If `z+1/z=2cos6^@` , then `z^1000+1/[z^1000]` +1 is equal to

Options

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

Explanation

To solve the problem \( z + \frac{1}{z} = 2 \cos 6^\circ \) and find \( z^{1000} + \frac{1}{z^{1000}} + 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ z + \frac{1}{z} = 2 \cos 6^\circ \]
We can multiply both sides by \( z \) to eliminate the fraction:
\[ z^2 + 1 = 2 \cos 6^\circ z \]

### Step 2: Rearrange into standard quadratic form
Rearranging gives us:
\[ z^2 - 2 \cos 6^\circ z + 1 = 0 \]

### Step 3: Identify coefficients for the quadratic formula
In the quadratic equation \( az^2 + bz + c = 0 \), we have:
- \( a = 1 \)
- \( b = -2 \cos 6^\circ \)
- \( c = 1 \)

### Step 4: Apply the quadratic formula
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ z = \frac{2 \cos 6^\circ \pm \sqrt{(2 \cos 6^\circ)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (2 \cos 6^\circ)^2 - 4 = 4 \cos^2 6^\circ - 4 = 4(\cos^2 6^\circ - 1) = 4(-\sin^2 6^\circ) \]
Thus, we have:
\[ z = \frac{2 \cos 6^\circ \pm 2i \sin 6^\circ}{2} = \cos 6^\circ \pm i \sin 6^\circ \]
This simplifies to:
\[ z = e^{i 6^\circ} \quad \text{or} \quad z = e^{-i 6^\circ} \]

### Step 5: Calculate \( z^{1000} \)
Now, we need to find \( z^{1000} \):
\[ z^{1000} = (e^{i 6^\circ})^{1000} = e^{i 6000^\circ} \]
We can reduce \( 6000^\circ \) by subtracting multiples of \( 360^\circ \):
\[ 6000 \div 360 = 16.6667 \quad \text{(approximately 16 full cycles)} \]
Calculating the remainder:
\[ 6000 - 16 \times 360 = 6000 - 5760 = 240^\circ \]
Thus,
\[ z^{1000} = e^{i 240^\circ} \]

### Step 6: Calculate \( \frac{1}{z^{1000}} \)
Similarly,
\[ \frac{1}{z^{1000}} = e^{-i 240^\circ} \]

### Step 7: Add \( z^{1000} + \frac{1}{z^{1000}} \)
Now we add:
\[ z^{1000} + \frac{1}{z^{1000}} = e^{i 240^\circ} + e^{-i 240^\circ} = 2 \cos 240^\circ \]
Using the cosine value:
\[ \cos 240^\circ = -\frac{1}{2} \]
Thus,
\[ z^{1000} + \frac{1}{z^{1000}} = 2 \left(-\frac{1}{2}\right) = -1 \]

### Step 8: Final calculation
Finally, we add 1:
\[ z^{1000} + \frac{1}{z^{1000}} + 1 = -1 + 1 = 0 \]

### Conclusion
The final answer is:
\[ \boxed{0} \]