If `(3/2+(isqrt(3))/2)^(50)=3^(25)(x+iy)`, where x and y are reals, then the ordered pair (x,y) is given by

Options

A. (0,3)
B. `(1//2,sqrt(3)//2)`
C. `(-3,0)`
D. `(0,-3)`

Explanation

To solve the problem, we start with the equation:

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

### Step 1: Rewrite the base in polar form

The complex number \(\frac{3}{2} + i\frac{\sqrt{3}}{2}\) can be expressed in polar form. We first find the modulus \(r\) and the argument \(\theta\).

The modulus \(r\) is given by:

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

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

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

Thus, we can express the complex number in polar form as:

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

### Step 2: Apply De Moivre's Theorem

Using De Moivre's Theorem, we can raise the polar form to the 50th power:

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

Calculating the modulus:

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

Now, we calculate the angle:

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

To simplify \(\frac{25\pi}{3}\), we can subtract \(8\pi\) (which is \(2\cdot 4\pi\)):

\[ \frac{25\pi}{3} - 8\pi = \frac{25\pi - 24\pi}{3} = \frac{\pi}{3} \]

### Step 3: Substitute back into the equation

Now we can substitute back into our expression:

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

Calculating \(\cos\frac{\pi}{3}\) and \(\sin\frac{\pi}{3}\):

\[ \cos\frac{\pi}{3} = \frac{1}{2}, \quad \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \]

Thus, we have:

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

### Step 4: Equate and find \(x\) and \(y\)

Now we equate this to \(3^{25}(x + iy)\):

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

Dividing both sides by \(3^{25}\):

\[ \frac{1}{2} + i\frac{\sqrt{3}}{2} = x + iy \]

From this, we can directly read off the values of \(x\) and \(y\):

\[ x = \frac{1}{2}, \quad y = \frac{\sqrt{3}}{2} \]

### Final Answer

The ordered pair \((x, y)\) is:

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