Animated Solution for Mathematics - Complex Numbers: Let $\omega =e^{i\pi /3}$, and $a,b,c,x,y,z$ be non-zero complex numbers such that $a+b+c=x,a+b\omega +c{\omega }^2=y,a+b{\omega }^2+c\omega =z$. Then the value of $\frac{|x{|}^2+|y{|}^2+|z{|}^2}{|a{|}^2+|b{|}^2+|c{|}^2}$ is .........

Welcome, fellow explorer of the mathematical universe! Today, we are diving into a problem that might look like a daunting algebraic mess at first glance, but beneath the surface lies a beautiful, symmetric structure.

We are given three complex numbers, $x,y,$ and $z$, defined as linear combinations of $a,b,$ and $c$ using the complex number $\omega =e^{i\pi /3}$. Our mission is to find the ratio of the sum of their squared magnitudes to the sum of the squared magnitudes of $a,b,$ and $c$.

The Brute Force Expansion

When you see expressions like $|x{|}^2,|y{|}^2,$ and $|z{|}^2$, your first instinct should be to use the fundamental property of complex numbers: $|z{|}^2=z\cdot \bar{z}$. This is our most powerful tool.

Let's start with $x=a+b+c$. Expanding $|x{|}^2$ is straightforward:

This gives us the sum of the squared magnitudes plus the cross-terms. Now, let's look at $y=a+b\omega +c{\omega }^2$.

The expansion of $|y{|}^2$ follows the same logic, but with a twist. Since $|\omega |=1$, we know that $\omega \bar{\omega }=1$. When we multiply $y$ by its conjugate $\bar{y}=\bar{a}+\bar{b}\bar{\omega }+\bar{c}{\bar{\omega }}^2$, we get:

Because $|\omega |=1$, the squared magnitudes of $b$ and $c$ remain $|b{|}^2$ and $|c{|}^2$. The cross-terms now carry the weight of $\omega$ and its conjugate. The pattern is emerging.

The Symmetry Revealed

Now, let's expand $|z{|}^2$ where $z=a+b{\omega }^2+c\omega$. Following the same procedure, we get:

Now, the magic happens. When we sum $|x{|}^2+|y{|}^2+|z{|}^2$, we add these three massive expressions together. The first part is easy: we get $3(|a{|}^2+|b{|}^2+|c{|}^2)$.

Let's group the cross-terms by their components, such as $a\bar{b}$. For the $a\bar{b}$ terms, we have:

The Grand Cancellation

This is where the JEE examiner rewards your patience. The problem states $\omega =e^{i\pi /3}$. Because of the symmetry of the expressions and the properties of $\omega$, these coefficients $(1+\omega +{\omega }^2)$ and $(1+\bar{\omega }+{\bar{\omega }}^2)$ evaluate to zero.

The cross-terms, which seemed so terrifying, perfectly cancel each other out! We are left with:

Finally, we calculate the ratio:

And there it is! The entire complex structure collapses into the simple integer 3. It is a beautiful reminder that in physics and mathematics, even the most complex-looking systems often obey simple, elegant laws.