Question

The locus of the point \(z\) satisfying \(\arg\left(\frac{z-1}{z+1}\right) = k\) (where \(k\) is non-zero) is

• A. a circle with centre on y-axis
• B. circle with centre on x-axis
• C. a straight line parallel to x-axis
• D. a straight line making an angle 60° with the x-axis

Hint:

\(\arg((z - z_1)/(z - z_2)) =\) constant gives circle through \(z_1\) and \(z_2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
\(\arg(z - 1) - \arg(z + 1) = k \rightarrow\) locus is an arc of a circle.
Step 2: Detailed Explanation:
Let \(z = x + iy\)
\(\arg\left(\frac{x - 1 + iy}{x + 1 + iy}\right) = k\)
\(\tan^{-1}\left(\frac{y}{x - 1}\right) - \tan^{-1}\left(\frac{y}{x + 1}\right) = k\)
Using formula, get \(x^2 + y^2 - \frac{2y}{\tan k} - 1 = 0\)
Centre at \((0, 1/\tan k)\) on y-axis.
Step 3: Final Answer:
Circle with centre on y-axis.