Animated Solution for Mathematics - Complex Numbers: Let $\alpha ,\beta$ be the roots of the quadratic equation $x^2+\sqrt{6}x+3=0$. Then $\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}$ is equal to

Taming Complex Roots

My dear student, take a deep breath. When you first look at an expression like

it is natural to feel a surge of panic. It looks like a mountain of algebra, a calculation that would take hours.

But in the world of JEE Advanced, we do not climb mountains by brute force; we climb them by finding the hidden path. This problem is not about calculation; it is about symmetry, geometry, and the elegant language of Euler.

The Quadratic Foundation

We begin with the quadratic equation $x^2+\sqrt{6}x+3=0$. Our first instinct is to find the roots, $\alpha$ and $\beta$.

Using the quadratic formula

we substitute our coefficients: $a=1$, $b=\sqrt{6}$, and $c=3$.

The discriminant is $6-12=-6$. This is the moment of truth. A negative discriminant tells us that our roots are not on the real number line; they are complex.

We write them as

The Complex Plane

Now, let us look at these roots with the eyes of a mathematician. If we factor out $\sqrt{3}$, we get

Do you recognize these values? They are the sine and cosine of $3\pi /4$ and $5\pi /4$.

This is the 'Spark' of the problem. We are not dealing with random complex numbers; we are dealing with points on a circle of radius $\sqrt{3}$ in the complex plane.

We can write $\alpha =\sqrt{3}{e}^{i3\pi /4}$ and $\beta =\sqrt{3}{e}^{i5\pi /4}$.

The Power of Euler

To handle high powers like $23$ or $15$, we define a general term $a_n={\alpha }^n+{\beta }^n$. By substituting our Euler forms, we get

With a little algebraic manipulation, factoring out $e^{in\pi }$, we arrive at the beautiful general form:

This formula is your weapon. It reduces any power of these roots to a simple cosine value.

The Elegant Cancellation

Now, look at the denominator of our original expression: ${\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}$. This is $a_{15}+a_{10}$.

When we calculate $a_{10}$, we need $\cos (10\pi /4)=\cos (5\pi /2)$, which is $0$. Similarly, $a_{14}$ involves $\cos (14\pi /4)=\cos (7\pi /2)$, which is also $0$.

The terms $a_{14}$ and $a_{10}$ vanish into thin air! We are left with the ratio $\frac{a_{23}}{a_{15}}$.

Finally, we compute the ratio. The terms $2(\sqrt{3}{)}^n(-1{)}^n$ partially cancel out, leaving us with

Since $23\pi /4$ and $15\pi /4$ are both coterminal with $-\pi /4$ in their respective cycles, the cosine terms are identical and cancel out perfectly.

We are left with $(\sqrt{3}{)}^8=3^4=81$. You see? The beast was not a monster; it was a masterpiece of symmetry. Keep this elegance in your heart as you solve your next problem.