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

To find the locus of the point \( z \) satisfying the condition \( \arg\left(\frac{z-1}{z+1}\right) = k \), we need to analyze the given equation. Here, \( \arg \) denotes the argument of a complex number, which is the angle the complex number makes with the positive real axis in the complex plane.

1. Let's express \( z \) as \( x + yi \), where \( x \) and \( y \) are real numbers and \( i \) is the imaginary unit.
2. The expression \( \frac{z-1}{z+1} \) can be evaluated as follows: \(\frac{z-1}{z+1} = \frac{(x-1) + yi}{(x+1) + yi}\).
3. Compute the argument: \(\arg\left(\frac{z-1}{z+1}\right) = \arg((x-1) + yi) - \arg((x+1) + yi)\).
4. From the given equation, this difference of arguments is a constant \( k \). This represents the angle subtended between \((x-1) + yi\) and \((x+1) + yi\).
5. The key geometric interpretation of this condition is that all points \( z \) such that \( \arg\left(\frac{z-1}{z+1}\right) = k \) trace out a circle, because a constant argument difference signifies that the angle remains constant for all such complex numbers.
6. This means the points form a circular arc in the complex plane where the circle's geometry ensures that its center lies somewhere where this condition is satisfied. Given that there is no particular constraint making the circle align or parallel with axes based on the symmetry dictated by the problem, the circle’s center is especially located along the y-axis.

Thus, the locus is a circle with the center on the y-axis.

The correct answer is: a circle with centre on y-axis