If `Z+1/Z =1, Z^64 +1/(Z^(64))` is equal to

Options

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

Answer

To solve the equation \( Z + \frac{1}{Z} = 1 \) and find the value of \( Z^{64} + \frac{1}{Z^{64}} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ Z + \frac{1}{Z} = 1 \]

### Step 2: Multiply both sides by \( Z \)
To eliminate the fraction, we multiply both sides by \( Z \):
\[ Z^2 + 1 = Z \]

### Step 3: Rearrange the equation
Rearranging gives us a standard quadratic equation:
\[ Z^2 - Z + 1 = 0 \]

### Step 4: Use the quadratic formula
We can apply the quadratic formula \( Z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = 1 \):
\[ Z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ Z = \frac{1 \pm \sqrt{1 - 4}}{2} \]
\[ Z = \frac{1 \pm \sqrt{-3}}{2} \]
\[ Z = \frac{1 \pm i\sqrt{3}}{2} \]

### Step 5: Find \( Z^{64} + \frac{1}{Z^{64}} \)
Let \( Z = \frac{1 + i\sqrt{3}}{2} \) and \( \frac{1}{Z} = \frac{1 - i\sqrt{3}}{2} \). We can use the property of powers of complex numbers.

Using De Moivre's theorem, we can express \( Z \) in polar form. The modulus \( r \) is:
\[ r = |Z| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]
The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}/2}{1/2}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Thus, we can express \( Z \) as:
\[ Z = e^{i\frac{\pi}{3}} \]
Then, \( Z^{64} \) can be calculated as:
\[ Z^{64} = \left(e^{i\frac{\pi}{3}}\right)^{64} = e^{i\frac{64\pi}{3}} = e^{i(21\pi + \frac{1\pi}{3})} = e^{i\frac{\pi}{3}} \quad (\text{since } e^{i2\pi k} = 1) \]

### Step 6: Calculate \( Z^{64} + \frac{1}{Z^{64}} \)
Now we find:
\[ Z^{64} + \frac{1}{Z^{64}} = e^{i\frac{\pi}{3}} + e^{-i\frac{\pi}{3}} = 2 \cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1 \]

### Final Answer
Thus, the value of \( Z^{64} + \frac{1}{Z^{64}} \) is:
\[ \boxed{1} \]