Animated Solution for Mathematics - Complex Numbers: If $x+iy=\sqrt{\frac{a+ib}{c+id}}$, prove that $(x^2+y^2{)}^2=\frac{a^2+b^2}{c^2+d^2}$.

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an algebraic equation; we are uncovering a hidden symmetry.

When you look at the problem $x+iy=\sqrt{\frac{a+ib}{c+id}}$, it is easy to feel overwhelmed. You see variables, you see square roots, and you see the imaginary unit $i$ lurking everywhere.

But look at the target: $(x^2+y^2{)}^2=\frac{a^2+b^2}{c^2+d^2}$. Notice anything? The $i$ is gone. The complexity has vanished. This is the hallmark of a problem that demands a change in perspective.

Your Most Powerful Tool

In the realm of complex numbers, the modulus is your greatest ally. Think of a complex number $z=x+iy$ as a vector in the Argand plane. The modulus, $|z|=\sqrt{x^2+y^2}$, is simply the length of that vector.

When we are asked to prove a relation involving $x^2+y^2$, we are essentially being asked to work with the square of the modulus.

Why is this so powerful? Because the modulus is a 'homomorphism' of sorts—it respects the structure of multiplication and division. We have two golden rules in our toolkit:

1. The modulus of a quotient is the quotient of the moduli: $\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$.

2. The modulus of a square root is the square root of the modulus: $|\sqrt{z}|=\sqrt{|z|}$.

These properties allow us to strip away the complex nature of the expression and focus purely on the magnitudes. Let us apply this to our problem.

We begin with our given equation: $x+iy=\sqrt{\frac{a+ib}{c+id}}$. We apply the modulus to both sides.

On the left, we have $|x+iy|$, which, by definition, is $\sqrt{x^2+y^2}$. On the right, we have the modulus of the entire square root expression: $\left|\sqrt{\frac{a+ib}{c+id}}\right|$.

Now, we invoke our second property. We can slide that modulus inside the square root:

Next, we use our first property to split the modulus of the fraction:

At this stage, the problem is practically solved. We just need to expand the moduli of the numerator and denominator. We know that $|a+ib|=\sqrt{a^2+b^2}$ and $|c+id|=\sqrt{c^2+d^2}$.

Substituting these back in, we get:

We have nested square roots, which might look intimidating, but remember: we are in control. We want to reach $(x^2+y^2{)}^2$. To get there, we must peel away these layers of roots.

First, square both sides to eliminate the outermost root:

Now, look at the target one last time. We need the square of the left side. So, we square both sides again:

When we square the right side, the square roots in the numerator and denominator vanish, leaving us with the elegant result:

Do you see the beauty in this? We didn't need to perform messy algebraic expansions or solve for $x$ and $y$ individually.

By recognizing the geometric nature of the problem—by seeing the modulus hidden within the algebraic structure—we navigated directly to the solution. This is the essence of JEE Advanced physics and mathematics: it is not about brute force; it is about choosing the right tool for the job.

Keep this modulus property in your arsenal; it will save you time and frustration in many problems to come. You have mastered this concept today. Keep that momentum going!