If `z(2-2sqrt(3i))^2=i(sqrt(3)+i)^4,` then `a r g(z)=`

Options

A. `(5pi)/6`
B. `-pi/6`
C. `pi/6`
D. `(7pi)/6`

Explanation

To solve the equation \( z(2 - 2\sqrt{3}i)^2 = i(\sqrt{3} + i)^4 \) and find the argument of \( z \), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding \( (2 - 2\sqrt{3}i)^2 \):
\[ (2 - 2\sqrt{3}i)^2 = 2^2 - 2 \cdot 2 \cdot 2\sqrt{3}i + (2\sqrt{3}i)^2 \]
Calculating each term:
- \( 2^2 = 4 \)
- \( -2 \cdot 2 \cdot 2\sqrt{3}i = -8\sqrt{3}i \)
- \( (2\sqrt{3}i)^2 = 4 \cdot 3 \cdot i^2 = 12(-1) = -12 \)

So, we have:
\[ (2 - 2\sqrt{3}i)^2 = 4 - 8\sqrt{3}i - 12 = -8 - 8\sqrt{3}i \]

### Step 2: Expand the right-hand side
Next, we expand \( (\sqrt{3} + i)^4 \):
Using the binomial theorem:
\[ (\sqrt{3} + i)^4 = \sum_{k=0}^{4} \binom{4}{k} (\sqrt{3})^{4-k} (i)^k \]
Calculating each term:
- For \( k = 0 \): \( \binom{4}{0} (\sqrt{3})^4 (i)^0 = 9 \)
- For \( k = 1 \): \( \binom{4}{1} (\sqrt{3})^3 (i)^1 = 4 \cdot 3\sqrt{3}i = 12\sqrt{3}i \)
- For \( k = 2 \): \( \binom{4}{2} (\sqrt{3})^2 (i)^2 = 6 \cdot 3 \cdot (-1) = -18 \)
- For \( k = 3 \): \( \binom{4}{3} (\sqrt{3})^1 (i)^3 = 4 \cdot \sqrt{3}(-i) = -4\sqrt{3}i \)
- For \( k = 4 \): \( \binom{4}{4} (i)^4 = 1 \)

Adding these together:
\[ (\sqrt{3} + i)^4 = 9 + 12\sqrt{3}i - 18 - 4\sqrt{3}i + 1 = -8 + 8\sqrt{3}i \]

### Step 3: Set the two sides equal
Now we set the two sides equal:
\[ z(-8 - 8\sqrt{3}i) = i(-8 + 8\sqrt{3}i) \]

### Step 4: Simplify the right-hand side
Calculating the right-hand side:
\[ i(-8 + 8\sqrt{3}i) = -8i + 8\sqrt{3}(-1) = -8i - 8\sqrt{3} \]

### Step 5: Solve for \( z \)
Now we have:
\[ z(-8 - 8\sqrt{3}i) = -8 - 8i \]
Thus,
\[ z = \frac{-8 - 8i}{-8 - 8\sqrt{3}i} \]

### Step 6: Multiply by the conjugate
To simplify \( z \), we multiply the numerator and denominator by the conjugate of the denominator:
\[ z = \frac{(-8 - 8i)(-8 + 8\sqrt{3}i)}{(-8 - 8\sqrt{3}i)(-8 + 8\sqrt{3}i)} \]

Calculating the denominator:
\[ (-8)^2 - (8\sqrt{3})^2 = 64 - 192 = -128 \]

Calculating the numerator:
\[ (-8)(-8) + (-8)(8\sqrt{3}i) + (-8i)(-8) + (-8i)(8\sqrt{3}i) = 64 - 64\sqrt{3}i + 64i - 64\sqrt{3}(-1) = 64 + 64\sqrt{3} + (64 - 64\sqrt{3})i \]

Thus,
\[ z = \frac{64 + 64\sqrt{3} + (64 - 64\sqrt{3})i}{-128} \]
This simplifies to:
\[ z = -\frac{1}{2} - \frac{1}{2}\sqrt{3} + \left(-\frac{1}{2} + \frac{1}{2}\sqrt{3}\right)i \]

### Step 7: Find the argument of \( z \)
Now we can find the argument of \( z \):
Let \( x = -\frac{1}{2} - \frac{1}{2}\sqrt{3} \) and \( y = -\frac{1}{2} + \frac{1}{2}\sqrt{3} \).

The argument \( \arg(z) \) is given by:
\[ \arg(z) = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-\frac{1}{2} + \frac{1}{2}\sqrt{3}}{-\frac{1}{2} - \frac{1}{2}\sqrt{3}}\right) \]

This simplifies to:
\[ \arg(z) = \tan^{-1}\left(\frac{1 - \sqrt{3}}{1 + \sqrt{3}}\right) \]

Using the known values, we find that this corresponds to \( -\frac{\pi}{6} \) in the fourth quadrant.

### Final Answer
Thus, the argument of \( z \) is:
\[ \arg(z) = -\frac{\pi}{6} \]