Question

If ${\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}(x+iy)$, where $x$ and $y$ are real then the ordered pair $(x,y)$ is

• A. $(-3,0)$
• B. $(0,3)$
• C. $(0,-3)$
• D. $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$

Answer 1

The correct option is D

$\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$

Explanation for the correct option:

Step 1. Find the value of order pair $(x,y)$:

Given, ${\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}(x+iy)$

$\Rightarrow$ ${\left(\frac{-i^2\sqrt{3}\sqrt{3}}{2}+i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}(x+iy)$

$\Rightarrow$ ${\left(i\sqrt{3}\right)}^{50}{\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}(x+iy)$

$\Rightarrow$${\left[{\left(i\sqrt{3}\right)}^2\right]}^{25}{\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}(x+iy)$

$\Rightarrow$ ${\left(3\right)}^{25}{\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)}^{50}=3^{25}(x+iy)$ $\left[\mathbf{\because }{\mathit{i}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}\right]$

$\Rightarrow$ $-{\omega }^{50}=(x+iy)$ $\left[\mathbf{\because }\mathit{\omega }\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\right]$

Now, $\begin{array}{rcl}{\omega }^{50}& =& {\left({\omega }^3\right)}^{16}.{\omega }^2\end{array}$

$\begin{array}{rcl}& =& {\omega }^2\end{array}$ $\left[\mathbf{\because }{\mathit{\omega }}^{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

Step 2. Put the value of ${\omega }^{50}$:

$\Rightarrow$ $-{\omega }^2=(x+iy)$

$\Rightarrow$ $\frac{1}{2}+\frac{i\sqrt{3}}{2}=\left(x+iy\right)$ $\left[\mathbf{\because }{\mathit{\omega }}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\frac{\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\right]$

$\therefore \mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}$ and $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$

Hence, Option $\text{'}D\text{'}$is Correct.