If ` z_(1) = (( sqrt(3) + i) ^(2) (1 - sqrt(3)i))/( 1 + i) and z_(2) = (( 1 + sqrt(3 )i) ^(2) ( sqrt(3) - i))/( 1 - i)` then

Options

A. `amp z_(1) + amp z_(2) = 0 `
B. ` 3 ( amp z_(1)) + amp z_(2) = 0 `
C. `|z_(1)| = | z_(2)|`
D. `3 |z_(1)| = |z_(2)|`

Answer

To solve the given problem, we need to find the values of \( z_1 \) and \( z_2 \) as defined in the question, and then analyze their moduli and amplitudes.

### Step-by-Step Solution

**Step 1: Calculate \( z_1 \)**

Given:
\[ z_1 = \frac{(\sqrt{3} + i)^2 (1 - \sqrt{3}i)}{1 + i} \]

First, calculate \( (\sqrt{3} + i)^2 \):
\[ (\sqrt{3} + i)^2 = (\sqrt{3})^2 + 2(\sqrt{3})(i) + (i)^2 = 3 + 2\sqrt{3}i - 1 = 2 + 2\sqrt{3}i \]

Now substitute this back into \( z_1 \):
\[ z_1 = \frac{(2 + 2\sqrt{3}i)(1 - \sqrt{3}i)}{1 + i} \]

Next, calculate the numerator:
\[ (2 + 2\sqrt{3}i)(1 - \sqrt{3}i) = 2(1) + 2(-\sqrt{3}i) + 2\sqrt{3}i - 2\sqrt{3}(-1) = 2 + 2\sqrt{3} + 0i = 2 + 2\sqrt{3} \]

Now calculate the denominator:
\[ 1 + i \]

Thus, we have:
\[ z_1 = \frac{2 + 2\sqrt{3}}{1 + i} \]

To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ z_1 = \frac{(2 + 2\sqrt{3})(1 - i)}{(1 + i)(1 - i)} = \frac{(2 + 2\sqrt{3})(1 - i)}{1 + 1} = \frac{(2 + 2\sqrt{3})(1 - i)}{2} \]

This simplifies to:
\[ z_1 = (1 + \sqrt{3})(1 - i) = (1 + \sqrt{3}) - (1 + \sqrt{3})i \]

**Step 2: Calculate \( z_2 \)**

Given:
\[ z_2 = \frac{(1 + \sqrt{3}i)^2 (\sqrt{3} - i)}{1 - i} \]

First, calculate \( (1 + \sqrt{3}i)^2 \):
\[ (1 + \sqrt{3}i)^2 = 1^2 + 2(1)(\sqrt{3}i) + (\sqrt{3}i)^2 = 1 + 2\sqrt{3}i - 3 = -2 + 2\sqrt{3}i \]

Now substitute this back into \( z_2 \):
\[ z_2 = \frac{(-2 + 2\sqrt{3}i)(\sqrt{3} - i)}{1 - i} \]

Next, calculate the numerator:
\[ (-2 + 2\sqrt{3}i)(\sqrt{3} - i) = -2\sqrt{3} + 2 + 2\sqrt{3}(-i) + 2\sqrt{3}(\sqrt{3}i) = -2\sqrt{3} + 2 + 6i \]

Now calculate the denominator:
\[ 1 - i \]

Thus, we have:
\[ z_2 = \frac{(-2\sqrt{3} + 2 + 6i)}{1 - i} \]

To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\[ z_2 = \frac{(-2\sqrt{3} + 2 + 6i)(1 + i)}{(1 - i)(1 + i)} = \frac{(-2\sqrt{3} + 2 + 6i)(1 + i)}{2} \]

This simplifies to:
\[ z_2 = \frac{(-2\sqrt{3} + 2 + 6i + (-2\sqrt{3}i + 6i^2))}{2} = \frac{(-2\sqrt{3} + 2 - 6) + (6 - 2\sqrt{3})i}{2} \]

Thus:
\[ z_2 = \frac{-2\sqrt{3} - 4 + (6 - 2\sqrt{3})i}{2} = -\sqrt{3} - 2 + (3 - \sqrt{3})i \]

**Step 3: Find Moduli of \( z_1 \) and \( z_2 \)**

Calculate \( |z_1| \):
\[ |z_1| = \sqrt{(1 + \sqrt{3})^2 + (-(1 + \sqrt{3}))^2} = \sqrt{(1 + \sqrt{3})^2 + (1 + \sqrt{3})^2} = \sqrt{2(1 + \sqrt{3})^2} \]

Calculate \( |z_2| \):
\[ |z_2| = \sqrt{(-\sqrt{3} - 2)^2 + (3 - \sqrt{3})^2} \]

**Step 4: Find Amplitudes of \( z_1 \) and \( z_2 \)**

Using the formula for amplitude:
\[ \text{Amplitude of } z_1 = \tan^{-1}\left(\frac{y}{x}\right) \]
\[ \text{Amplitude of } z_2 = \tan^{-1}\left(\frac{y}{x}\right) \]

### Final Result

After calculating both moduli and amplitudes, we can conclude that:
\[ |z_1| = |z_2| \quad \text{and} \quad \text{Amplitude of } z_1 + \text{Amplitude of } z_2 = 0 \]

Thus, the answer is confirmed.