If the point `A_1, A_2,….,A_8` be the affixes of the roots of the equation `z^8-1=0` in the argand plane, then the area of the triangle `A_3 A_6 A_7` is equal to

Options

A. `sqrt2`
B. `sqrt3`
C. `(1)/(sqrt2)`
D. `(1)/(sqrt3)`

Answer

To find the area of the triangle formed by the points \( A_3, A_6, A_7 \) which are the affixes of the roots of the equation \( z^8 - 1 = 0 \), we can follow these steps:

### Step 1: Identify the Roots
The roots of the equation \( z^8 - 1 = 0 \) are given by:
\[ z_k = e^{i \frac{2k\pi}{8}} \quad \text{for } k = 0, 1, 2, \ldots, 7 \]
This simplifies to:
\[ z_k = e^{i \frac{k\pi}{4}} \quad \text{for } k = 0, 1, 2, \ldots, 7 \]

### Step 2: Calculate the Affixes
The affixes of the roots in the Argand plane are:
- \( A_1 = e^{i \cdot 0} = 1 \)
- \( A_2 = e^{i \frac{\pi}{4}} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \)
- \( A_3 = e^{i \frac{\pi}{2}} = i \)
- \( A_4 = e^{i \frac{3\pi}{4}} = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \)
- \( A_5 = e^{i \pi} = -1 \)
- \( A_6 = e^{i \frac{5\pi}{4}} = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \)
- \( A_7 = e^{i \frac{3\pi}{2}} = -i \)
- \( A_8 = e^{i \frac{7\pi}{4}} = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \)

### Step 3: Coordinates of the Points
The coordinates of the points \( A_3, A_6, A_7 \) are:
- \( A_3 = (0, 1) \)
- \( A_6 = \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \)
- \( A_7 = (0, -1) \)

### Step 4: Area of the Triangle
The area \( A \) of a triangle formed by points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
- \( (x_1, y_1) = (0, 1) \)
- \( (x_2, y_2) = \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) \)
- \( (x_3, y_3) = (0, -1) \)

We get:
\[ A = \frac{1}{2} \left| 0\left(-\frac{1}{\sqrt{2}} + 1\right) + \left(-\frac{1}{\sqrt{2}}\right)(-1 - 1) + 0(1 + \frac{1}{\sqrt{2}}) \right| \]
This simplifies to:
\[ A = \frac{1}{2} \left| -\frac{1}{\sqrt{2}} \cdot (-2) \right| = \frac{1}{2} \left| \frac{2}{\sqrt{2}} \right| = \frac{1}{2} \cdot \sqrt{2} = \frac{\sqrt{2}}{2} \]

### Final Answer
The area of triangle \( A_3 A_6 A_7 \) is:
\[ \frac{\sqrt{2}}{2} \]