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 values of `'theta'` is : (a) 6 (b) 7 (c) 8 (d) 9

Options

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

Answer

To solve the problem, we need to find the number of values of \( \theta \) for which \( z = re^{i\theta} \) is a root of the given polynomial equation:

\[ z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = 0 \]

### Step 1: Rewrite the equation
We can rewrite the polynomial equation as follows:

\[ z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 - z + 1 = 0 \]

### Step 2: Multiply and divide by \( z + 1 \)
To simplify the equation, we can multiply and divide by \( z + 1 \):

\[ \frac{z^8 - z^7 + z^6 - z^5 + z^4 - z^3 + z^2 - z + 1}{z + 1} = 0 \]

This is valid as long as \( z \neq -1 \).

### Step 3: Expand the numerator
Now, we can expand the numerator:

\[ (z^8 + 1) - (z^7 + z) + (z^6 + z^2) - (z^5 + z^3) + (z^4 - z^4) = 0 \]

This simplifies to:

\[ z^9 + 1 = 0 \]

### Step 4: Solve for \( z \)
From the equation \( z^9 + 1 = 0 \), we can rewrite it as:

\[ z^9 = -1 \]

### Step 5: Find the roots
The roots of \( z^9 = -1 \) can be expressed in exponential form as:

\[ z = e^{i(\pi + 2k\pi)/9} \quad \text{for } k = 0, 1, 2, \ldots, 8 \]

This gives us 9 distinct roots.

### Step 6: Exclude the root \( z = -1 \)
However, we need to exclude the root \( z = -1 \) (which corresponds to \( \theta = \pi \)), since we divided by \( z + 1 \). Therefore, we have:

\[ \text{Total roots} = 9 - 1 = 8 \]

### Conclusion
Thus, the number of values of \( \theta \) is \( 8 \).

The correct answer is (c) 8.

---