JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Complex Numbers

Since $|z|=1$, write $z=e^{i\theta }$, so $\bar{z}=e^{-i\theta }$ and


The given condition becomes


For $\cos 2\theta =\frac{1}{2}$: in $[0,2\pi )$, $2\theta \in \{\frac{\pi }{3},\frac{5\pi }{3},\frac{7\pi }{3},\frac{11\pi }{3}\}$, giving $\theta \in \{\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\}$ — 4 values.

For $\cos 2\theta =-\frac{1}{2}$: in $[0,2\pi )$, $2\theta \in \{\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}\}$, giving $\theta \in \{\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}\}$ — 4 more values.

In total there are 8 distinct values of $\theta$, hence 8 complex numbers $z$.