For `n=6k, k in z, ((1-isqrt(3))/(2))^(n)+((-1-isqrt(3))/(2))^(n)` has the value

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to evaluate the expression

\[ \left( \frac{1 - i\sqrt{3}}{2} \right)^{n} + \left( \frac{-1 - i\sqrt{3}}{2} \right)^{n} \]

for \( n = 6k \), where \( k \) is an integer.

### Step-by-step Solution:

1. **Identify the complex numbers**:
Let
\[ \omega = \frac{-1 - i\sqrt{3}}{2} \]
and
\[ \omega^2 = \frac{-1 + i\sqrt{3}}{2} \]
We note that \( \omega \) is one of the cube roots of unity.

2. **Cube roots of unity**:
The cube roots of unity are \( 1, \omega, \omega^2 \) where:
\[ \omega^3 = 1 \]
and
\[ \omega + \omega^2 + 1 = 0 \]

3. **Substituting \( n = 6k \)**:
We substitute \( n = 6k \) into our expression:
\[ \left( \frac{1 - i\sqrt{3}}{2} \right)^{6k} + \left( \frac{-1 - i\sqrt{3}}{2} \right)^{6k} \]
We can rewrite this as:
\[ \left( \omega^2 \right)^{6k} + \left( \omega \right)^{6k} \]

4. **Simplifying the powers**:
Since \( \omega^3 = 1 \), we have:
\[ \left( \omega^2 \right)^{6k} = \omega^{12k} = (\omega^3)^{4k} = 1^{4k} = 1 \]
and
\[ \left( \omega \right)^{6k} = \omega^{6k} = (\omega^3)^{2k} = 1^{2k} = 1 \]

5. **Final calculation**:
Now, we can combine the results:
\[ 1 + 1 = 2 \]

Thus, the value of the expression is

\[ \boxed{2} \]