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

We need to find $$A \cap B$$ where $$A = \left\{z \in \mathbb{C} : \left|\frac{z+1}{z-1}\right| < 1\right\}$$ and $$B = \left\{z \in \mathbb{C} : \arg\left(\frac{z-1}{z+1}\right) = \frac{2\pi}{3}\right\}$$.

Analyze set A.

$$\left|\frac{z+1}{z-1}\right| < 1 \implies |z+1| < |z-1|$$

This means the point $$z$$ is closer to $$-1$$ than to $$1$$, which gives us $$\text{Re}(z) < 0$$ (the left half of the complex plane).

Analyze set B.

$$\arg\left(\frac{z-1}{z+1}\right) = \frac{2\pi}{3}$$

The locus of points where $$\arg\left(\frac{z-1}{z+1}\right) = \alpha$$ (constant) is an arc of a circle passing through the points $$1$$ and $$-1$$.

Let $$z = x + iy$$. Then:

$$\frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy} = \frac{[(x-1)+iy][(x+1)-iy]}{|z+1|^2}$$

$$= \frac{(x^2+y^2-1) + i \cdot 2y}{(x+1)^2+y^2}$$

For the argument to be $$\frac{2\pi}{3}$$:

$$\tan\left(\frac{2\pi}{3}\right) = \frac{2y}{x^2+y^2-1} = -\sqrt{3}$$

This gives: $$x^2 + y^2 + \frac{2y}{\sqrt{3}} - 1 = 0$$

$$x^2 + \left(y + \frac{1}{\sqrt{3}}\right)^2 = 1 + \frac{1}{3} = \frac{4}{3}$$

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

Determine the relevant arc.

Since $$\arg = \frac{2\pi}{3}$$ (which is in the second quadrant of argument values), we need the real part of $$\frac{z-1}{z+1}$$ to be negative, i.e., $$x^2+y^2 < 1$$, and the imaginary part $$2y$$ must be positive, so $$y > 0$$.

But set A requires $$\text{Re}(z) < 0$$, i.e., $$x < 0$$.

Points on the arc with $$y > 0$$ and $$x < 0$$ lie in the second quadrant only.

Find $$A \cap B$$.

$$A \cap B$$ is a portion of the circle centred at $$\left(0, -\frac{1}{\sqrt{3}}\right)$$ that lies in the second quadrant only.

The correct answer is Option B.