Let `alpha and beta` be real numbers and z be a complex number. If `z^(2)+alphaz+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. `betain(0,1)`

Explanation

To solve the problem step by step, we need to analyze the given quadratic equation \( z^2 + \alpha z + \beta = 0 \) under the conditions specified.

### Step 1: Set up the complex number
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Since we know that the real part of \( z \) is 1, we have:
\[ x = 1 \]
Thus, we can rewrite \( z \) as:
\[ z = 1 + iy \]

### Step 2: Substitute \( z \) into the equation
Substituting \( z \) into the quadratic equation gives:
\[ (1 + iy)^2 + \alpha(1 + iy) + \beta = 0 \]
Expanding this, we get:
\[ 1^2 + 2(1)(iy) + (iy)^2 + \alpha + \alpha(iy) + \beta = 0 \]
This simplifies to:
\[ 1 + 2iy - y^2 + \alpha + \alpha iy + \beta = 0 \]
Combining like terms, we separate the real and imaginary parts:
\[ (1 - y^2 + \alpha + \beta) + i(2y + \alpha y) = 0 \]

### Step 3: Set real and imaginary parts to zero
For the equation to hold, both the real part and the imaginary part must be equal to zero:
1. Real part:
\[ 1 - y^2 + \alpha + \beta = 0 \quad \text{(1)} \]
2. Imaginary part:
\[ 2y + \alpha y = 0 \quad \text{(2)} \]

### Step 4: Solve the imaginary part equation
From equation (2):
\[ y(2 + \alpha) = 0 \]
This gives us two cases:
1. \( y = 0 \) (which would imply \( z \) is real, contradicting the problem statement).
2. \( 2 + \alpha = 0 \) leading to:
\[ \alpha = -2 \]

### Step 5: Substitute \( \alpha \) back into the real part equation
Substituting \( \alpha = -2 \) into equation (1):
\[ 1 - y^2 - 2 + \beta = 0 \]
This simplifies to:
\[ \beta - y^2 - 1 = 0 \]
Thus, we have:
\[ \beta = 1 + y^2 \quad \text{(3)} \]

### Step 6: Analyze the value of \( \beta \)
Since \( y^2 \) is always non-negative (\( y^2 \geq 0 \)), we can conclude:
\[ \beta \geq 1 \]
This means that \( \beta \) must be greater than or equal to 1.

### Conclusion
The necessary condition for \( \beta \) is:
\[ \beta \geq 1 \]