SATHEE: Complex Numbers 4 Question 8

Complex Numbers 4 Question 8

8.

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 $∠ z_{1} O z_{2}$ is equal to

(2013 JEE Adv.)

(a) $\frac{π}{2}$

(b) $\frac{π}{6}$

(d) $\frac{5 π}{6}$

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Answer:

Correct Answer: 8. (c, d)

Solution:

PLAN: It is the simple representation of points on Argand plane and to find the angle between the points

Here, $P = W^{n} = (\cos \frac{π}{6} + i \sin {\frac{π}{6}}^{n}) = \cos \frac{n π}{6} + i \sin \frac{n π}{6}$

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

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

Hence, it belongs to I or IV quadrant.

$\begin{aligned} \Rightarrow z_{1} = P \cap H_{1} = \cos \frac{π}{6} + i \sin \frac{π}{6} or \cos \frac{11 π}{6} + i \sin \frac{11 π}{6} \\ ∴ z_{1} = \frac{\sqrt{3}}{2} + \frac{i}{2} or \frac{\sqrt{3}}{2} - \frac{i}{2} \end{aligned}$

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

$\cos \frac{n π}{6} < 0$

$∴ z_{2} = \cos π + i \sin π, \cos \frac{5 π}{6} + i \sin \frac{5 π}{6}, \frac{\cos 7 π}{6}$

$+ i \sin \frac{7 π}{6}$

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

Thus, $∠ z_{1} O z_{2} = \frac{2 π}{3}, \frac{5 π}{6}, π$

Complex Numbers 4 Question 8

8.

Answer:

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Complex Numbers 4 Question 8

8.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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