Let `A={z inC:|(z+1)/(z-1)|lt1}`
and `B={z inC: arg((z-1)/(z+1))=(2pi)/(3)}`. Then A∩B

Options

A. a portionof a circle centred at `(0,-(1)/(sqrt(3)))` that lies in the second and third quadrants only
B. a portion of a circle centred at `(0,-(1)/(sqrt(3)))` that lies in the second quadrant only
C. an empty set
D. a portion of a circle of radius `(2)/(sqrt(3))` that lies in the third quadrant only

Answer

To solve the problem, we need to analyze the sets \( A \) and \( B \) given in the question.

### Step 1: Analyze Set \( A \)

Set \( A \) is defined as:
\[ A = \{ z \in \mathbb{C} : \left| \frac{z+1}{z-1} \right| < 1 \} \]

This inequality can be rewritten as:
\[ |z + 1| < |z - 1| \]

This means that the distance from the point \( z \) to \( -1 \) is less than the distance from \( z \) to \( 1 \). Geometrically, this describes the region of the complex plane that is closer to the point \( -1 \) than to the point \( 1 \).

### Step 2: Interpret the Geometric Condition

To find the boundary of this region, we can square both sides:
\[ |z + 1|^2 < |z - 1|^2 \]

Let \( z = x + iy \) where \( x \) and \( y \) are real numbers. Then we have:
\[ |z + 1|^2 = (x + 1)^2 + y^2 \]
\[ |z - 1|^2 = (x - 1)^2 + y^2 \]

Setting up the inequality:
\[ (x + 1)^2 + y^2 < (x - 1)^2 + y^2 \]

### Step 3: Simplify the Inequality

Cancel \( y^2 \) from both sides:
\[ (x + 1)^2 < (x - 1)^2 \]

Expanding both sides:
\[ x^2 + 2x + 1 < x^2 - 2x + 1 \]

Subtract \( x^2 + 1 \) from both sides:
\[ 2x < -2x \]

Combining like terms gives:
\[ 4x < 0 \quad \Rightarrow \quad x < 0 \]

### Step 4: Conclusion for Set \( A \)

Thus, the set \( A \) consists of all complex numbers \( z \) such that the real part \( x < 0 \). This corresponds to the left half of the complex plane.

### Step 5: Analyze Set \( B \)

Set \( B \) is defined as:
\[ B = \{ z \in \mathbb{C} : \arg\left(\frac{z-1}{z+1}\right) = \frac{2\pi}{3} \} \]

This means that the argument of the complex number \( \frac{z-1}{z+1} \) is \( \frac{2\pi}{3} \), which corresponds to a specific direction in the complex plane.

### Step 6: Rewrite the Argument Condition

The condition can be rewritten as:
\[ \frac{z-1}{z+1} = re^{i\frac{2\pi}{3}} \quad \text{for some } r > 0 \]

Cross-multiplying gives:
\[ z - 1 = re^{i\frac{2\pi}{3}}(z + 1) \]

### Step 7: Rearranging the Equation

Rearranging this equation leads to:
\[ z - re^{i\frac{2\pi}{3}}z = re^{i\frac{2\pi}{3}} + 1 \]
\[ z(1 - re^{i\frac{2\pi}{3}}) = re^{i\frac{2\pi}{3}} + 1 \]

### Step 8: Solve for \( z \)

From the above equation, we can express \( z \):
\[ z = \frac{re^{i\frac{2\pi}{3}} + 1}{1 - re^{i\frac{2\pi}{3}}} \]

This represents a line in the complex plane.

### Step 9: Find the Intersection \( A \cap B \)

To find \( A \cap B \), we need to find the points that satisfy both conditions:
1. \( x < 0 \) (from set \( A \))
2. The line described by the argument condition (from set \( B \))

### Step 10: Conclusion

The intersection \( A \cap B \) will be the set of points in the left half-plane that also lie on the line described by the argument condition. This can be visualized graphically, where the line intersects the left half-plane.