Q.Let $z$ satisfy $|z| = 1$ and $\operatorname{Im}\left( \frac{z - 1}{z + 1} \right) = 0$ . Then $z$ lies on

the real axis

the imaginary axis

the line

y = 1

the line

x = 1

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.Let zzz satisfy ∣z∣=1|z| = 1∣z∣=1 and Im⁡(z−1z+1)=0\operatorname{Im}\left( \frac{z - 1}{z + 1} \right) = 0Im(z+1z−1​)=0. Then zzz lies on