If z is a complex number satisfying the equation `z^6 +z^3 + 1 = 0`. If this equation has a root `re^(itheta)` with `90^@<0<180^@` then the value of `theta` is

Options

A. `100^(@)`
B. `110^(@)`
C. `160^(@)`
D. `170^(@)`

Explanation

To solve the equation \( z^6 + z^3 + 1 = 0 \) for a complex number \( z \) in the form \( re^{i\theta} \) where \( 90^\circ < \theta < 180^\circ \), we can follow these steps:

### Step 1: Substitute \( t = z^3 \)
We start by substituting \( t = z^3 \) into the equation. This transforms the original equation into a quadratic equation:
\[ t^2 + t + 1 = 0 \]

### Step 2: Solve the quadratic equation
We can solve the quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \). Plugging in these values gives:
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ t = \frac{-1 \pm i\sqrt{3}}{2} \]

### Step 3: Identify the roots
The two roots for \( t \) are:
\[ t_1 = \frac{-1 + i\sqrt{3}}{2}, \quad t_2 = \frac{-1 - i\sqrt{3}}{2} \]

### Step 4: Convert \( t \) to polar form
We will convert \( t_1 \) to polar form. The modulus \( r \) is calculated as:
\[ r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]
Next, we find the argument \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \tan^{-1}(-\sqrt{3}) = 120^\circ \quad (\text{since } t_1 \text{ is in the second quadrant}) \]
Thus, we can express \( t_1 \) as:
\[ t_1 = e^{i \cdot 120^\circ} \]

### Step 5: Find \( z \)
Since \( t = z^3 \), we have:
\[ z^3 = e^{i \cdot 120^\circ} \]
To find \( z \), we take the cube root:
\[ z = e^{i \cdot \frac{120^\circ + 360^\circ n}{3}} \quad (n \in \mathbb{Z}) \]
This simplifies to:
\[ z = e^{i \cdot (40^\circ + 120^\circ n)} \]

### Step 6: Determine suitable \( n \)
We need \( \theta \) such that \( 90^\circ < \theta < 180^\circ \). We can test values of \( n \):
- For \( n = 0 \):
\[ \theta = 40^\circ \]
- For \( n = 1 \):
\[ \theta = 40^\circ + 120^\circ = 160^\circ \]
- For \( n = 2 \):
\[ \theta = 40^\circ + 240^\circ = 280^\circ \quad (\text{not valid as it exceeds } 180^\circ) \]

### Conclusion
The only valid solution for \( \theta \) that satisfies \( 90^\circ < \theta < 180^\circ \) is:
\[ \theta = 160^\circ \]

### Final Answer
The value of \( \theta \) is \( 160^\circ \).