Question

The locus of point \( z \) which satisfies:

\[ \arg\left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{3} \] is:

• A. \( x^2+y^2-2y+1 = 0 \)
• B. \( 3x^2+3y^2+10x+3 \ge 0 \)
• C. \( 3x^2+3y^2+10x+3 = 0 \)
• D. \( x^2+y^2 - \frac{2}{\sqrt{3}}y - 1 = 0 \)

Hint:

Geometrically, \( \arg(z-z_1) - \arg(z-z_2) = \theta \) represents the locus of points \(z\) such that the angle \( \angle z_2 z z_1 = \theta \). This locus is an arc of a circle passing through the points \(z_1\) and \(z_2\). In this problem, \(z_1=1\) and \(z_2=-1\), so the locus is an arc of a circle passing through \( (1,0) \) and \( (-1,0) \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The argument of a complex number, \( \arg(w) \), represents the angle it makes with the positive real axis. The condition \( \arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{3} \) has a geometric interpretation: the angle subtended by the line segment from \(z=-1\) to \(z=1\) at the point \(z\) is \( \pi/3 \). The locus of points satisfying such a condition is an arc of a circle passing through the points -1 and 1. We can find the equation of this circle algebraically.

Step 2: Key Formula or Approach:
Let \( z = x+iy \). We will substitute this into the expression, simplify the complex fraction \( \frac{z-1}{z+1} \) into the form \( A+iB \), and then use the property that \( \arg(A+iB) = \tan^{-1}(B/A) \).

Step 3: Detailed Explanation:
First, substitute \( z = x+iy \): \[ \frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy} \] Multiply the numerator and denominator by the conjugate of the denominator: \[ = \frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy} = \frac{((x-1)(x+1)+y^2) + i(y(x+1)-y(x-1))}{(x+1)^2+y^2} \] \[ = \frac{(x^2-1+y^2) + i(xy+y-xy+y)}{(x+1)^2+y^2} = \frac{(x^2+y^2-1) + i(2y)}{(x+1)^2+y^2} \] The argument of this complex number is \( \arg\left(\frac{\text{Imaginary Part}}{\text{Real Part}}\right) \). We are given that this is \( \pi/3 \). \[ \tan\left(\arg\left(\frac{z-1}{z+1}\right)\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \] Therefore, we have: \[ \frac{2y}{x^2+y^2-1} = \sqrt{3} \] Rearrange the equation to find the locus: \[ 2y = \sqrt{3}(x^2+y^2-1) \] \[ \sqrt{3}x^2 + \sqrt{3}y^2 - 2y - \sqrt{3} = 0 \] Divide by \( \sqrt{3} \) to get a standard form for a circle: \[ x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0 \] This is the equation of a circle. It matches option (D) exactly.

Step 4: Final Answer:
The locus of the point z is the circle given by the equation \( x^2+y^2 - \frac{2}{\sqrt{3}}y - 1 = 0 \).