Question

If We have an expression as ${{\left( \dfrac{3}{2}+\dfrac{\sqrt{3}}{2} \right)}^{50}}={{3}^{25}}\left( x-iy \right)$ where $x,y$ are real, then the ordered pair $\left( x,y \right)$ given by
A. $\left( 0,3 \right)$
B. $\left( \dfrac{1}{2},-\dfrac{\sqrt{3}}{2} \right)$
C. $\left( -3,0 \right)$
D. $\left( 0,-3 \right)$

Answer 1

Hint: To obtain the ordered pair of the value given we will use complex numbers rule to simplify it. Firstly we will simplify the value given on the left side by introducing iota in it so that some term can be taken as common. Then depending on the right side we will take the term common in the left side and compare the equation to get our desired answer.

Complete step-by-step solution:
The value given is:
${{\left( \dfrac{3}{2}+\dfrac{\sqrt{3}}{2} \right)}^{50}}={{3}^{25}}\left( x-iy \right)$
As we can see, we have to get the term ${{3}^{25}}$ common on the left side for comparing them.
So we will change the term 3 inside the bracket in left side by introducing iota term as:
${{\left( \dfrac{-{{i}^{2}}\sqrt{3}\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2} \right)}^{50}}={{3}^{25}}\left( x-iy \right)$
As ${{i}^{2}}=-1$
Now we will simplify it further by taking the term common as:
$\begin{align}
  & \Rightarrow {{\left( \dfrac{-{{i}^{2}}\sqrt{3}\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2} \right)}^{50}}={{3}^{25}}\left( x-iy \right) \\
 & \Rightarrow {{\left( i\sqrt{3} \right)}^{50}}\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)={{3}^{25}}\left( x-iy \right) \\
 & \Rightarrow {{\left( {{\left( i\sqrt{3} \right)}^{2}} \right)}^{25}}\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)={{3}^{25}}\left( x-iy \right) \\
 & \Rightarrow {{\left( {{i}^{2}}\times 3 \right)}^{25}}\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)={{3}^{25}}\left( x-iy \right) \\
\end{align}$
As ${{i}^{2}}=-1$ so we get,
$-{{\left( 3 \right)}^{25}}{{\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)}^{50}}={{3}^{25}}\left( x-iy \right)$
On comparing Right and Left side we get,
$-{{\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)}^{50}}=\left( x-iy \right)$…..$\left( 1 \right)$
Using property of complex number:
$\begin{align}
  & {{w}^{50}}={{\left( {{w}^{3}} \right)}^{16}}\times {{w}^{2}} \\
 & {{w}^{50}}={{w}^{2}} \\
\end{align}$
Here $w={{\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)}^{50}}$
Using the above property in equation (1) we get,
$\begin{align}
  & \Rightarrow -{{\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} \right)}^{2}}=x-iy \\
 & \Rightarrow -\left( \dfrac{1}{2}-2\times \dfrac{1}{2}\times \dfrac{i\sqrt{3}}{2}+{{\left( \dfrac{i\sqrt{3}}{2} \right)}^{2}} \right)=x-iy \\
 & \Rightarrow -\left( \dfrac{1}{2}-\dfrac{i\sqrt{3}}{2}-\dfrac{3}{4} \right)=x-iy \\
 & \therefore \dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}=x-iy \\
\end{align}$
On comparing right and left side of above equation we get,
$\begin{align}
  & x=\dfrac{1}{2} \\
 & y=\dfrac{-\sqrt{3}}{2} \\
\end{align}$
So the ordered pair is $\left( \dfrac{1}{2},-\dfrac{\sqrt{3}}{2} \right)$
Hence, option (B) is correct.

Note: The complex numbers are those that can be expressed in the form of $x+iy$ where $a,b$ are real numbers and $i$ denotes the imaginary values. The real number $a$ is known as the real part and $b$ is known as the imaginary part. The value of $i=\sqrt{-1}$ and by using it the other value of iota with power is calculated. Complex numbers give solutions to all polynomial functions whose otherwise solution is not possible.