Let w⁡ = 3 + i 2 and P = w⁡ n : n = 1, 2, 3, … .Further H 1 = z ∈ C : Re z > 1 2 and H 2 = z ∈ C : 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 =

JEE Advanced · Mathematics · 3. Complex Numbers

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 $∠ z_{1} O z_{2} =$

A $\frac{π}{2}$
B $\frac{π}{6}$
C $\frac{2 π}{3}$
D $\frac{5 π}{6}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{5 π}{6}$

Step-by-step Solution

Detailed explanation

P = ω^{n} = cos \frac{n π}{6} + i sin \frac{n π}{6}, H_{1} = Rez > 1 / 2

z_{1} = P \cap H_{1} = \frac{\sqrt{3} + i}{2}, \frac{\sqrt{3} - i}{2}

z_{2} = P \cap H_{2} = - 1, \frac{- \sqrt{3} + i}{2}, \frac{- \sqrt{3} - i}{2}

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List-I with List - II and select the correct answer using the code given below the lists

	List-I		List-II
A.	Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $2$ . Then the volume of the parallelepiped determined by vectors $2 (\vec{a} \times \vec{b}), 3 (\vec{b} \times \vec{c})$ and $(\vec{c} \times \vec{a})$ is	P.	$100$
B.	Volume of parallel piped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $5$ . Then the volume of the parallelepiped determined by vectors $3 (\vec{a} + \vec{b}), (\vec{b} + \vec{c})$ and $2 (\vec{c} + \vec{a})$ is	Q.	$30$
C	Area of a triangle with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $20$ .Then the area of the triangle with adjacent sides determined by vectors $(2 \vec{a} + 3 \vec{b})$ and $(\vec{a} - \vec{b})$ is	R.	$24$
D	Area of a parallelogram with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $30$ . Then the area of the parallelogram with adjacent sides determined by vectors $(\vec{a} + \vec{b})$ and $\vec{a}$ is	S.	$60$

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From JEE Advanced

LIST-

I

contains metal species and LIST-

II

contains their properties.

List - I	List - II
(I) $\left[ Cr ( CN )_6\right]^{4-}$	(P) $t _{2 g}$ orbitals contain 4 electrons
(II) $\left[ RuCl _6\right]^{2-}$	(Q) $\mu$ spin - only $=4.9$ BM
(III) $\left[ Cr \left( H _2 O \right)_6\right]^{2+}$	(R) low spin complex ion
(IV) $\left[ Fe \left( H _2 O \right)_6\right]^{2+}$	(S) metal ion in 4 + oxidation state
	(T) $d ^4$ species

[Given: Atomic number of

Cr = 24, Ru = 44, Fe = 26

]
Match each metal species in LIST-

I

with their properties in LIST-

II

, and choose the correct option

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List - I	List - II
(I) \(\left[ Cr ( CN )_6\right]^{4-}\)	(P) \(t _{2 g}\) orbitals contain 4 electrons
(II) \(\left[ RuCl _6\right]^{2-}\)	(Q) \(\mu\) spin - only \(=4.9\) BM
(III) \(\left[ Cr \left( H _2 O \right)_6\right]^{2+}\)	(R) low spin complex ion
(IV) \(\left[ Fe \left( H _2 O \right)_6\right]^{2+}\)	(S) metal ion in 4 + oxidation state
	(T) \(d ^4\) species