Animated Solution for Mathematics - Complex Numbers: If $\alpha ,\beta \in R$ are such that $1-2i$ (here $i^2=-1$) is a root of $z^2+\alpha z+\beta =0$, then $(\alpha -\beta )$ is equal to:

Welcome, future engineer. Today, we are not just solving a quadratic equation; we are uncovering a fundamental law of algebra.

When you look at the equation $z^2+\alpha z+\beta =0$, it might seem like a simple, standard problem. However, the condition $\alpha ,\beta \in \mathbb{R}$ is a beacon of information. It tells us that this polynomial is anchored in the real number system, which fundamentally dictates the behavior of its roots.

Imagine you are standing in the complex plane. You have a root at $z_1=1-2i$, which sits in the fourth quadrant.

Because our coefficients $\alpha$ and $\beta$ are real, the polynomial possesses a beautiful, inherent symmetry known as the Conjugate Root Theorem. It dictates that complex roots of polynomials with real coefficients must appear in conjugate pairs.

Think of it as a mirror reflection across the real axis. If $z_1=1-2i$ is a root, its 'mirror image', $z_2=1+2i$, must also be a root. They are inseparable partners, bound by the reality of the coefficients.

Now that we have both roots, $z_1=1-2i$ and $z_2=1+2i$, we could laboriously multiply $(z-z_1)(z-z_2)$ to find the equation. But why take the long road when we have the elegant shortcut of Vieta's formulas?

Vieta's formulas act as a bridge between the roots and the coefficients of a polynomial. For a quadratic equation $a{z}^2+bz+c=0$, the sum of the roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$.

In our case, the equation is $z^2+\alpha z+\beta =0$, where $a=1$, $b=\alpha$, and $c=\beta$. Therefore:

A Dance of Numbers

Let's perform the sum first:

Notice the magic here? The imaginary parts, $-2i$ and $+2i$, cancel each other out perfectly, leaving us with $1+1=2$. Thus, $2=-\alpha$, which gives us:

Next, we calculate the product:

This is a classic difference of squares: $(a-b)(a+b)=a^2-b^2$. So, we get $1^2-(2i{)}^2$. Since $i^2=-1$, we have $(2i{)}^2=4i^2=-4$. This transforms our expression into $1-(-4)=5$. Thus:

We have arrived at the finish line. We need to calculate $(\alpha -\beta )$. Substituting our values:

It is easy to rush through the final subtraction and lose a sign, but look at the elegance of the result. By understanding the symmetry of the roots and the power of Vieta's formulas, we bypassed the complexity and found the answer with precision.

The final result is $-7$.