Let w = 3 + i 2 and P = { w n : n = 1 , 2 , 3 , … } . Further, H 1 = { z : Re ( z ) 1 2 } and H 2 = { z : Re ( z ) 1 2 } , where c is the set of all complex numbers. If z 1 P H 1 , z 2 P H 2 and O represents the origin, then z 1 O z 2

Question

Let $w = \frac{\sqrt{3} + i}{2}$ and $P = {w^{n} : n = 1, 2, 3, \dots}$ . Further, $H_{1} = {z \in ℂ : Re (z) > \frac{1}{2}} and H_{2} = {z \in ℂ : Re (z) < - \frac{1}{2}},$

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 $∠ z_{1} O z_{2} =$ [2013]

Smart Achievers · Accepted Answer

(3, 4)

$We have w = \frac{\sqrt{3} + i}{2} = \cos \frac{π}{6} + i \sin \frac{π}{6}$

$\Rightarrow w^{n} = \cos \frac{n π}{6} + i \sin \frac{n π}{6}$

$∴$ $P contains all those points which lie on unit circle and have arguments \frac{π}{6}, \frac{2 π}{6}, \frac{3 π}{6} and so on.$

$Since, z_{1} \in P \cap H_{1} and z_{2} \in P \cap H_{2}, therefore z_{1} and z_{2} can have possible positions as shown in the figure.$

$∴$ $∠ z_{1} O z_{2} can be \frac{2 π}{3} or \frac{5 π}{6}$

Solution

1 $\frac{π}{2}$

2 $\frac{π}{6}$

3 $\frac{2 π}{3}$

4 $\frac{5 π}{6}$

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Solution

Q. Let w=3+i2 and P={wn:n=1,2,3,…}. Further, H1={z∈ℂ:Re(z)>12} and H2={z∈ℂ:Re(z)<-12}, where c is the set of all complex numbers. If z1∈P∩H1, z2∈P∩H2 and O represents the origin, then ∠z1Oz2= [2013]

1 π2

2 π6

3 2π3

4 5π6

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1 $\frac{π}{2}$

2 $\frac{π}{6}$

3 $\frac{2 π}{3}$

4 $\frac{5 π}{6}$