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 $(2x, 2y)$ is

• A. \((-6, 0)\)
• B. \((0, 6)\)
• C. \((0, -6)\)
• D. \((1, \sqrt{3})\)

Hint:

When solving problems with complex numbers in polar form, recall that \( \text{cis} \theta = \cos \theta + i \sin \theta \), and use De Moivre’s Theorem for raising complex numbers to powers.

Answer 1

The Correct Option is D

Given the complex number \( z = \left( \frac{3}{2} + i \frac{\sqrt{3}}{2} \right) \), we want to compute \( z^{50} = 3^{25}(x + iy) \) and find the ordered pair \( (2x, 2y) \).

Firstly, identify \( z \) in polar form. The modulus of \( z \) is calculated as:

\[ |z| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \]

The argument \( \theta \) is determined by:

\[ \tan\theta = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3} \]

which gives \(\theta = \frac{\pi}{6}\). Therefore, the polar form is:

\[ z = \sqrt{3} \left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right) \]

Using De Moivre’s Theorem, \( z^{50} \) is:

\[ z^{50} = (\sqrt{3})^{50} \left( \cos\left(50 \times \frac{\pi}{6}\right) + i\sin\left(50 \times \frac{\pi}{6}\right) \right) \]

Now, compute the powers:

\[ (\sqrt{3})^{50} = (3^{0.5})^{50} = 3^{25} \]

Next, simplify the angle:

\[ 50 \times \frac{\pi}{6} = \frac{50\pi}{6} = \frac{25\pi}{3} \]

Find the equivalent angle within \( 2\pi \) by subtracting \( 8\pi \) (since \(\frac{25\pi}{3} - 8\pi = \frac{\pi}{3}\)):

The angle simplifies to:

\[ \frac{25\pi}{3} \equiv \frac{\pi}{3} \text{ (mod } 2\pi) \]

Thus,

\[ z^{50} = 3^{25} \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right) \]

Substitute \(\cos \frac{\pi}{3} = \frac{1}{2}\) and \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\):

\[ z^{50} = 3^{25} \left( \frac{1}{2} + i\frac{\sqrt{3}}{2} \right) \]

Thus, \( x = \frac{1}{2} \) and \( y = \frac{\sqrt{3}}{2} \), giving \( (2x, 2y) = (1, \sqrt{3}) \).

The ordered pair is \((1, \sqrt{3})\).

Answer 2

We begin by expressing \( \frac{3}{2} + i \frac{\sqrt{3}}{2} \) in polar form. 
The modulus is: \[ r = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3} \] The argument \( \theta \) is: \[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \] Thus, \( \frac{3}{2} + i \frac{\sqrt{3}}{2} = \sqrt{3} \text{cis} \frac{\pi}{6} \), where \( \text{cis} \theta = \cos \theta + i \sin \theta \). Now, raising both sides to the power of 50: \[ \left( \sqrt{3} \text{cis} \frac{\pi}{6} \right)^{50} = \left( \sqrt{3} \right)^{50} \text{cis} \left( 50 \times \frac{\pi}{6} \right) = 3^{25} \text{cis} \left( \frac{50\pi}{6} \right) \] Since \( \frac{50\pi}{6} = 8\pi + \frac{\pi}{3} \), the argument simplifies to \( \frac{\pi}{3} \). Therefore, the expression becomes: \[ 3^{25} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) \] This corresponds to: \[ 3^{25} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \] Now comparing this with the given expression \( 3^{25}(x + iy) \), we have: \[ x = \frac{1}{2}, \quad y = \frac{\sqrt{3}}{2} \] Therefore, the ordered pair \( (2x, 2y) \) is: \[ (2x, 2y) = (1, \sqrt{3}) \]