Animated Solution for Mathematics - Complex Numbers: Let $\alpha ,\beta$ be real and $z$ be a complex number. If $z^2+\alpha z+\beta =0$ has two distinct roots on the line $Rez=1$, then it is necessary that

Imagine you are standing on the complex plane, a vast, two-dimensional landscape where the horizontal axis represents the real part of a number and the vertical axis represents the imaginary part.

We are given a quadratic equation, $z^2+\alpha z+\beta =0$, and we are told that its roots are trapped on a very specific, vertical highway: the line $Re(z)=1$. This is not just a random constraint; it is a geometric invitation to explore the symmetry of complex numbers.

Before we dive into the algebra, let us pause and appreciate the power of the coefficients. We are told that $\alpha$ and $\beta$ are real numbers.

In the world of polynomials with real coefficients, complex roots are never lonely; they always travel in pairs. If $z_1$ is a root, its conjugate $\overline{z_1}$ must also be a root. They are perfect mirror images across the real axis.

Since our roots are confined to the line $Re(z)=1$, this symmetry forces them to be $z_1=1+iy$ and $z_2=1-iy$, where $y$ is some real number. This is the core geometric reality of our problem.

Now, let us invoke the wisdom of Vieta. For any quadratic equation $z^2+\alpha z+\beta =0$, the sum of the roots is given by $-\alpha$, and the product is given by $\beta$.

Let us calculate these using our defined roots:

Equating this to $-\alpha$, we immediately find that $\alpha =-2$. This is a satisfyingly simple result, but the real treasure lies in the product.

The product is $(1+iy)(1-iy)$. Using the difference of squares identity, this simplifies as follows:

Thus, we have established the fundamental relationship: $\beta =1+y^2$.

We are almost at the finish line, but we must be careful. The problem explicitly states that the roots are distinct.

If $y$ were zero, both roots would collapse into the single point $z=1$, which would be a repeated root. To keep our roots distinct, we must insist that $y eq 0$.

Since $y$ is a non-zero real number, its square $y^2$ must be strictly greater than zero. If we take our expression $\beta =1+y^2$ and apply this inequality, we get:

And there it is! The range of $\beta$ is the open interval $(1,\infty )$.

It is a beautiful result, born from the marriage of geometry and algebra. We started with a simple line on the complex plane and, by respecting the symmetry of conjugate roots and the constraints of the equation, we uncovered the hidden behavior of the constant term $\beta$.

Remember, in JEE Advanced, the math is not just about calculation; it is about visualizing the story behind the numbers. The final result is $\beta \in (1,\infty )$.