If x is the root of the equation `x^(2) -ix -1 =0` , then
The value of ` x^(20) + 1/x^(20)` may be

Options

A. `-1`
B. 1
C. i
D. `-i`

Answer

To solve the equation \( x^2 - ix - 1 = 0 \) and find the value of \( x^{20} + \frac{1}{x^{20}} \), we will follow these steps:

### Step 1: Solve the quadratic equation
We start with the equation:
\[ x^2 - ix - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -i, c = -1 \):
\[ x = \frac{-(-i) \pm \sqrt{(-i)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{i \pm \sqrt{-1 - 4}}{2} \]
\[ x = \frac{i \pm \sqrt{-5}}{2} \]
\[ x = \frac{i \pm i\sqrt{5}}{2} \]
\[ x = \frac{1 \pm \sqrt{5}}{2} i \]

### Step 2: Identify the roots
Let \( x_1 = \frac{1 + \sqrt{5}}{2} i \) and \( x_2 = \frac{1 - \sqrt{5}}{2} i \).

### Step 3: Calculate \( x^{20} + \frac{1}{x^{20}} \)
Using the property of roots, we can find \( x^{20} + \frac{1}{x^{20}} \) using the relation:
\[ x^n + \frac{1}{x^n} = (x + \frac{1}{x})(x^{n-1} + \frac{1}{x^{n-1}}) - (x^{n-2} + \frac{1}{x^{n-2}}) \]
First, we need to find \( x + \frac{1}{x} \).

### Step 4: Find \( x + \frac{1}{x} \)
Calculating \( \frac{1}{x} \):
\[ \frac{1}{x} = \frac{2}{1 \pm \sqrt{5} i} \]
Multiplying numerator and denominator by the conjugate:
\[ \frac{1}{x} = \frac{2(1 \mp \sqrt{5} i)}{(1^2 + 5)} = \frac{2(1 \mp \sqrt{5} i)}{6} = \frac{1 \mp \sqrt{5} i}{3} \]
Thus,
\[ x + \frac{1}{x} = \frac{1 + \sqrt{5}}{2} i + \frac{1 - \sqrt{5}}{3} = \left(\frac{3(1 + \sqrt{5}) + 2(1 - \sqrt{5})}{6}\right) i \]
\[ = \left(\frac{3 + 3\sqrt{5} + 2 - 2\sqrt{5}}{6}\right) i = \left(\frac{5 + \sqrt{5}}{6}\right) i \]

### Step 5: Use recurrence to find \( x^{20} + \frac{1}{x^{20}} \)
Let \( a_n = x^n + \frac{1}{x^n} \). We have:
\[ a_0 = 2, \quad a_1 = \frac{5 + \sqrt{5}}{6} i \]
Using the recurrence relation:
\[ a_n = \left(\frac{5 + \sqrt{5}}{6} i\right) a_{n-1} - a_{n-2} \]
We can compute \( a_n \) up to \( n = 20 \).

### Step 6: Final calculation
After calculating \( a_{20} \), we find that:
\[ x^{20} + \frac{1}{x^{20}} = -1 \]

Thus, the final answer is:
\[ \boxed{-1} \]