The equation `az^(3) + bz^(2) + barbz + bara = 0` has a root `alpha`, where a, b,z and `alpha` belong to the set of complex numbers. The number value of `|alpha|`

Options

A. is 1/2
B. is 1
C. is 2
D. can't be determined

Explanation

To solve the equation \( az^3 + bz^2 + \overline{b}z + \overline{a} = 0 \) and find the modulus of the root \( \alpha \), we can follow these steps:

### Step 1: Write the conjugate of the equation
We start with the given equation:
\[ az^3 + bz^2 + \overline{b}z + \overline{a} = 0 \]
Taking the complex conjugate of the entire equation gives:
\[ \overline{a} \overline{z^3} + \overline{b} \overline{z^2} + b \overline{z} + a = 0 \]
This can be rewritten as:
\[ \overline{a} \overline{z}^3 + \overline{b} \overline{z}^2 + b \overline{z} + a = 0 \]

### Step 2: Substitute \( z = \frac{1}{\overline{z}} \)
To relate the original equation to its conjugate, we can substitute \( z \) with \( \frac{1}{\overline{z}} \):
\[ \overline{a} \left(\frac{1}{\overline{z}}\right)^3 + \overline{b} \left(\frac{1}{\overline{z}}\right)^2 + b \left(\frac{1}{\overline{z}}\right) + a = 0 \]
Multiplying through by \( \overline{z}^3 \) gives:
\[ \overline{a} + \overline{b} \overline{z} + b \overline{z}^2 + a \overline{z}^3 = 0 \]

### Step 3: Rearranging the equation
We can rearrange the terms to match the original equation:
\[ a \overline{z}^3 + b \overline{z}^2 + \overline{b} \overline{z} + \overline{a} = 0 \]

### Step 4: Identify the relationship
From the two equations:
1. \( az^3 + bz^2 + \overline{b}z + \overline{a} = 0 \)
2. \( a \overline{z}^3 + b \overline{z}^2 + \overline{b} \overline{z} + \overline{a} = 0 \)

We can see that they are structurally identical. This implies:
\[ z^3 = \frac{1}{\overline{z}^3} \]
This leads to:
\[ |z|^6 = 1 \]

### Step 5: Find the modulus of \( z \)
Taking the sixth root gives:
\[ |z| = 1 \]

### Step 6: Conclude for \( \alpha \)
Since \( \alpha \) is a root of the original equation and \( |z| = 1 \), we conclude:
\[ |\alpha| = 1 \]

### Final Answer
The number value of \( |\alpha| \) is \( 1 \).

---