If all the roots of `z^3+az^2+bz+c=0` are of unit modulus, then (A) `|a|le3` (B) `|b|le3` (C) `|c|=1` (D) none of these

Options

A. `|a| le 3`
B. `|b | le 3`
C. `|c | =1`
D. All of the above

Answer

To solve the problem, we need to analyze the polynomial \( z^3 + az^2 + bz + c = 0 \) under the condition that all its roots \( z_1, z_2, z_3 \) have unit modulus, meaning \( |z_1| = |z_2| = |z_3| = 1 \).

### Step 1: Sum of Roots
Using Vieta's formulas, we know:
\[ z_1 + z_2 + z_3 = -a \]
Taking the modulus, we have:
\[ |z_1 + z_2 + z_3| = | -a | = |a| \]
Since \( |z_1| = |z_2| = |z_3| = 1 \), we can apply the triangle inequality:
\[ |z_1 + z_2 + z_3| \leq |z_1| + |z_2| + |z_3| = 1 + 1 + 1 = 3 \]
Thus, we conclude:
\[ |a| \leq 3 \]

### Step 2: Sum of Products of Roots
Next, we consider the sum of the products of the roots taken two at a time:
\[ z_1 z_2 + z_2 z_3 + z_3 z_1 = b \]
Taking the modulus, we have:
\[ |z_1 z_2 + z_2 z_3 + z_3 z_1| = |b| \]
Again applying the triangle inequality:
\[ |z_1 z_2 + z_2 z_3 + z_3 z_1| \leq |z_1 z_2| + |z_2 z_3| + |z_3 z_1| = 1 + 1 + 1 = 3 \]
Thus, we conclude:
\[ |b| \leq 3 \]

### Step 3: Product of Roots
Finally, we consider the product of the roots:
\[ z_1 z_2 z_3 = -c \]
Taking the modulus, we have:
\[ |z_1 z_2 z_3| = | -c | = |c| \]
Since \( |z_1| = |z_2| = |z_3| = 1 \):
\[ |z_1 z_2 z_3| = |z_1| \cdot |z_2| \cdot |z_3| = 1 \cdot 1 \cdot 1 = 1 \]
Thus, we conclude:
\[ |c| = 1 \]

### Conclusion
From the analysis, we have:
- \( |a| \leq 3 \) (Option A is correct)
- \( |b| \leq 3 \) (Option B is correct)
- \( |c| = 1 \) (Option C is correct)

Thus, the correct options are A, B, and C.