Animated Solution for Mathematics - Complex Numbers: Let the curve $z(1+i)+\overline{z}(1-i)=4,z\in C$, divide the region $|z-3|\le 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha -\beta |$ equals:

Welcome, fellow traveler, to the beautiful intersection of complex numbers and Euclidean geometry. Often, when we see a problem involving $z$ and $\overline{z}$, our instinct is to dive into algebraic manipulation.

But today, I want you to pause. Look at the equation $|z-3|\le 1$.

You see a circle. Not just any circle, but a solid disk living on the Argand plane, centered at the point $(3,0)$ with a radius of $1$. This is our canvas.

Now, let us look at the curve: $z(1+i)+\overline{z}(1-i)=4$. It looks intimidating, doesn't it? But let us strip away the complexity.

We substitute $z=x+iy$ and $\overline{z}=x-iy$. When we expand this, something truly elegant happens. The imaginary parts, those pesky terms involving $i$, vanish into thin air.

We are left with $2(x-y)=4$, or simply $x-y=2$. Just like that, the complex curve reveals its true identity: a straight line cutting through our disk.

Now, we must determine how this line interacts with our circle. We need the perpendicular distance $d$ from the center of our circle, $C(3,0)$, to the line $x-y-2=0$.

Using the standard distance formula, we find:

Since our radius $r$ is $1$, and $d<r$, we know for certain that this line is a secant, slicing our disk into two distinct segments. This is the moment where the physics of the problem becomes tangible—we are literally cutting a slice of the pie.

To find the area of these segments, we need the central angle $\theta$. Imagine the triangle formed by the center of the circle and the two points where the line intersects the circumference.

This is an isosceles triangle. By dropping a perpendicular from the center to the chord, we create two right-angled triangles. In these triangles, the cosine of the half-angle is the ratio of the adjacent side (the distance $d$) to the hypotenuse (the radius $r$).

Thus:

This tells us that $\frac{\theta }{2}=\frac{\pi }{4}$, so the total central angle is $\theta =\frac{\pi }{2}$, or $90^{\circ }$. We are dealing with a perfect quadrant-like slice.

Now, we calculate the area of the smaller segment, $\beta$. As we discussed, this is the area of the sector minus the area of the triangle:

Substituting our values, we get:

The larger area, $\alpha$, is simply the remaining area of the circle:

Finally, we seek the absolute difference $|\alpha -\beta |$:

There it is. The complexity of the initial equation has dissolved into a clean, elegant result. This is the essence of JEE Advanced mathematics: finding the simple, geometric truth hidden beneath a layer of complex notation.