If ` ( sqrt( 3) + i) ^(10) = a + ib` then, a and b are respectively

Options

A. `128 and 128 sqrt(3)`
B. `64 and - 64 sqrt(3)`
C. `512 and - 512 sqrt(3)`
D. none of these

Answer

To solve the problem \( ( \sqrt{3} + i )^{10} = a + ib \), we will convert the complex number into polar form and then apply De Moivre's theorem. Here is the step-by-step solution:

### Step 1: Convert to Polar Form
First, we need to express \( \sqrt{3} + i \) in polar form. The modulus \( r \) of the complex number is given by:
\[ r = |z| = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

Next, we find the argument \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

Thus, we can express \( \sqrt{3} + i \) in polar form as:
\[ \sqrt{3} + i = 2 \left( \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} \right) = 2 e^{i \frac{\pi}{6}} \]

### Step 2: Apply De Moivre's Theorem
Now we raise the polar form to the power of 10:
\[ ( \sqrt{3} + i )^{10} = \left( 2 e^{i \frac{\pi}{6}} \right)^{10} = 2^{10} e^{i \frac{10\pi}{6}} = 1024 e^{i \frac{5\pi}{3}} \]

### Step 3: Convert Back to Rectangular Form
Next, we convert \( e^{i \frac{5\pi}{3}} \) back to rectangular form:
\[ e^{i \frac{5\pi}{3}} = \cos\left(\frac{5\pi}{3}\right) + i \sin\left(\frac{5\pi}{3}\right) \]
Calculating the cosine and sine:
\[ \cos\left(\frac{5\pi}{3}\right) = \cos\left(2\pi - \frac{\pi}{3}\right) = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} \]
\[ \sin\left(\frac{5\pi}{3}\right) = \sin\left(2\pi - \frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \]

Thus,
\[ e^{i \frac{5\pi}{3}} = \frac{1}{2} - i \frac{\sqrt{3}}{2} \]

Now, substituting back, we get:
\[ 1024 e^{i \frac{5\pi}{3}} = 1024 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = 512 - 512\sqrt{3} i \]

### Step 4: Identify \( a \) and \( b \)
From the expression \( 512 - 512\sqrt{3} i \), we can identify:
\[ a = 512, \quad b = -512\sqrt{3} \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ \boxed{(512, -512\sqrt{3})} \]