Key Concepts and Formulas

• Quadratic Equations with Real Coefficients: If a quadratic equation $az^2 + bz + c = 0$ has real coefficients $a, b, c$, then its non-real roots occur in conjugate pairs.
• Vieta's Formulas: For a quadratic equation $z^2 + \alpha z + \beta = 0$, the sum of the roots $z_1 + z_2 = -\alpha$ and the product of the roots $z_1 z_2 = \beta$.
• Complex Conjugates: If $z = a + bi$, then its conjugate is $\bar{z} = a - bi$. Also, $z\bar{z} = a^2 + b^2$.

Step-by-Step Solution

Step 1: Define the Roots and Apply the Given Condition

Since the quadratic equation $z^2 + \alpha z + \beta = 0$ has real coefficients $\alpha$ and $\beta$, and it has two distinct roots with $Re(z) = 1$, these roots must be complex conjugates. Let the roots be $z_1 = 1 + pi$ and $z_2 = 1 - pi$, where $p$ is a real number and $p \ne 0$ (because the roots are distinct).

Explanation: We are defining the general form of the complex conjugate roots based on the given condition that their real part is 1. The condition $p \ne 0$ ensures that the roots are distinct.

Step 2: Apply Vieta's Formulas

We use Vieta's formulas to relate the roots to the coefficients $\alpha$ and $\beta$.

• Sum of Roots: $z_1 + z_2 = -\alpha$ $(1 + pi) + (1 - pi) = -\alpha$ $2 = -\alpha$ $\alpha = -2$

Explanation: The sum of the roots gives us the value of $\alpha$. The imaginary parts cancel out, confirming that $\alpha$ is indeed real.*

• Product of Roots: $z_1 z_2 = \beta$ $(1 + pi)(1 - pi) = \beta$ $1^2 - (pi)^2 = \beta$ $1 - p^2i^2 = \beta$ Since $i^2 = -1$, we have $1 - p^2(-1) = \beta$ $1 + p^2 = \beta$

Explanation: The product of the roots gives us the value of $\beta$. The product of complex conjugates results in a real number.*

Step 3: Determine the Range of $\beta$

Since $p$ is a real number and $p \ne 0$, then $p^2 > 0$. Therefore, $\beta = 1 + p^2 > 1$.

Explanation: Because the roots are distinct ($p\ne 0$), $p^2$ is strictly positive, making $\beta$ strictly greater than 1.*

Thus, $\beta \in (1, \infty)$.

Step 4: Check the discriminant

The discriminant of the quadratic equation is given by $D = \alpha^2 - 4\beta$. Substituting $\alpha = -2$ and $\beta = 1 + p^2$, we get: $D = (-2)^2 - 4(1 + p^2) = 4 - 4 - 4p^2 = -4p^2$

Since $p \ne 0$, $p^2 > 0$, so $D = -4p^2 < 0$. This confirms that the roots are indeed complex and distinct.

Explanation: We verify that the discriminant is negative, which is consistent with having complex conjugate roots.*

Common Mistakes & Tips

• Forgetting the distinctness condition: If the roots were not distinct, then $p=0$, and $\beta = 1$.
• Assuming all roots are real: Complex roots are possible when the coefficients are real, and in this case, necessary for the condition Re(z) = 1 to hold with distinct roots.
• Incorrectly applying Vieta's formulas: Ensure the correct signs are used when relating the roots to the coefficients.

Summary

Given the quadratic equation $z^2 + \alpha z + \beta = 0$ with real coefficients $\alpha$ and $\beta$, and two distinct roots on the line $Re(z) = 1$, the roots must be complex conjugates of the form $1 \pm pi$, where $p \ne 0$. Applying Vieta's formulas, we find that $\beta = 1 + p^2$, which implies $\beta > 1$, so $\beta \in (1, \infty)$.

Final Answer

The final answer is $\beta \, \in (1,\infty )$, which corresponds to option (C).