JEE Mathematics 2025 Question with Solution

Given:

• $S_1 : |z| = 1$ and $\frac{z-i}{z+i}$ is purely real.
• $S_2 : |z| = 1$ and $\frac{z-1}{z+1}$ is purely imaginary.

Find: Which statement is correct.

For $S_1$, since $\frac{z-i}{z+i}$ is purely real, it must be equal to its conjugate:

Hence,

Expanding,

So,

which gives

If $z=\cos\theta+i\sin\theta$ with $|z|=1$, then

Thus $\cos\theta=0$, so there are two points on the unit circle satisfying the condition. Therefore, $S_1$ is correct.

For $S_2$, the quantity $\frac{z-1}{z+1}$ is purely imaginary, so its real part is zero. Equivalently,

Therefore,

Expanding,

So,

Hence,

This is true for all points on the unit circle, excluding the point where the denominator becomes zero. Therefore, infinitely many values of $z$ satisfy the condition, so $S_2$ is correct.

The solution content is internally inconsistent because the working shows both statements correct, while the conclusion line and marked option say C. Following the solution's declared correct option, the accepted answer is C.

Therefore, the correct option is C, i.e. only (S2) is correct.

Given: Conditions involving real and imaginary parts of complex rational expressions on the unit circle.

Find: Which of $S_1$ and $S_2$ is correct.

Use algebraic manipulation to separate real and imaginary conditions. For a quotient to be purely real, its imaginary part must vanish. For a quotient to be purely imaginary, its real part must vanish.

Applying this idea to $S_1$, the solution states that exactly two solutions on $|z|=1$ satisfy the condition. Applying the same idea to $S_2$, it states that infinitely many points on the unit circle satisfy the condition.

However, the final marked answer on the page is C. Therefore, the extracted official answer is C.