If x is the root of the equation `x^(2) -ix -1 =0` , then
The value of ` x^(51)` is

Options

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

Answer

To find the value of \( x^{51} \) where \( x \) is a root of the equation \( x^2 - ix - 1 = 0 \), we will follow these steps:

### Step 1: Solve the quadratic equation
The given quadratic equation is:
\[ x^2 - ix - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -i, c = -1 \):
\[ x = \frac{-(-i) \pm \sqrt{(-i)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{i \pm \sqrt{-1 - 4(-1)}}{2} \]
\[ x = \frac{i \pm \sqrt{-1 + 4}}{2} \]
\[ x = \frac{i \pm \sqrt{3}}{2} \]

### Step 2: Identify the roots
The roots of the equation are:
\[ x_1 = \frac{i + \sqrt{3}}{2}, \quad x_2 = \frac{i - \sqrt{3}}{2} \]

### Step 3: Convert to polar form
We can express \( x_1 \) in polar form. The modulus \( r \) and argument \( \theta \) can be calculated as follows:
\[ r = |x_1| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]
The argument \( \theta \) is given by:
\[ \theta = \tan^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
Thus, we can write:
\[ x_1 = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) \]

### Step 4: Calculate \( x^{51} \)
Using De Moivre's theorem:
\[ x^{51} = \left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right)^{51} = \cos\left(51 \cdot \frac{\pi}{6}\right) + i\sin\left(51 \cdot \frac{\pi}{6}\right) \]
Calculating \( 51 \cdot \frac{\pi}{6} \):
\[ 51 \cdot \frac{\pi}{6} = \frac{51\pi}{6} = \frac{50\pi}{6} + \frac{\pi}{6} = \frac{25\pi}{3} + \frac{\pi}{6} \]
Now, we can reduce \( \frac{25\pi}{3} \) modulo \( 2\pi \):
\[ \frac{25\pi}{3} = 8\pi + \frac{\pi}{3} \quad (\text{since } 8\pi = 2\pi \cdot 4) \]
Thus:
\[ \frac{25\pi}{3} \equiv \frac{\pi}{3} \mod 2\pi \]
Now adding \( \frac{\pi}{6} \):
\[ \frac{25\pi}{3} + \frac{\pi}{6} = \frac{50\pi}{6} + \frac{\pi}{6} = \frac{51\pi}{6} \]
Reducing \( \frac{51\pi}{6} \) modulo \( 2\pi \):
\[ \frac{51\pi}{6} = 8\pi + \frac{3\pi}{6} = 8\pi + \frac{\pi}{2} \equiv \frac{\pi}{2} \mod 2\pi \]

### Step 5: Final calculation
Thus:
\[ x^{51} = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) = 0 + i \cdot 1 = i \]

### Conclusion
The value of \( x^{51} \) is:
\[ \boxed{i} \]