JEE Main 2025 Apr 7 Shift 1, Mathematics Q11: Introduction to Complex Numbers And Their Conjugates

Analysing $(S1)$: The expression $\frac{z-i}{z+i}$ is real $\iff \frac{z-i}{z+i}=\overline{\left(\frac{z-i}{z+i}\right)}=\frac{\bar{z}+i}{\bar{z}-i}$.

Cross-multiplying: $(z-i)(\bar{z}-i)=(z+i)(\bar{z}+i)$.

Expanding: $|z{|}^2-i(z+\bar{z})-1=|z{|}^2+i(z+\bar{z})-1$.

So $2i(z+\bar{z})=0\Rightarrow z+\bar{z}=0\Rightarrow \text{Re}(z)=0$.

Combined with $|z|=1$: $z=\pm i$. But $z=-i$ is excluded. So the set is $\{i\}$, which has exactly one element, not two.

$\therefore (S1)$ is incorrect.

Analysing $(S2)$: The expression $\frac{z-1}{z+1}$ is purely imaginary $\iff \frac{z-1}{z+1}+\overline{\left(\frac{z-1}{z+1}\right)}=0$.

$(z-1)(\bar{z}+1)+(z+1)(\bar{z}-1)=0$

Expanding: $[|z{|}^2+z-\bar{z}-1]+[|z{|}^2-z+\bar{z}-1]=0\Rightarrow 2|z{|}^2-2=0\Rightarrow |z{|}^2=1$.

So every $z$ on the unit circle (except $z=-1$) satisfies this; that is infinitely many points.

$\therefore (S2)$ is correct.

Overall: only $(S2)$ is correct $\Rightarrow$ Option (3).