If the cubic equation `z^(3)+az^(2)+bz+c=0 AA a, b, c in R, c ne0` has a purely imaginary root, then (where `i^(2)=-1`)

Options

A. `c=ab`
B. `b=ac`
C. the imaginary root is equal to `pmic`
D. the imaginary root is equal to `pmia`

Answer

To solve the problem, we need to analyze the cubic equation \( z^3 + az^2 + bz + c = 0 \) under the condition that it has a purely imaginary root. Let's denote this root as \( z = i\alpha \), where \( \alpha \) is a real number.

### Step-by-Step Solution:

1. **Substituting the Imaginary Root**:
We substitute \( z = i\alpha \) into the cubic equation:
\[ (i\alpha)^3 + a(i\alpha)^2 + b(i\alpha) + c = 0 \]

2. **Calculating Each Term**:
We calculate each term:
- \( (i\alpha)^3 = i^3 \alpha^3 = -i\alpha^3 \) (since \( i^3 = -i \))
- \( a(i\alpha)^2 = a(i^2 \alpha^2) = a(-\alpha^2) = -a\alpha^2 \)
- \( b(i\alpha) = bi\alpha \)
- The constant term remains \( c \).

Thus, the equation becomes:
\[ -i\alpha^3 - a\alpha^2 + bi\alpha + c = 0 \]

3. **Separating Real and Imaginary Parts**:
We can separate the real and imaginary parts of the equation:
- Real part: \( -a\alpha^2 + c = 0 \)
- Imaginary part: \( -\alpha^3 + b\alpha = 0 \)

4. **Setting Up the Equations**:
From the real part:
\[ c = a\alpha^2 \quad \text{(1)} \]
From the imaginary part:
\[ -\alpha^3 + b\alpha = 0 \quad \Rightarrow \quad \alpha(-\alpha^2 + b) = 0 \]
Since \( \alpha \neq 0 \) (as the root is purely imaginary), we have:
\[ -\alpha^2 + b = 0 \quad \Rightarrow \quad b = \alpha^2 \quad \text{(2)} \]

5. **Substituting Back**:
Now, substituting equation (2) into equation (1):
\[ c = a\alpha^2 = ab \]

6. **Conclusion**:
Therefore, we have established the relationship:
\[ c = ab \]
This means that if the cubic equation has a purely imaginary root, then the relationship between the coefficients is \( c = ab \).