For any integer k, let αk = cos(kπ/7) + i sin(kπ/7), where i = √-1. The value of the expression (Σ k

Q: Wait, why is the distance between $\alpha_{k+1}$ and $\alpha_k$ constant? Shouldn't it change as we move around the circle?

That is a brilliant intuition, but look closer at the geometry! Because the points $\alpha_k$ are equally spaced by an angle of $\frac{\pi}{7}$ along the circumference, the chord connecting any two consecutive points subtends the same central angle of $\frac{\pi}{7}$. In a circle, equal central angles always subtend equal chords. Algebraically, the factor $e^{i\frac{k\pi}{7}}$ that we pull out has a modulus of 1, which effectively 'rotates' the chord but preserves its length.

Q: How do I derive $|e^{i\theta} - 1| = 2\sin(\frac{\theta}{2})$? I always forget this identity.

Don't memorize it—derive it! Write $e^{i\theta} - 1$ as $(\cos\theta - 1) + i\sin\theta$. Use the half-angle identities: $\cos\theta - 1 = -2\sin^2(\frac{\theta}{2})$ and $\sin\theta = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$. Factor out $2\sin(\frac{\theta}{2})$ and you'll see the magnitude is simply $2\sin(\frac{\theta}{2})$.

Q: Does the index $4k-1$ and $4k-2$ in the denominator change the distance? It looks like a different chord.

It looks intimidating, but test it! For $k=1$, the indices are $3$ and $2$. For $k=2$, they are $7$ and $6$. In every single case, the difference between the indices is $(4k-1) - (4k-2) = 1$. Since the difference is always 1, you are always calculating the distance between consecutive points. The 'gap' is constant, so the chord length is constant.

Question

For any integer $k$, let $\alpha_k = \cos\left(\frac{k\pi}{7}\right) + i \sin\left(\frac{k\pi}{7}\right)$, where $i = \sqrt{-1}$. The value of the expression $\frac{\sum_{k=1}^{12} |\alpha_{k+1} - \alpha_k|}{\sum_{k=1}^3 |\alpha_{4k-1} - \alpha_{4k-2}|}$ is .........

Sigma Percentile · Accepted Answer

### The Geometry of Complex Numbers: A Journey into Symmetry

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a hidden symmetry.

When you first look at the expression involving $\alpha_k = \cos\left(\frac{k\pi}{7}ight) + i \sin\left(\frac{k\pi}{7}ight)$, it is easy to feel overwhelmed. You see summations, complex indices, and a ratio that looks like it might lead to a nightmare of trigonometric expansion.

But I want you to take a deep breath. In JEE Advanced, the most complex-looking problems often hide the most elegant, simple truths. Let us peel back the layers together.

### Phase 1: The Unit Circle Visualization

First, let us demystify $\alpha_k$. By Euler's formula, we know that $\alpha_k = e^{i\frac{k\pi}{7}}$.

This is not just an algebraic expression; it is a geometric instruction. It tells us that every single $\alpha_k$ lies on the unit circle in the complex plane. The angle of each point is simply $	heta_k = \frac{k\pi}{7}$.

Imagine standing at the origin and watching these points appear one by one on the circle. They are like beads on a necklace, perfectly spaced. Because the angle between any $\alpha_{k+1}$ and $\alpha_k$ is constant—specifically, $\frac{(k+1)\pi}{7} - \frac{k\pi}{7} = \frac{\pi}{7}$—the distance between any two consecutive beads is identical.

This is our first breakthrough: the distance is invariant.

### Phase 2: The Algebra of the Chord

Now, let us tackle the term $|\alpha_{k+1} - \alpha_k|$. In the complex plane, the modulus of the difference of two numbers is the Euclidean distance between them. We are looking for the length of the chord connecting these points.

Let us write it out:

$$|\alpha_{k+1} - \alpha_k| = |e^{i\frac{(k+1)\pi}{7}} - e^{i\frac{k\pi}{7}}|$$

Here is where the magic happens. We factor out the common term $e^{i\frac{k\pi}{7}}$:

$$|\alpha_{k+1} - \alpha_k| = |e^{i\frac{k\pi}{7}} \cdot (e^{i\frac{\pi}{7}} - 1)|$$

Using the property that the modulus of a product is the product of the moduli, and knowing that $|e^{i\frac{k\pi}{7}}| = 1$, the variable $k$ simply vanishes! We are left with $|e^{i\frac{\pi}{7}} - 1|$.

This is a constant. It does not matter if $k$ is 1, 10, or 100; the distance between consecutive points is always the same. Using the half-angle identity, we find this distance is $2\sin\left(\frac{\pi}{14}ight)$.

### Phase 3: The Summation and the Cancellation

Now, look at the numerator: $\sum_{k=1}^{12} |\alpha_{k+1} - \alpha_k|$. We are summing the same constant value 12 times.

The numerator becomes:

$$12 	imes 2\sin\left(\frac{\pi}{14}ight)$$

Now, look at the denominator: $\sum_{k=1}^3 |\alpha_{4k-1} - \alpha_{4k-2}|$. Do not let the indices $4k-1$ and $4k-2$ scare you.

Calculate the difference between the indices: $(4k-1) - (4k-2) = 1$. Every single term in this sum is also the distance between consecutive points! So, we are summing the same constant value 3 times.

The denominator becomes:

$$3 	imes 2\sin\left(\frac{\pi}{14}ight)$$

### The Final Victory

When we place these into our ratio, the trigonometric term $2\sin\left(\frac{\pi}{14}ight)$—which we spent time deriving—simply cancels out.

We are left with:

$$\frac{12}{3} = 4$$

This is the beauty of JEE Advanced mathematics. It rewards those who look past the complexity to find the underlying symmetry. You didn't need to calculate sines or cosines; you needed to understand the geometry of the circle.

You have conquered this problem not by brute force, but by insight. Keep this mindset, and no problem will ever be too difficult for you. The final answer is **4**.

Enter Numerical Value:

Visualized Solution

Visualizing $α_{k}$ on the Unit Circle

Geometric Meaning of $∣ α_{k + 1} - α_{k} ∣$

Algebraic Setup of the Chord Length

Factoring the Exponential Expression

Simplifying the Modulus

Applying the Half-Angle Formula

Analyzing the Numerator Sum

Analyzing the Denominator Sum

Calculating the Final Ratio

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

Phase 1

Phase 2

Phase 3

The Final Victory

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