If $3^{49}(x+iy)=(\frac{3}{2}+\frac{\sqrt{3}}{2}i)^{100}$ and $x=ky,$ then k is:

Question

Accepted Answer

To solve the problem, we need to analyze the given equation and determine the value of $ k $ such that $ x = ky $.

The equation given is:

$$ 3^{49}(x + iy) = \left(\frac{3}{2} + \frac{\sqrt{3}}{2}i\right)^{100} $$

First, let's express the complex number $\frac{3}{2} + \frac{\sqrt{3}}{2}i$ in polar form. The modulus of this complex number is:

$$ \left|\frac{3}{2} + \frac{\sqrt{3}}{2}i\right| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} $$

The argument (angle) $\theta$ of the complex number is given by:

$$ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} $$

Thus, the polar form of the complex number is:

$$ \frac{3}{2} + \frac{\sqrt{3}}{2}i = \sqrt{3} \left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) $$

Raising this to the 100th power, we use De Moivre's Theorem:

$$ \left(\sqrt{3}\right)^{100} \left(\cos\left(\frac{100\pi}{6}\right) + i\sin\left(\frac{100\pi}{6}\right)\right) $$

Simplifying the angle:

$$ \frac{100\pi}{6} = \frac{50\pi}{3} = 16\pi + \frac{2\pi}{3} $$

Since $16\pi$ is a multiple of $2\pi$, it does not affect the angle, so we have:

$$ \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2} $$

Thus, the expression becomes:

$$ 3^{50} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) $$

Equating this to $3^{49}(x + iy)$, we have:

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

From this, we can equate the real and imaginary parts:

$$ x = -\frac{3}{2}, \quad y = \frac{3\sqrt{3}}{2} $$

Given $ x = ky $, we substitute:

$$ -\frac{3}{2} = k \left(\frac{3\sqrt{3}}{2}\right) $$

Solving for $ k $:

$$ k = \frac{-\frac{3}{2}}{\frac{3\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} $$

Thus, the correct value of $ k $ is $-\frac{1}{\sqrt{3}}$, which corresponds to Option 4.

**Explanation for Incorrect Options:**

1. **Option 1: $-\frac{1}{3}$** - This value does not satisfy the equation derived from equating the real and imaginary parts of the complex numbers.

2. **Option 2: $\sqrt{3}$** - This value would imply a positive relationship between $ x $ and $ y $, which contradicts the negative sign in the real part of the complex number.

3. **Option 3: $-\sqrt{3}$** - This value is incorrect as it does not match the derived ratio of the real and imaginary parts.

Therefore, Option 4 is the only correct choice, as it accurately reflects the relationship between $ x $ and $ y $ in the given equation.