Discussion Forum

1)

Let   $w=\frac{\sqrt{3}+i}{2}$  and   $P= \left\{ W^{n}:n=1,2,3...\right\}$   Further

  $H_{1}= \left\{ z \epsilon C;Rez>\frac{1}{2}\right\}$     and

$H_{2}= \left\{ z \epsilon C;Rez<-\frac{1}{2}\right\}$

where C is the set of all complex numbers, if   $z_{1}\in P\cap H_{1},z_{2}\in P\cap H_{2}$    and O represents the origin, then   $\angle z_{1}Oz_{2}$ is equal to 

Answer:

Option C,D

Explanation:

Concept involved

 It is simple representation of points on argand plane and to find the angle between the points

 here,  $P=W^{n}=\left( \cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)^{n}$

  $= \cos\frac{n\pi}{6}+i\sin\frac{n\pi}{6}$

 $H_{1}=\left\{Z\in C: Re (z)>\frac{1}{2}\right\}$

 $\therefore$   $P\cap H_{1}$ represents those points for which $\cos\frac{n \pi}{6}$ is +ve

$\therefore$ it belongs to I to IV quadrant

$\Rightarrow$    $z_{1}=P\cap H_{1}=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}$

 or    $\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}$

 $\therefore$        $z_{1}=\frac{\sqrt{3}}{2}+\frac{i}{2} or\frac{\sqrt{3}}{2}-\frac{i}{2}$    .....(i)

 Similarly ,   $z_{2}=P\cap H_{2}$ i,e.   those points for which 

 $\cos \frac{n\pi}{6}<0$

 $\therefore$     $z_{2}=\cos\pi+i\sin\pi,\cos\frac{5\pi}{6}$

                                     $+i\sin\frac{5\pi}{6},\frac{\cos 7\pi}{6}+i\sin\frac{7\pi}{6}$

 $\Rightarrow z_{2}=-1, \frac{-\sqrt{3}}{2}+\frac{i}{2},\frac{-\sqrt{3}}{2}-\frac{i}{2}$

 Thus,  $\angle z{1}Oz_{2}=\frac{2\pi}{3},\frac{5\pi}{6},\pi$