Consider the quadratic equation `az^(2)+bz+c=0` where `a,b,c` are non-zero complex numbers. Now answer the following.
The condition that the equation has both roots purely imaginary is

Options

A. `(a)/(a)=(b)/(b)=(c)/(c)`
B. `(a)/(a)=(b)/(b)=(c)/(c)`
C. `(a)/(a)=(b)/(b)=-(c)/(c)`
D. `(a)/(a)=-(b)/(b)=(c)/(c)`

Answer

To determine the condition under which the quadratic equation \( az^2 + bz + c = 0 \) has both roots purely imaginary, we can follow these steps:

### Step 1: Assume the Roots
Let the roots of the quadratic equation be \( z_1 \) and \( z_2 \). Since we want both roots to be purely imaginary, we can express them as:
\[ z_1 = iy \quad \text{and} \quad z_2 = -iy \]
where \( y \) is a real number.

### Step 2: Use Vieta's Formulas
According to Vieta's formulas, for the quadratic equation \( az^2 + bz + c = 0 \):
- The sum of the roots \( z_1 + z_2 = -\frac{b}{a} \)
- The product of the roots \( z_1 z_2 = \frac{c}{a} \)

Substituting the values of \( z_1 \) and \( z_2 \):
\[ z_1 + z_2 = iy + (-iy) = 0 \]
This implies:
\[ -\frac{b}{a} = 0 \quad \Rightarrow \quad b = 0 \]
However, since \( b \) is given as a non-zero complex number, we need to explore the product of the roots.

### Step 3: Calculate the Product of the Roots
The product of the roots is:
\[ z_1 z_2 = (iy)(-iy) = -i^2y^2 = y^2 \]
According to Vieta's formulas:
\[ z_1 z_2 = \frac{c}{a} \]
Thus, we have:
\[ y^2 = \frac{c}{a} \]

### Step 4: Analyze the Conditions
Since \( y^2 \) is a real number (as \( y \) is real), the right-hand side \( \frac{c}{a} \) must also be a real number. Therefore, we can conclude that for the equation to have both roots purely imaginary, the condition is:
\[ \frac{c}{a} \text{ is real} \]

### Step 5: Final Condition
To express this condition algebraically, we can say:
\[ c \overline{a} = \overline{c} a \]
This means that the product of \( c \) and the conjugate of \( a \) must equal the product of \( \overline{c} \) and \( a \).

### Summary
The condition that the quadratic equation \( az^2 + bz + c = 0 \) has both roots purely imaginary is:
\[ c \overline{a} = \overline{c} a \]

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