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Maths Question 2 – JEE-MAIN 2025

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

A complex number w is purely real if w=w¯, and purely imaginary if w=−w¯.

Step 1: Analyze Statement (S1)✦ Active

For w=z−iz+i to be purely real, we must have w=w¯. This implies z−iz+i=z¯+iz¯−i. Cross-multiplying gives (z−i)(z¯−i)=(z+i)(z¯+i), which simplifies to zz¯−zi−iz¯−1=zz¯+zi+iz¯−1. This further reduces to 2zi+2iz¯=0, or 2i(z+z¯)=0. Since 2i≠0, we must have z+z¯=0. If z=x+iy, then z+z¯=2x, so 2x=0⟹x=0. This means z must be purely imaginary.

Given the condition |z|=1, the purely imaginary numbers satisfying this are z=i and z=−i. However, the set for (S1) excludes z=−i. Therefore, the set contains only one element, z=i. Statement (S1) claims "exactly two elements", which is false.

Step 2: Analyze Statement (S2)○ Expand

For w=z−1z+1 to be purely imaginary, we must have w=−w¯. This implies z−1z+1=−z¯−1z¯+1. Cross-multiplying gives (z−1)(z¯+1)=−(z+1)(z¯−1), which expands to zz¯+z−z¯−1=−(zz¯−z+z¯−1). This simplifies to |z|2+z−z¯−1=−|z|2+z−z¯+1.

Given the condition |z|=1, we substitute |z|2=1 into the equation: 1+z−z¯−1=−1+z−z¯+1. This simplifies to z−z¯=z−z¯, which is an identity. This means the condition that z−1z+1 is purely imaginary holds for all z such that |z|=1 and z≠−1 (to avoid division by zero). The set of points z with |z|=1 (the unit circle) contains infinitely many elements. Excluding a single point (z=−1) still leaves infinitely many elements. Therefore, statement (S2) claims "infinitely many elements", which is true.

Step 3: Conclusion○ Expand

Since Statement (S1) is incorrect and Statement (S2) is correct, the option "only (S2) is correct" is the right answer.

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