If `(1+ i sqrt(3))^(1999)=a+ib`, then

Options

A. `a=2^(1998),b=2^(1998)sqrt(3)`
B. `a=2^(1999),b=2^(1999)sqrt(3)`
C. `a=-2^(1998),b=-2^(1998)sqrt(3)`
D. None of these

Answer

To solve the problem \((1 + i \sqrt{3})^{1999} = a + ib\), we will convert the complex number into polar form and then apply De Moivre's theorem.

### Step 1: Convert to Polar Form
First, we need to find the modulus \( r \) and the argument \( \theta \) of the complex number \( 1 + i \sqrt{3} \).

1. **Calculate the modulus \( r \)**:
\[ r = |1 + i \sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

2. **Calculate the argument \( \theta \)**:
\[ \theta = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

Thus, we can express \( 1 + i \sqrt{3} \) in polar form as:
\[ 1 + i \sqrt{3} = 2 \left( \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) \right) \]

### Step 2: Apply De Moivre's Theorem
Now we can raise this polar form to the power of 1999:
\[ (1 + i \sqrt{3})^{1999} = \left(2 \left( \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\right)\right)^{1999} \]
Using De Moivre's theorem:
\[ = 2^{1999} \left( \cos\left(1999 \cdot \frac{\pi}{3}\right) + i \sin\left(1999 \cdot \frac{\pi}{3}\right) \right) \]

### Step 3: Simplify the Argument
Next, we need to simplify \( 1999 \cdot \frac{\pi}{3} \):
\[ 1999 \cdot \frac{\pi}{3} = \frac{1999\pi}{3} \]
To find the equivalent angle in the range \( [0, 2\pi) \), we can subtract multiples of \( 2\pi \):
\[ \frac{1999\pi}{3} = 666\pi + \frac{\pi}{3} \quad (\text{since } 666 \cdot 3 = 1998) \]
Thus, the angle simplifies to:
\[ \frac{1999\pi}{3} \equiv \frac{\pi}{3} \quad (\text{mod } 2\pi) \]

### Step 4: Substitute Back
Now we substitute back into the expression:
\[ (1 + i \sqrt{3})^{1999} = 2^{1999} \left( \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) \right) \]
Using the values of cosine and sine:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
So we have:
\[ = 2^{1999} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \]
\[ = 2^{1999} \cdot \frac{1}{2} + i \cdot 2^{1999} \cdot \frac{\sqrt{3}}{2} \]
\[ = 2^{1998} + i \cdot 2^{1998} \sqrt{3} \]

### Final Result
Thus, we can express \( a \) and \( b \) as:
\[ a = 2^{1998}, \quad b = 2^{1998} \sqrt{3} \]