If x is the root of the equation `x^(2) -ix -1 =0` , then
`x^(2013) - 1/x^(2013)` may be

Options

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

Answer

To solve the problem, we need to find the value of \( x^{2013} - \frac{1}{x^{2013}} \) given that \( x \) is a root of the equation \( x^2 - ix - 1 = 0 \).

### Step 1: Find the roots of the equation
The equation is given as:
\[ x^2 - ix - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -i, c = -1 \):
\[ x = \frac{-(-i) \pm \sqrt{(-i)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
Calculating \( (-i)^2 \):
\[ (-i)^2 = -1 \]
Thus, we have:
\[ x = \frac{i \pm \sqrt{-1 + 4}}{2} = \frac{i \pm \sqrt{3}}{2} \]
This gives us the roots:
\[ x = \frac{\sqrt{3}}{2} + \frac{1}{2}i \quad \text{or} \quad x = \frac{\sqrt{3}}{2} - \frac{1}{2}i \]

### Step 2: Express \( x \) in polar form
The root \( x = \frac{\sqrt{3}}{2} + \frac{1}{2}i \) can be expressed in polar form. The modulus \( r \) is given by:
\[ r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]
The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
Thus, we can write:
\[ x = e^{i\frac{\pi}{6}} \]

### Step 3: Calculate \( x^{2013} \)
Using the polar form:
\[ x^{2013} = \left(e^{i\frac{\pi}{6}}\right)^{2013} = e^{i\frac{2013\pi}{6}} = e^{i335.5\pi} \]
To simplify \( 335.5\pi \):
\[ 335.5 \mod 2 = 1.5 \quad \text{(since \( 335.5 = 167 \cdot 2 + 1.5 \))} \]
Thus:
\[ x^{2013} = e^{i\frac{3\pi}{2}} = -i \]

### Step 4: Calculate \( \frac{1}{x^{2013}} \)
Now, we find:
\[ \frac{1}{x^{2013}} = \frac{1}{-i} = i \]

### Step 5: Find \( x^{2013} - \frac{1}{x^{2013}} \)
Now we compute:
\[ x^{2013} - \frac{1}{x^{2013}} = -i - i = -2i \]

### Final Answer
Thus, the value of \( x^{2013} - \frac{1}{x^{2013}} \) is:
\[ \boxed{-2i} \]