For a complex number Z, if all the roots of the equation `Z^3 + aZ^2 + bZ + c = 0` are unit modulus, then

Options

A. `|a| gt 3 and |c| = 1`
B. `|a | le 3 and |c| = 3`
C. `|a| gt 3 and |c| = 1/3`
D. `|a| le 3 & |c| = 1`

Answer

To solve the problem, we need to analyze the properties of the roots of the polynomial equation \( Z^3 + aZ^2 + bZ + c = 0 \) given that all roots have unit modulus.

### Step 1: Define the roots
Let the roots of the polynomial be \( \alpha, \beta, \gamma \). Since the roots are of unit modulus, we have:
\[ |\alpha| = |\beta| = |\gamma| = 1 \]

### Step 2: Apply Vieta's Formulas
According to Vieta's formulas, we know:
1. \( \alpha + \beta + \gamma = -a \)
2. \( \alpha\beta + \beta\gamma + \gamma\alpha = b \)
3. \( \alpha\beta\gamma = -c \)

### Step 3: Analyze the modulus of the sum of the roots
Using the property of the modulus, we have:
\[ |\alpha + \beta + \gamma| \leq |\alpha| + |\beta| + |\gamma| = 1 + 1 + 1 = 3 \]
Thus, we can write:
\[ |-a| \leq 3 \implies |a| \leq 3 \]

### Step 4: Analyze the modulus of the product of the roots
From Vieta's formulas, we also know:
\[ |\alpha\beta\gamma| = |\alpha| \cdot |\beta| \cdot |\gamma| = 1 \cdot 1 \cdot 1 = 1 \]
Thus, we can write:
\[ |-c| = |c| = 1 \]

### Conclusion
From the analysis, we have derived two important conditions:
1. \( |a| \leq 3 \)
2. \( |c| = 1 \)

### Final Answer
The correct conditions derived from the problem statement are:
- \( |a| \leq 3 \)
- \( |c| = 1 \)