Key Concepts and Formulas

• Modulus of a Complex Number: $|z - a|$ represents the distance between the complex number $z$ and the complex number $a$ in the complex plane.
• Argument of a Complex Number: $\arg\left(\frac{z-a}{z-b}\right)$ represents the angle subtended at $z$ by the points $a$ and $b$. It is the angle measured counter-clockwise from the vector $\vec{zb}$ to the vector $\vec{za}$.
• Cartesian Form of a Complex Number: A complex number $z$ can be written as $z = x + iy$, where $x$ and $y$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$).

Step-by-Step Solution

Step 1: Analyze Set A and simplify the modulus condition.

We are given $A = \left\{ {z \in C:\left| {{{z + 1} \over {z - 1}}} \right| < 1} \right\}$. Our goal is to find a simpler condition for $z$ to be in $A$.

$\left| {{{z + 1} \over {z - 1}}} \right| < 1$ $|z + 1| < |z - 1|$ Explanation: We separated the modulus of the numerator and denominator. This inequality means the distance between $z$ and $-1$ is less than the distance between $z$ and $1$.

Step 2: Convert to algebraic form using $z = x + iy$ and simplify.

We want to express the condition in terms of the real and imaginary parts of $z$. $|z + 1| < |z - 1|$ $|z + 1|^2 < |z - 1|^2$ Substituting $z = x + iy$: $|(x + 1) + iy|^2 < |(x - 1) + iy|^2$ $(x + 1)^2 + y^2 < (x - 1)^2 + y^2$ $x^2 + 2x + 1 + y^2 < x^2 - 2x + 1 + y^2$ $2x < -2x$ $4x < 0$ $x < 0$ Explanation: Squaring both sides allows us to use the property $|a+bi|^2 = a^2 + b^2$. This simplifies the inequality and allows us to isolate the real part $x$.

Step 3: Conclusion for Set A.

The set $A$ consists of all complex numbers $z$ such that $\operatorname{Re}(z) < 0$. Geometrically, this is the open left half-plane.

Step 4: Analyze Set B and express the argument condition geometrically.

We are given $B = \left\{ {z \in C:\arg \left( {{{z - 1} \over {z + 1}}} \right) = {{2\pi } \over 3}} \right\}$.

The condition $\arg \left( {{{z - 1} \over {z + 1}}} \right) = {{2\pi } \over 3}$ means that the angle between the vectors from $z$ to $1$ and $z$ to $-1$ is $\frac{2\pi}{3}$. This implies that $z$ lies on an arc of a circle passing through $1$ and $-1$.

Explanation: The argument condition defines the angle subtended at $z$ by the points $1$ and $-1$. Points that subtend the same angle lie on a circle.

Step 5: Convert the argument condition to algebraic form.

Let $z = x + iy$. Then $\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy} = \frac{((x - 1) + iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)}$ $= \frac{(x^2 - 1 + y^2) + i(y(x + 1) - y(x - 1))}{(x + 1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + i(2y)}{(x + 1)^2 + y^2}$ Let $w = \frac{z - 1}{z + 1} = a + bi$, where $a = \frac{x^2 + y^2 - 1}{(x + 1)^2 + y^2}$ and $b = \frac{2y}{(x + 1)^2 + y^2}$. We are given that $\arg(w) = \frac{2\pi}{3}$. This means $\tan(\frac{2\pi}{3}) = \frac{b}{a} = -\sqrt{3}$. So, $\frac{2y}{x^2 + y^2 - 1} = -\sqrt{3}$. $2y = -\sqrt{3}(x^2 + y^2 - 1)$ $\sqrt{3}x^2 + \sqrt{3}y^2 + 2y - \sqrt{3} = 0$ $x^2 + y^2 + \frac{2}{\sqrt{3}}y - 1 = 0$ $x^2 + \left(y + \frac{1}{\sqrt{3}}\right)^2 - \frac{1}{3} - 1 = 0$ $x^2 + \left(y + \frac{1}{\sqrt{3}}\right)^2 = \frac{4}{3}$ Explanation: This step converts the argument condition into an algebraic equation by rationalizing the denominator and using the tangent of the argument. Completing the square allows us to identify the circle's center and radius.

Step 6: Identify the center and radius of the circle and the relevant arc.

The equation $x^2 + \left(y + \frac{1}{\sqrt{3}}\right)^2 = \frac{4}{3}$ represents a circle with center $\left(0, -\frac{1}{\sqrt{3}}\right)$ and radius $\frac{2}{\sqrt{3}}$. Since $\arg\left(\frac{z - 1}{z + 1}\right) = \frac{2\pi}{3}$, the imaginary part of $\frac{z - 1}{z + 1}$ must be positive. This implies that $\frac{2y}{(x + 1)^2 + y^2} > 0$, which means $y > 0$. Thus, we are considering the arc of the circle that lies in the upper half-plane.

Explanation: The derived equation matches the standard form of a circle. The argument condition restricts the solution set to the portion of the circle where $y > 0$.

Step 7: Find the intersection $A \cap B$.

We need to find the region where both $x < 0$ (from set $A$) and $x^2 + \left(y + \frac{1}{\sqrt{3}}\right)^2 = \frac{4}{3}$ with $y > 0$ (from set $B$) are satisfied. This means we are looking for the portion of the circle centered at $\left(0, -\frac{1}{\sqrt{3}}\right)$ with radius $\frac{2}{\sqrt{3}}$ that lies in the upper half-plane ($y > 0$) and the left half-plane ($x < 0$). This corresponds to the arc of the circle in the second quadrant. The point $(-1,0)$ lies on the negative real axis, which forms the boundary between the second and third quadrants.

Explanation: We combine the conditions derived for sets $A$ and $B$ to identify the region that satisfies both.

Common Mistakes & Tips

• Remember that $|z+1| < |z-1|$ represents the set of points closer to $-1$ than to $1$, which is the open left half-plane.
• When dealing with arguments, always consider the quadrant in which the complex number lies.
• Be careful with strict inequalities. The boundary is not included in the set if the inequality is strict.

Summary

The set A consists of complex numbers with a negative real part ($x < 0$). The set B consists of complex numbers lying on an arc of a circle centered at $\left(0, - \frac{1}{\sqrt{3}}\right)$ with radius $\frac{2}{\sqrt{3}}$ and $y>0$. The intersection $A \cap B$ is the portion of this circle that lies in the second quadrant and bordering the third quadrant at $(-1,0)$.

The final answer is \boxed{a portion of a circle centred at $\left( {0, - {1 \over {\sqrt 3 }}} \right)$ that lies in the second and third quadrants only}, which corresponds to option (A).