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 follow these steps:

### Step 1: Convert to Polar Form
First, we need to express \( 1 + i \sqrt{3} \) in polar form. We can find the modulus and the argument of the complex number.

- **Modulus**:
\[ r = |1 + i \sqrt{3}| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

- **Argument**:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]

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

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

### Step 3: Simplify the Angle
Next, we 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 reduce \( \frac{1999}{3} \):
\[ 1999 \div 3 = 666 \quad \text{(remainder 1)} \]
Thus,
\[ \frac{1999\pi}{3} = 666 \cdot 2\pi + \frac{\pi}{3} \]
This means:
\[ \cos\left(\frac{1999\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
\[ \sin\left(\frac{1999\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

### Step 4: Substitute Back
Now substituting back into our expression:
\[ (1 + i \sqrt{3})^{1999} = 2^{1999} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \]
\[ = 2^{1999} \cdot \frac{1}{2} + i \cdot 2^{1999} \cdot \frac{\sqrt{3}}{2} \]
\[ = 2^{1998} + i \cdot 2^{1998} \sqrt{3} \]

### Step 5: Identify \( a \) and \( b \)
From this, we can identify:
\[ a = 2^{1998}, \quad b = 2^{1998} \sqrt{3} \]

### Final Answer
Thus, the final result is:
\[ (1 + i \sqrt{3})^{1999} = 2^{1998} + i \cdot 2^{1998} \sqrt{3} \]