Let `alpha` and `beta` be real number and z be a complex number. If `z^(2) + alpha z + beta = 0` has two distinct non-real roots with Re(z) =1, then it is necessary that

Options

A. `beta in ( -1,0)`
B. `| beta | =1`
C. ` beta in ( 1, oo)`
D. ` beta in ( 0,1)`

Answer

To solve the problem, we need to analyze the quadratic equation given by:

\[ z^2 + \alpha z + \beta = 0 \]

where \( \alpha \) and \( \beta \) are real numbers, and \( z \) is a complex number. We are given that this equation has two distinct non-real roots with the real part of \( z \) equal to 1.

### Step 1: Use the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:

\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case, \( a = 1 \), \( b = \alpha \), and \( c = \beta \). Therefore, the roots are:

\[ z = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\beta}}{2} \]

### Step 2: Determine the real part of the roots
The real part of the roots is given by:

\[ \text{Re}(z) = \frac{-\alpha}{2} \]

We know from the problem statement that the real part of \( z \) is equal to 1. Therefore, we set up the equation:

\[ \frac{-\alpha}{2} = 1 \]

### Step 3: Solve for \( \alpha \)
Multiplying both sides by -2 gives us:

\[ \alpha = -2 \]

### Step 4: Analyze the discriminant for distinct non-real roots
For the roots to be distinct and non-real, the discriminant must be less than zero:

\[ \alpha^2 - 4\beta < 0 \]

Substituting \( \alpha = -2 \) into the discriminant condition:

\[ (-2)^2 - 4\beta < 0 \]

This simplifies to:

\[ 4 - 4\beta < 0 \]

### Step 5: Solve the inequality
Rearranging the inequality gives:

\[ 4 < 4\beta \]

Dividing both sides by 4 results in:

\[ 1 < \beta \]

or equivalently:

\[ \beta > 1 \]

### Conclusion
Thus, we conclude that for the quadratic equation \( z^2 + \alpha z + \beta = 0 \) to have two distinct non-real roots with the real part equal to 1, it is necessary that:

\[ \beta \in (1, \infty) \]

### Final Answer
The correct option is \( \beta \) belongs to the set \( (1, \infty) \).

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