If `z=re^(itheta)` ( r gt 0 & `0 le theta lt 2pi`) is a root of the equation `z^8-z^7+z^6-z^5+z^4-z^3+z^2 -z + 1=0` then number of value of `'theta'` is :

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the equation \( z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = 0 \) where \( z = re^{i\theta} \) (with \( r > 0 \) and \( 0 \leq \theta < 2\pi \)), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given polynomial equation:
\[ z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = 0 \]

### Step 2: Factor the polynomial
Notice that we can factor the polynomial by multiplying and dividing by \( z + 1 \):
\[ (z^9 - 1) = (z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 - z + 1)(z + 1) \]
This means:
\[ z^9 - 1 = 0 \quad \text{for } z \neq -1 \]

### Step 3: Find the roots of \( z^9 - 1 = 0 \)
The roots of the equation \( z^9 - 1 = 0 \) are the 9th roots of unity:
\[ z_k = e^{i\frac{2\pi k}{9}} \quad \text{for } k = 0, 1, 2, \ldots, 8 \]

### Step 4: Exclude the root \( z = -1 \)
The root \( z = -1 \) corresponds to \( k = 4 \) (since \( e^{i\pi} = -1 \)). Since we are excluding this root, we only consider the remaining roots.

### Step 5: Count the valid values of \( \theta \)
The valid roots from \( z^9 - 1 = 0 \) that we keep are:
- \( z_0 = e^{i\frac{2\pi \cdot 0}{9}} \)
- \( z_1 = e^{i\frac{2\pi \cdot 1}{9}} \)
- \( z_2 = e^{i\frac{2\pi \cdot 2}{9}} \)
- \( z_3 = e^{i\frac{2\pi \cdot 3}{9}} \)
- \( z_5 = e^{i\frac{2\pi \cdot 5}{9}} \)
- \( z_6 = e^{i\frac{2\pi \cdot 6}{9}} \)
- \( z_7 = e^{i\frac{2\pi \cdot 7}{9}} \)
- \( z_8 = e^{i\frac{2\pi \cdot 8}{9}} \)

This gives us a total of 8 valid roots.

### Conclusion
Thus, the number of values of \( \theta \) is:
\[ \text{Number of values of } \theta = 8 \]

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