Animated Solution for Mathematics - Complex Numbers: Let $z=x+iy$ be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation $z{\bar{z}}^3+\bar{z}{z}^3=350$ is

Welcome, future engineer. Today, we are not just solving an equation; we are embarking on a journey through the Argand plane.

When you first look at the equation $z{\bar{z}}^3+\bar{z}{z}^3=350$, it might seem intimidating. It looks like a high-degree polynomial, and the thought of expanding it might make you want to reach for a calculator.

But hold on. In the world of JEE Advanced, the most complex-looking problems often hide the most elegant, simple solutions. Our goal is to peel back the layers of this equation until we find the geometric truth hidden underneath.

Let us start by observing the structure. We have $z{\bar{z}}^3+\bar{z}{z}^3=350$.

Notice the symmetry? Both terms contain $z$ and $\bar{z}$. This is a classic invitation to factor.

If we pull out the common term $z\bar{z}$, the equation transforms into:

Suddenly, the problem feels much lighter. We have reduced a high-degree expression into a product of two simpler components. This is the first step in mastering complex numbers: look for the structure before you start calculating.

Now, we introduce the Cartesian form, $z=x+iy$. But wait, look at the problem statement again. It tells us that $x$ and $y$ are integers.

This is the most critical piece of information. It changes the nature of the problem entirely. We are no longer solving for arbitrary complex numbers; we are hunting for integer coordinates.

Let us evaluate our pieces. We know that $z\bar{z}=|z{|}^2=x^2+y^2$. This is a fundamental property that you should always keep in your toolkit.

Next, consider the term $z^2+{\bar{z}}^2$. When we expand $z^2=(x+iy{)}^2$, we get $x^2-y^2+2ixy$. When we expand ${\bar{z}}^2=(x-iy{)}^2$, we get $x^2-y^2-2ixy$.

When we add them together, the imaginary parts $2ixy$ and $-2ixy$ vanish into thin air. We are left with $2(x^2-y^2)$. This cancellation is the beauty of complex conjugates.

Now, let us substitute these back into our factored equation:

Dividing both sides by $2$, we get $(x^2+y^2)(x^2-y^2)=175$. If you recognize the difference of squares, you will see that this simplifies to:

This is where the integer constraint becomes our best friend. We are looking for two perfect fourth powers that differ by $175$.

Let us list them: $1^4=1$, $2^4=16$, $3^4=81$, $4^4=256$, $5^4=625$. Look at $256$ and $81$. The difference is $256-81=175$.

It fits perfectly! This implies $x^4=256$ and $y^4=81$, which means $x^2=16$ and $y^2=9$. Thus, $x=\pm 4$ and $y=\pm 3$.

We have found our coordinates. The roots are the four points $(4,3),(4,-3),(-4,3),$ and $(-4,-3)$.

If you plot these on the complex plane, you will see a rectangle. The length of this rectangle, parallel to the real axis, is the distance between $x=-4$ and $x=4$, which is $8$.

The width, parallel to the imaginary axis, is the distance between $y=-3$ and $y=3$, which is $6$. The area of a rectangle is simply length times width.

So, $8\times 6=48$. We have arrived at the solution. It was not about brute force; it was about identifying the symmetry, respecting the constraints, and letting the algebra guide us to the geometric reality.