Ratio of Modulus Sums of Complex Roots of Unity

To find the value of the given expression, we first analyze the definition of $\alpha_k$: $\alpha_k = \cos\left(\frac{k\pi}{7}\right) + i \sin\left(\frac{k\pi}{7}\right) = e^{i \frac{k\pi}{7}}$

Since $|e^{i\theta}| = 1$ for any real $\theta$, the magnitude of $\alpha_k$ is always $1$.

Step 1: Simplify the general term $|\alpha_{m+1} - \alpha_m|$

For any integer $m$, the difference between consecutive terms can be written as: $|\alpha_{m+1} - \alpha_m| = \left| e^{i \frac{(m+1)\pi}{7}} - e^{i \frac{m\pi}{7}} \right|$

Factoring out $e^{i \frac{m\pi}{7}}$: $|\alpha_{m+1} - \alpha_m| = \left| e^{i \frac{m\pi}{7}} \left( e^{i \frac{\pi}{7}} - 1 \right) \right| = \left| e^{i \frac{m\pi}{7}} \right| \cdot \left| e^{i \frac{\pi}{7}} - 1 \right| = 1 \cdot \left| e^{i \frac{\pi}{7}} - 1 \right|$

Using the trigonometric identity $|e^{i\theta} - 1| = 2\sin\left(\frac{\theta}{2}\right)$: $|\alpha_{m+1} - \alpha_m| = 2 \sin\left(\frac{\pi}{14}\right)$

This value is constant and completely independent of $m$.

Step 2: Calculate the Numerator

The numerator is given by: $N = \sum_{k=1}^{12} |\alpha_{k+1} - \alpha_k|$

Since each term in the sum is equal to $\left| e^{i \frac{\pi}{7}} - 1 \right|$, and there are $12$ terms in the summation: $N = 12 \cdot \left| e^{i \frac{\pi}{7}} - 1 \right| = 24 \sin\left(\frac{\pi}{14}\right)$

Step 3: Calculate the Denominator

The denominator is given by: $D = \sum_{k=1}^{3} |\alpha_{4k-1} - \alpha_{4k-2}|$

Notice that the difference in indices for each term is $(4k-1) - (4k-2) = 1$. By setting $m = 4k-2$, we apply the same simplification from Step 1: $|\alpha_{4k-1} - \alpha_{4k-2}| = \left| e^{i \frac{\pi}{7}} - 1 \right| = 2 \sin\left(\frac{\pi}{14}\right)$

Since there are $3$ terms in this summation: $D = 3 \cdot \left| e^{i \frac{\pi}{7}} - 1 \right| = 6 \sin\left(\frac{\pi}{14}\right)$

Step 4: Evaluate the Ratio

Dividing the numerator by the denominator: $\frac{\sum_{k=1}^{12} |\alpha_{k+1} - \alpha_k|}{\sum_{k=1}^{3} |\alpha_{4k-1} - \alpha_{4k-2}|} = \frac{12 \cdot \left| e^{i \frac{\pi}{7}} - 1 \right|}{3 \cdot \left| e^{i \frac{\pi}{7}} - 1 \right|} = \frac{12}{3} = 4$