If `z = ((1)/(sqrt(3)) + (1)/(2)i)^(7) + ((1)/(sqrt(3))-(1)/(2)i)^(7)`, then

Options

A. Re (z) = 0
B. Im (z) = 0
C. Re (z) `gt` 0, Im (z) `lt` 0
D. Re (z) `lt` 0, Im (z) `gt` 0

Answer

To solve the problem \( z = \left( \frac{1}{\sqrt{3}} + \frac{1}{2} i \right)^{7} + \left( \frac{1}{\sqrt{3}} - \frac{1}{2} i \right)^{7} \), we can use the binomial theorem and properties of complex numbers.

### Step-by-Step Solution:

1. **Identify the terms**: Let \( a = \frac{1}{\sqrt{3}} \) and \( b = \frac{1}{2} i \). Then we can rewrite \( z \) as:
\[ z = (a + b)^{7} + (a - b)^{7} \]

2. **Apply the Binomial Theorem**: According to the binomial theorem:
\[ (x + y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \]
We can apply this to both \( (a + b)^{7} \) and \( (a - b)^{7} \).

3. **Expand \( (a + b)^{7} \)**:
\[ (a + b)^{7} = \sum_{k=0}^{7} \binom{7}{k} a^{7-k} b^{k} \]

4. **Expand \( (a - b)^{7} \)**:
\[ (a - b)^{7} = \sum_{k=0}^{7} \binom{7}{k} a^{7-k} (-b)^{k} = \sum_{k=0}^{7} \binom{7}{k} a^{7-k} (-1)^{k} b^{k} \]

5. **Combine the expansions**:
\[ z = \sum_{k=0}^{7} \binom{7}{k} a^{7-k} b^{k} + \sum_{k=0}^{7} \binom{7}{k} a^{7-k} (-1)^{k} b^{k} \]
This simplifies to:
\[ z = \sum_{k=0}^{7} \binom{7}{k} a^{7-k} b^{k} (1 + (-1)^{k}) \]

6. **Identify terms that contribute to \( z \)**: The expression \( (1 + (-1)^{k}) \) equals:
- 2 when \( k \) is even (because \( (-1)^{k} = 1 \))
- 0 when \( k \) is odd (because \( (-1)^{k} = -1 \))

Therefore, only the even \( k \) terms contribute to \( z \).

7. **Sum the even terms**:
\[ z = 2 \sum_{k \text{ even}} \binom{7}{k} a^{7-k} b^{k} \]
The even values of \( k \) are \( 0, 2, 4, 6 \).

8. **Calculate each term**:
- For \( k = 0 \): \( 2 \cdot \binom{7}{0} a^{7} b^{0} = 2 \cdot 1 \cdot \left( \frac{1}{\sqrt{3}} \right)^{7} \)
- For \( k = 2 \): \( 2 \cdot \binom{7}{2} a^{5} b^{2} = 2 \cdot 21 \cdot \left( \frac{1}{\sqrt{3}} \right)^{5} \cdot \left( \frac{1}{2} i \right)^{2} = 2 \cdot 21 \cdot \left( \frac{1}{\sqrt{3}} \right)^{5} \cdot \left( -\frac{1}{4} \right) \)
- For \( k = 4 \): \( 2 \cdot \binom{7}{4} a^{3} b^{4} = 2 \cdot 35 \cdot \left( \frac{1}{\sqrt{3}} \right)^{3} \cdot \left( \frac{1}{2} i \right)^{4} = 2 \cdot 35 \cdot \left( \frac{1}{\sqrt{3}} \right)^{3} \cdot \left( \frac{1}{16} \right) \)
- For \( k = 6 \): \( 2 \cdot \binom{7}{6} a^{1} b^{6} = 2 \cdot 7 \cdot \left( \frac{1}{\sqrt{3}} \right)^{1} \cdot \left( \frac{1}{2} i \right)^{6} = 2 \cdot 7 \cdot \left( \frac{1}{\sqrt{3}} \right) \cdot \left( -\frac{1}{64} \right) \)

9. **Combine all contributions**: After calculating, we will find that all contributions are real, and the imaginary parts cancel out.

10. **Conclusion**: Therefore, the imaginary part of \( z \) is zero, and \( z \) is purely real.

### Final Answer:
The value of \( z \) is a real number, and the imaginary part of \( z \) is \( 0 \).