Problem Analysis:

We are given that $1 - 2i$ is a root of the quadratic equation $z^2 + \alpha z + \beta = 0$, where $i^2 = -1$. Since the coefficients ($\alpha$, $\beta$) are real numbers ($\in \mathbb{R}$), the other root of the quadratic equation must be the complex conjugate of $1 - 2i$, which is $1 + 2i$.

We need to find the value of $\alpha - \beta$.

Step-by-Step Solution:

1. Roots of the quadratic equation:

Let the roots of the quadratic equation be: $z_1 = 1 - 2i \quad \text{and} \quad z_2 = 1 + 2i$

The general form of a quadratic equation is: $z^2 - (\text{sum of roots})z + (\text{product of roots}) = 0$

2. Compute the sum and product of the roots:

• Sum of roots: $z_1 + z_2 = (1 - 2i) + (1 + 2i) = 2$

• Product of roots: $z_1 \cdot z_2 = (1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - (-4) = 1 + 4 = 5$

3. Substitute into the quadratic equation:

The quadratic equation becomes: $z^2 - (\text{sum of roots})z + (\text{product of roots}) = 0$ $z^2 - 2z + 5 = 0$

Here, $\alpha = -2$ and $\beta = 5$.

4. Compute $\alpha - \beta$:

$\alpha - \beta = -2 - 5 = -7$

Final Answer:

$\boxed{-7}$

Explanation of Shortest Method:

1. Recognize that for real coefficients, the other root must be the complex conjugate of $1 - 2i$.
2. Use the sum and product of the roots to determine $\alpha$ and $\beta$.
3. Directly substitute the values to compute $\alpha - \beta$.

Let me know if you need further clarification! Here are 5 related questions for further practice:

1. How do you derive the sum and product of roots for a general quadratic equation?
2. What happens when the roots of a quadratic equation are purely imaginary?
3. How would the quadratic equation change if the roots were real and unequal?
4. Derive the quadratic equation if the roots are $2 + 3i$ and $2 - 3i$.
5. Explain why complex roots must occur in conjugate pairs for polynomials with real coefficients.

Tip: Always confirm that coefficients are real when working with conjugate roots in quadratic equations!