Let ω = (√3 + i)/2 and P = {ωⁿ : n = 1, 2, 3, ...}. Further H1 = {z ∈ ℂ : Re z 1/2} and H2 = {z ∈

Q: Wait, why do we represent $\omega$ as $e^{i\pi/6}$?

Great observation! Any complex number $z = x + iy$ can be written in polar form as $r(\cos\theta + i\sin\theta)$, where $r = |z|$ and $\theta = \text{arg}(z)$. For $\omega = \frac{\sqrt{3}}{2} + i\frac{1}{2}$, the modulus is $\sqrt{(\sqrt{3}/2)^2 + (1/2)^2} = 1$. The angle $\theta$ satisfies $\cos\theta = \sqrt{3}/2$ and $\sin\theta = 1/2$, which corresponds to $\pi/6$ radians. Using Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta$, we get $\omega = e^{i\pi/6}$.

Q: How do we know the set $P$ only contains 12 points?

The set $P$ is defined by powers of $\omega$. Since $\omega = e^{i\pi/6}$, then $\omega^n = e^{in\pi/6}$. This is a periodic sequence. Specifically, $\omega^{12} = e^{i12\pi/6} = e^{i2\pi} = 1$. Any power beyond $n=12$ will just repeat the values already found (e.g., $\omega^{13} = \omega^{12} \cdot \omega = 1 \cdot \omega = \omega$). Thus, there are exactly 12 distinct points evenly spaced around the unit circle.

Q: Why are there multiple possible answers for the angle?

The problem states $z_1 \in P \cap H_1$ and $z_2 \in P \cap H_2$. Because there are multiple points in $P$ that satisfy the conditions for $H_1$ and $H_2$, there are multiple valid pairs of $(z_1, z_2)$. Each pair subtends a specific angle at the origin. The question asks for the angle $\angle z_1Oz_2$, and since the options provided are specific values, we must identify which of our calculated angles match the given choices.

Question

Let $\omega = \frac{\sqrt{3} + i}{2}$ and $P = \{\omega^n : n = 1, 2, 3, \dots\}$. Further $H_1 = \{z \in \mathbb{C} : Re z > 1/2\}$ and $H_2 = \{z \in \mathbb{C} : Re z < -1/2\}$, where $\mathbb{C}$ 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_1Oz_2 = $

Options:

(A) $\pi/2$

(B) $\pi/6$

(C) $2\pi/3$

(D) $5\pi/6$

Sigma Percentile · Accepted Answer

### The Geometry of Complex Rotations Welcome, fellow traveler on the path of JEE mastery. Today, we are not just solving a problem; we are embarking on a journey through the elegant, circular world of complex numbers. Imagine you are standing at the origin of the Argand plane, looking out at a unit circle. This is where our story begins. ### Decoding the Complex Number $\omega$ We are given: $$\omega = \frac{\sqrt{3} + i}{2}$$ At first glance, this might look like just another fraction, but look closer. It is $\frac{\sqrt{3}}{2} + i\frac{1}{2}$. These are the cosine and sine values of $30^\circ$, or $\pi/6$ radians. By Euler's beautiful identity, we can write this as: $$\omega = e^{i\pi/6}$$ This is not just a number; it is a rotation operator. Every time we multiply by $\omega$, we are effectively rotating a vector by $30^\circ$ counter-clockwise on the unit circle. ### The Clockwork of Set $P$ Now, consider the set $P = \{\omega^n : n = 1, 2, 3, \dots\}$. Because $\omega = e^{i\pi/6}$, the $n$-th power is simply $e^{in\pi/6}$. This is like a clock with 12 positions. As $n$ increases, we step around the unit circle in increments of $30^\circ$. Since $12 imes 30^\circ = 360^\circ$, or $2\pi$, the sequence repeats every 12 steps. Thus, $P$ consists of exactly 12 distinct points: $$P = \{e^{i\pi/6}, e^{i2\pi/6}, \dots, e^{i12\pi/6} = 1\}$$ ### The Half-Plane Constraints Next, we encounter the regions $H_1$ and $H_2$. $H_1$ is the set of complex numbers where $ ext{Re}(z) > 1/2$. Geometrically, this is the vertical strip to the right of the line $x = 1/2$. On the unit circle, the real part is $\cos( heta)$, so we require $\cos( heta) > 1/2$. This inequality holds when $ heta$ is between $-\pi/3$ and $\pi/3$. Looking at our 12 points, only $e^{i\pi/6}$ ($30^\circ$) and $e^{-i\pi/6}$ ($330^\circ$) fall into this region. These are our candidates for $z_1$. Similarly, $H_2$ is defined by $ ext{Re}(z) < -1/2$, which means $\cos( heta) < -1/2$. This occurs when $ heta$ is between $2\pi/3$ ($120^\circ$) and $4\pi/3$ ($240^\circ$). The points in $P$ that satisfy this are $e^{i5\pi/6}$ ($150^\circ$), $e^{i\pi}$ ($180^\circ$), and $e^{i7\pi/6}$ ($210^\circ$). These are our candidates for $z_2$. ### The Final Synthesis We are looking for the angle $\angle z_1Oz_2$, which is the difference in arguments between our chosen $z_1$ and $z_2$. Let's test the pairs. If $z_1 = e^{i\pi/6}$ ($30^\circ$), then pairing with $z_2$ gives angles of: $$|150^\circ - 30^\circ| = 120^\circ ext{ (or } 2\pi/3)$$ $$|180^\circ - 30^\circ| = 150^\circ ext{ (or } 5\pi/6)$$ $$|210^\circ - 30^\circ| = 180^\circ ext{ (or } \pi)$$ If $z_1 = e^{-i\pi/6}$ ($-30^\circ$), we get similar results. The possible angles are **$2\pi/3$, $5\pi/6$, and $\pi$**. You have successfully navigated the geometry of the complex plane!

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Visualized Solution

$A na l y z in g ω$

$P o l a r F or m o f ω$

$T h e S e tP$

$E l e m e n t so f P$

$R e g i o n H_{1}$

$V i s u a l i z in g H_{1}$

$F in d in g z_{1}$

$R e g i o n H_{2}$

$V i s u a l i z in g H_{2}$

$F in d in g z_{2}$

$A n g l e ∠ z_{1} O z_{2}$

$V ec t or s z_{1} an d z_{2}$

$C a l c u l a t in g A n g l es : C a se 1$

$C a l c u l a t in g A n g l es : C a se 2$

$F ina l A n s w er$

The Sigma Insight: Argand Plane and Polar Representation

The Geometry of Complex Rotations

Decoding the Complex Number $ω$

The Clockwork of Set $P$

The Half-Plane Constraints

The Final Synthesis

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzingω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PolarFormofω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheSetP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ElementsofP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RegionH1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VisualizingH1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingz1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RegionH2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VisualizingH2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingz2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle∠z1​Oz2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vectorsz1​andz2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingAngles:Case1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingAngles:Case2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalAnswer