Animated Solution for Mathematics - Complex Numbers: Let $A=\left\{z\in \mathbb{C}:\left|\frac{z+1}{z-1}\right|<1\right\}$ and $B=\left\{z\in \mathbb{C}:\arg \left(\frac{z-1}{z+1}\right)=\frac{2\pi }{3}\right\}$. Then $A\cap B$ is :

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are painting a picture on the Argand plane. We are going to explore two sets, $A$ and $B$, and discover where they meet.

This is a classic JEE Advanced problem that tests your ability to bridge the gap between algebraic manipulation and geometric intuition.

Let us begin with Set $A$, defined by the condition:

When you see a modulus inequality like this, do not panic. Think of it as a distance relationship. We can rewrite this as $|z-(-1)|<|z-1|$.

Imagine you are standing on the complex plane. You have two fixed beacons: one at $-1$ and one at $1$. The condition tells us that your distance to the beacon at $-1$ must be strictly less than your distance to the beacon at $1$.

If you were exactly equidistant, you would be standing on the perpendicular bisector of the segment connecting $-1$ and $1$, which is the imaginary axis ($x=0$). Since you must be closer to $-1$, you are restricted to the entire left half-plane.

Algebraically, if we set $z=x+iy$, we find:

This simplifies beautifully to $4x<0$, or simply $x<0$. Set $A$ is the vast, open territory of the left half-plane.

Now, let us turn our attention to Set $B$, defined by:

This is where the beauty of complex numbers truly shines. The argument of a quotient is the difference of the arguments: $\arg (z-1)-\arg (z+1)=\frac{2\pi }{3}$.

This is a classic locus problem. The set of points $z$ that subtend a constant angle at two fixed points is an arc of a circle. We are looking for the path where the angle between the vectors from $-1$ and $1$ to $z$ is exactly $120^{\circ }$.

To find the equation of this circle, we use the tangent subtraction formula:

By substituting $z=x+iy$, we transform this into the equation:

This is a circle centered at $\left(0,-\frac{1}{\sqrt{3}}\right)$ with a radius of $\frac{2}{\sqrt{3}}$.

We have our two regions. Set $A$ is the left half-plane ($x<0$). Set $B$ is an arc of a circle.

Because the argument is positive ($\frac{2\pi }{3}>0$), we know the arc must lie in the upper half-plane ($y>0$).

Now, we bring them together. We need points that are in the left half-plane ($x<0$) AND on the upper arc of the circle ($y>0$).

If you visualize the circle centered at $\left(0,-\frac{1}{\sqrt{3}}\right)$, it crosses the imaginary axis and extends into both the left and right sides. However, the condition $x<0$ forces us to discard everything to the right of the imaginary axis.

The condition $y>0$ forces us to discard the lower portion of the circle. What remains? Only the portion of the circle that resides in the second quadrant.

It is a precise, elegant slice of the circle. You have successfully navigated the constraints, visualized the geometry, and arrived at the solution. This is the essence of JEE Advanced mathematics—not just calculation, but the ability to see the structure behind the symbols.