Understanding the Problem: Quadratic Equations with Complex Roots

The question asks for the condition on the real coefficient \(\beta\) for the quadratic equation \(z^2 + \alpha z + \beta = 0\) to have two distinct non-real roots, given that the real part of these roots is 1.

A fundamental property of quadratic equations with real coefficients is that if they have non-real roots, these roots must occur in complex conjugate pairs. Let the roots be \(z_1\) and \(z_2\).

Properties of Complex Conjugate Roots

Since the coefficients \(\alpha\) and \(\beta\) are real, and the roots are non-real and distinct, the roots must be of the form \(z_1 = x + iy\) and \(z_2 = x - iy\), where \(x\) and \(y\) are real numbers and \(y \neq 0\) (for distinct non-real roots). The question states that the real part of the roots is 1, meaning \(x = 1\).

So, the two distinct non-real roots are \(z_1 = 1 + iy\) and \(z_2 = 1 - iy\), with the condition that \(y \neq 0\).

Applying Vieta's Formulas

For a quadratic equation of the form \(az^2 + bz + c = 0\), Vieta's formulas give the relationships between the roots and the coefficients:

• Sum of roots: \(z_1 + z_2 = -\frac{b}{a}\)
• Product of roots: \(z_1 z_2 = \frac{c}{a}\)

In our equation, \(z^2 + \alpha z + \beta = 0\), we have \(a=1\), \(b=\alpha\), and \(c=\beta\).

Using the roots \(z_1 = 1 + iy\) and \(z_2 = 1 - iy\):

• Sum of roots: \((1 + iy) + (1 - iy) = 2 = -\frac{\alpha}{1} = -\alpha\). This implies \(\alpha = -2\). Since \(\alpha\) is given as a real number, this is consistent.
• Product of roots: \((1 + iy)(1 - iy) = \beta\). Let's calculate the product:
  \((1 + iy)(1 - iy) = 1^2 - (iy)^2 = 1 - i^2 y^2\)
  Since \(i^2 = -1\), this becomes \(1 - (-1)y^2 = 1 + y^2\).
  So, \(\beta = 1 + y^2\).

Determining the Condition for β

We found that \(\beta = 1 + y^2\), where \(y\) is the imaginary part of the roots. The problem states that the roots are non-real and distinct, which requires that the imaginary part \(y\) is non-zero (\(y \neq 0\)).

If \(y \neq 0\), then \(y^2\) must be a positive real number.
Mathematically, \(y^2 > 0\) for any real \(y \neq 0\).

Substituting this into the expression for \(\beta\):
\(\beta = 1 + y^2\)
Since \(y^2 > 0\), we have \(1 + y^2 > 1 + 0\).
Therefore, \(\beta > 1\).

In interval notation, the condition \(\beta > 1\) is expressed as \(\beta \varepsilon (1, \infty)\).

Comparing with Options

Let's examine the given options based on our finding that \(\beta \varepsilon (1, \infty)\):

The necessary condition for \(\beta\) is that \(\beta\) must be strictly greater than 1.

Revision Table: Quadratic Roots and Coefficients

Additional Information: Discriminant and Complex Roots

Alternatively, we could also use the discriminant. For \(z^2 + \alpha z + \beta = 0\), the discriminant is \(\Delta = \alpha^2 - 4\beta\).

For distinct non-real roots, the discriminant must be negative: \(\alpha^2 - 4\beta < 0\).

We found that \(\alpha = -2\) from the sum of roots. Substituting this into the discriminant condition:

\((-2)^2 - 4\beta < 0\)
\(4 - 4\beta < 0\)
\(4 < 4\beta\)
Dividing by 4 (which is positive, so the inequality direction doesn't change):
\(1 < \beta\)

This confirms our previous finding that \(\beta > 1\), or \(\beta \varepsilon (1, \infty)\).

This approach using the discriminant provides an alternative way to reach the same conclusion regarding the necessary condition for \(\beta\).