Among the statements
(S1) : The set {z ∈ ℂ-{-i}:|z|=1 and (z-i)/(z+i) is purely real } contains exactly two elements, and
(S2) : The set {z ∈ ℂ-{-1}:|z|=1 and (z-1)/(z+1) is purely imaginary } contains infinitely many elements.

Solution

Maths sir · 4 steps

On the unit circle, x²+y²=1 wipes out the real part of both ratios. Then just count the points.

1. Set up. Write z = x + iy with x² + y² = 1. Multiply each ratio by the conjugate of its denominator, then use x² + y² = 1 to simplify the numerator.

   (z-i)/(z+i) = (-2x i)/(x²+(y+1)²), (z-1)/(z+1) = (2y i)/((x+1)²+y²)

2. Check S1. The first ratio has no real part at all, so it is real only when it equals 0, that is when x = 0. On the circle that gives z = i or z = -i, and -i is excluded. The set is {i}, one element, so S1 is false.
3. Check S2. The second ratio is purely imaginary for every z on the circle except -1. A circle has infinitely many points, so S2 is true.
4. So. S1 is incorrect and S2 is correct. The answer is (c).

The idea to keep. For z on a circle, (z-a)/(z-b) with a, b at the ends of a diameter is always purely imaginary, because the angle in a semicircle is 90°. It can be real only when it is 0.

Answer from the NTA final answer key. Solution by PrepWiser.