For a complex number Z, if all the roots of the equation `Z^3 + aZ^2 + bZ + c = 0` are unimodular, 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 are unimodular (i.e., they lie on the unit circle in the complex plane).

### Step-by-step Solution:

1. **Understanding Unimodular Roots**:
- Let the roots of the polynomial be \( Z_1, Z_2, Z_3 \).
- Since they are unimodular, we have:
\[ |Z_1| = |Z_2| = |Z_3| = 1 \]

2. **Using Vieta's Formulas**:
- According to Vieta's formulas for a cubic polynomial \( Z^3 + aZ^2 + bZ + c = 0 \):
- The sum of the roots \( Z_1 + Z_2 + Z_3 = -a \)
- The sum of the products of the roots taken two at a time \( Z_1Z_2 + Z_2Z_3 + Z_3Z_1 = b \)
- The product of the roots \( Z_1Z_2Z_3 = -c \)

3. **Finding the Modulus of the Sum of Roots**:
- Since \( |Z_1| = |Z_2| = |Z_3| = 1 \), we can use the triangle inequality:
\[ |Z_1 + Z_2 + Z_3| \leq |Z_1| + |Z_2| + |Z_3| = 1 + 1 + 1 = 3 \]
- Therefore, we have:
\[ |-a| \leq 3 \implies |a| \leq 3 \]

4. **Finding the Modulus of the Product of Roots**:
- The product of the roots is given by \( Z_1Z_2Z_3 = -c \).
- Since \( |Z_1| = |Z_2| = |Z_3| = 1 \), we have:
\[ |Z_1Z_2Z_3| = |Z_1| \cdot |Z_2| \cdot |Z_3| = 1 \cdot 1 \cdot 1 = 1 \]
- Thus, we find:
\[ |-c| = 1 \implies |c| = 1 \]

5. **Conclusion**:
- From our analysis, we conclude:
- \( |a| \leq 3 \)
- \( |c| = 1 \)

### Final Answer:
The conditions derived from the problem state that if all roots of the polynomial are unimodular, then:
- \( |a| \leq 3 \)
- \( |c| = 1 \)