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Question #40671

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha = e^{i \frac{2\pi}{11}}$ . Evaluate the sum $S = \sum_{k=1}^{10} \sqrt{ (\alpha^k - \alpha^{k+1})^2 }$ where $\sqrt{z}$ denotes the principal square root of $z$ (i.e., for $z = re^{i\theta}$ with $\theta \in (-\pi, \pi]$ , $\sqrt{z} = \sqrt{r}e^{i\theta/2}$ ).

Options

A.
$\alpha - 1$
B.
$\alpha - 2\alpha^6 + 1$
C.
$\alpha - 2\alpha^7 + 1$
D.
$2\alpha^6 - \alpha - 1$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

The Telescopic Series Trap|Roots of Unity Properties|Square Root of Complex Numbers

Why This Question Is Tricky

Common Mistakes

This question tests the 'Ghost Sum' trap in telescopic series involving complex numbers.

Students often blindly simplify $\sqrt{Z^2} = Z$ , leading to a standard telescoping sum $\alpha - 1$ .

However, the principal square root of a complex number requires strict branch cut analysis.

Because the argument of the terms rotates cyclically with the roots of unity, the principal square root forces a sign change halfway through the series, breaking the continuous telescoping nature and requiring the sum to be split into two distinct collapsing parts.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	The Telescopic Series Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free