JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Complex Numbers

First, note that $w=\frac{\sqrt{3}+\mathrm{i}}{2}=e^{i\pi /6}$ has modulus 1 and argument $\pi /6$.

Powers of $w$ lie on the unit circle at angles that are multiples of $\pi /6$: $\pi /6,\pi /3,\pi /2,2\pi /3,5\pi /6,\pi ,7\pi /6,\dots$

For $H_1$: $\mathrm{Re}(z)>1/2$ means $\cos (\theta )>1/2$, so $\theta \in (-\pi /3,\pi /3)$. From $P$, this gives points at angles $\pi /6$ (and also $-\pi /6=11\pi /6$ if we consider negative powers).

For $H_2$: $\mathrm{Re}(z)<-1/2$ means $\cos (\theta )<-1/2$, so $\theta \in (2\pi /3,4\pi /3)$. From $P$, this gives points at angles $2\pi /3,5\pi /6,\pi ,7\pi /6$.

The angle between $z_1$ and $z_2$ can be:

• Between $\pi /6$ and $2\pi /3$: angle = $2\pi /3-\pi /6=\pi /2$
• Between $\pi /6$ and $5\pi /6$: angle = $5\pi /6-\pi /6=2\pi /3$
• Between $\pi /6$ and $\pi$: angle = $\pi -\pi /6=5\pi /6$

Therefore, possible values are $\frac{2\pi }{3}$ and $\frac{5\pi }{6}$.