z is a complex number such that `z+1/z=2cos 3^0` ,then the value of `z^(2000)+ 1/z^(2000)` is equal to

Options

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

Answer

To solve the problem, we are given that \( z + \frac{1}{z} = 2 \cos 3^\circ \). We need to find the value of \( z^{2000} + \frac{1}{z^{2000}} \).

### Step 1: Start with the given equation
We have:
\[ z + \frac{1}{z} = 2 \cos 3^\circ \]

### Step 2: Square both sides
Squaring both sides gives:
\[ \left(z + \frac{1}{z}\right)^2 = (2 \cos 3^\circ)^2 \]
Expanding the left side:
\[ z^2 + 2 + \frac{1}{z^2} = 4 \cos^2 3^\circ \]
Rearranging gives:
\[ z^2 + \frac{1}{z^2} = 4 \cos^2 3^\circ - 2 \]

### Step 3: Use the double angle formula
Using the double angle formula for cosine, we know:
\[ 4 \cos^2 3^\circ - 2 = 2 \cos 6^\circ \]
Thus, we have:
\[ z^2 + \frac{1}{z^2} = 2 \cos 6^\circ \]

### Step 4: Cube both sides
Next, we will find \( z^3 + \frac{1}{z^3} \). We use the identity:
\[ z^3 + \frac{1}{z^3} = \left(z + \frac{1}{z}\right)\left(z^2 + \frac{1}{z^2}\right) - \left(z + \frac{1}{z}\right) \]
Substituting the known values:
\[ z^3 + \frac{1}{z^3} = (2 \cos 3^\circ)(2 \cos 6^\circ) - 2 \cos 3^\circ \]
This simplifies to:
\[ z^3 + \frac{1}{z^3} = 2 \cos 3^\circ (2 \cos 6^\circ - 1) \]

### Step 5: Use the triple angle formula
Using the triple angle formula for cosine:
\[ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \]
We can express \( z^3 + \frac{1}{z^3} \) in terms of \( \cos 9^\circ \):
\[ z^3 + \frac{1}{z^3} = 2 \cos 9^\circ \]

### Step 6: Generalize for \( z^n + \frac{1}{z^n} \)
We can see a pattern forming:
- \( z + \frac{1}{z} = 2 \cos 3^\circ \)
- \( z^2 + \frac{1}{z^2} = 2 \cos 6^\circ \)
- \( z^3 + \frac{1}{z^3} = 2 \cos 9^\circ \)

From this, we can generalize:
\[ z^n + \frac{1}{z^n} = 2 \cos(3n^\circ) \]

### Step 7: Calculate \( z^{2000} + \frac{1}{z^{2000}} \)
Now substituting \( n = 2000 \):
\[ z^{2000} + \frac{1}{z^{2000}} = 2 \cos(6000^\circ) \]

### Step 8: Simplify \( \cos(6000^\circ) \)
To simplify \( \cos(6000^\circ) \), we can reduce it:
\[ 6000 \div 360 = 16.6667 \quad \text{(which means 16 complete cycles and a remainder)} \]
Calculating the remainder:
\[ 6000 - 360 \times 16 = 240^\circ \]
Thus:
\[ \cos(6000^\circ) = \cos(240^\circ) = -\frac{1}{2} \]

### Final Step: Substitute back
Finally, substituting back:
\[ z^{2000} + \frac{1}{z^{2000}} = 2 \left(-\frac{1}{2}\right) = -1 \]

### Conclusion
The value of \( z^{2000} + \frac{1}{z^{2000}} \) is:
\[ \boxed{-1} \]