cube root of 1 are 1, \[\omega = \dfrac{{ - 1 + i\sqrt 3 }}{2}\] and \[{\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}\].
\[{\omega ^3} = 1\]
\[{i^2} = - 1\]

Complete step by step solution
Given equation is \[{\left( {\dfrac{3}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\]
Take common \[i\sqrt 3 \]
\[{\left( {i\sqrt 3 } \right)^{50}}{\left( {\dfrac{{\sqrt 3 }}{{2i}} + \dfrac{1}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\]
Multiply \[i\] with numerator and denominator of \[\dfrac{{\sqrt 3 }}{{2i}}\]
\[ \Rightarrow {\left( {i\sqrt 3 } \right)^{50}}{\left( {\dfrac{{i\sqrt 3 }}{{2{i^2}}} + \dfrac{1}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\]
Putting \[{i^2} = - 1\]
\[ \Rightarrow {\left( {i\sqrt 3 } \right)^{50}}{\left( { - \dfrac{{i\sqrt 3 }}{2} + \dfrac{1}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\]
\[ \Rightarrow {i^{50}}{\left( {{3^{\dfrac{1}{2}}}} \right)^{50}}{\left( {\dfrac{{1 - i\sqrt 3 }}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\]
\[ \Rightarrow {i^{50}}{\left( {{3^{\dfrac{1}{2}}}} \right)^{50}}{\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\]
Now putting \[\omega = \dfrac{{ - 1 + i\sqrt 3 }}{2}\]
\[ \Rightarrow {i^{50}}{\left( {{3^{\dfrac{1}{2}}}} \right)^{50}}{\omega ^{50}} = {3^{25}}\left( {x + iy} \right)\]
\[ \Rightarrow {i^{48}} \cdot {i^2} \cdot {3^{25}} \cdot {\omega ^{50}} = {3^{25}}\left( {x + iy} \right)\] [Since \[{\left( {{3^{\dfrac{1}{2}}}} \right)^{50}} = {3^{\dfrac{1}{2} \cdot 50}} = {3^{25}}\]]
\[ \Rightarrow {\left( {{i^4}} \right)^{12}} \cdot {i^2} \cdot {3^{25}} \cdot {\omega ^{50}} = {3^{25}}\left( {x + iy} \right)\]
Putting \[{i^4} = 1\] and \[{i^2} = - 1\]
\[ \Rightarrow - {3^{25}} \cdot {\omega ^{50}} = {3^{25}}\left( {x + iy} \right)\]
Now divide both sides by \[{3^{25}}\]
\[ \Rightarrow - {\omega ^{50}} = \left( {x + iy} \right)\]
\[ \Rightarrow - {\omega ^{48}} \cdot {\omega ^2} = \left( {x + iy} \right)\]
\[ \Rightarrow - {\left( {{\omega ^3}} \right)^{16}} \cdot {\omega ^2} = \left( {x + iy} \right)\]
Putting \[{\omega ^3} = 1\] in the above equation
\[ \Rightarrow - {\left( 1 \right)^{16}} \cdot {\omega ^2} = \left( {x + iy} \right)\]
\[ \Rightarrow - {\omega ^2} = \left( {x + iy} \right)\]
Putting \[{\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}\] in the above equation
\[ \Rightarrow - \dfrac{{ - 1 - i\sqrt 3 }}{2} = \left( {x + iy} \right)\]
\[ \Rightarrow \dfrac{1}{2} + \dfrac{{i\sqrt 3 }}{2} = \left( {x + iy} \right)\]
Now comparing the real part and imaginary part
\[x = \dfrac{1}{2}\] and \[y = \dfrac{{\sqrt 3 }}{2}\]
So, the ordered pair is \[\left( {\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}} \right)\].
Hence option D is the correct option.

Note: Students often do a common mistake to solve the equation \[{\left( {\dfrac{3}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{50}} = {3^{25}}\left( {x + iy} \right)\]. They used binomial expansion to solve the incorrect way. We should use cube roots of 1 to solve the equation.