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: Concept Overview:
The argument \( \arg(w) \) of a complex number \( w \) is its angle with the positive real axis. The condition \( \arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{3} \) geometrically means the angle formed at point \( z \) by the line segment connecting \( z=-1 \) and \( z=1 \) is \( \pi/3 \). The solution set is an arc of a circle passing through -1 and 1. We will derive this circle's equation.

Step 2: Method:
Let \( z = x+iy \). Substitute this into the expression and simplify the complex fraction \( \frac{z-1}{z+1} \) into \( A+iB \). Then, use \( \arg(A+iB) = \tan^{-1}(B/A) \).

Step 3: Detailed Solution:
Substitute \( z = x+iy \): \[ \frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy} \]Multiply numerator and denominator by the conjugate: \[ = \frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy} = \frac{((x-1)(x+1)+y^2) + i(y(x+1)-y(x-1))}{(x+1)^2+y^2} \]\[ = \frac{(x^2-1+y^2) + i(xy+y-xy+y)}{(x+1)^2+y^2} = \frac{(x^2+y^2-1) + i(2y)}{(x+1)^2+y^2} \]The argument is \( \arg\left(\frac{\text{Imaginary Part}}{\text{Real Part}}\right) \), which is \( \pi/3 \). \[ \tan\left(\arg\left(\frac{z-1}{z+1}\right)\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]Therefore: \[ \frac{2y}{x^2+y^2-1} = \sqrt{3} \]Rearrange 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 \( \sqrt{3} \): \[ x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0 \]This is a circle's equation, matching option (D).

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