The value of Σ k=1 to 10 (sin 2πk/11 + i cos 2πk/11) is (A) i (B) 1 (C) -1 (D) -i

Q: Wait, why do we factor out $i$? Can't we just use the formula directly?

Great question! The formula $e^{i\theta} = \cos \theta + i \sin \theta$ is rigid. If you try to use it on $\sin \theta + i \cos \theta$, you will get the wrong phase. Factoring out $i$ is the algebraic 'key' that unlocks the standard form, allowing us to use the power of Euler's identity.

Q: Why is the sum of all $n^{th}$ roots of unity zero?

Think geometrically! The $n^{th}$ roots of unity are vectors pointing from the origin to the vertices of a regular polygon inscribed in the unit circle. By symmetry, they perfectly balance each other out. The center of mass of these vertices is the origin, so their vector sum must be zero.

Q: Why did we split the sum at $k=0$?

The theorem states that the sum of *all* roots (from $k=0$ to $n-1$) is zero. Our problem only asks for the sum from $k=1$ to $10$. By adding and subtracting the $k=0$ term, we bridge the gap between our specific problem and the powerful theorem we know. It is a classic 'add and subtract' trick used to complete a series.

Question

The value of $\sum_{k=1}^{10} \left(\sin \frac{2\pi k}{11} + i \cos \frac{2\pi k}{11}\right)$ is

Options:

(A) i

(B) 1

(C) -1

(D) -i

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dissect a problem that, at first glance, seems like a standard summation, but it is actually a masterclass in the geometry of complex numbers. We are tasked with evaluating the sum:

$$S = \sum_{k=1}^{10} \left(\sin \frac{2\pi k}{11} + i \cos \frac{2\pi k}{11}ight)$$

The first thing you might notice is a slight sense of unease. We are all trained to look for the beautiful Euler form, $\cos 	heta + i \sin 	heta$. But here, the imaginary unit $i$ is attached to the cosine term! This is a classic JEE trap designed to test your attention to detail.

### The Algebraic Transformation

To fix this, we must force the expression into the correct form. We factor out $i$ from the entire bracket. When we pull $i$ outside, the cosine term becomes isolated, which is exactly what we want.

Since $\frac{1}{i} = -i$, the sine term transforms accordingly. Our expression becomes:

$$S = i \sum_{k=1}^{10} \left(\cos \frac{2\pi k}{11} - i \sin \frac{2\pi k}{11}ight)$$

Suddenly, the fog clears. We have successfully manipulated the expression into a form where we can apply Euler's formula, $e^{-i	heta} = \cos 	heta - i \sin 	heta$. By substituting our angle $	heta = \frac{2\pi k}{11}$, the expression condenses into:

$$S = i \sum_{k=1}^{10} e^{-i \frac{2\pi k}{11}}$$

### The Geometry of Unity

Now, let us pause and visualize what these terms represent. These are the $11^{th}$ roots of unity. Because of the minus sign, they are technically the complex conjugates, but they form the exact same set of eleven points spaced equally around the unit circle on the complex plane.

Imagine them as vertices of a regular eleven-sided polygon. There is a powerful theorem you must remember: the sum of all $n^{th}$ roots of unity is always exactly zero. Because they are symmetrically distributed around the origin, their vectors perfectly cancel each other out.

### The Final Piece of the Puzzle

Let us write down this complete sum mathematically:

$$\sum_{k=0}^{10} e^{-i \frac{2\pi k}{11}} = 0$$

Do not rush, as our summation starts from $k=1$, not $k=0$. We must separate the first term ($k=0$) from the rest of the summation. When $k=0$, the angle is zero, which corresponds to the root $e^0 = 1$.

Thus, we have:

$$1 + \sum_{k=1}^{10} e^{-i \frac{2\pi k}{11}} = 0$$

To isolate the summation from $k=1$ to $10$, we shift the $1$ to the right side, yielding a sum of $-1$. Finally, we multiply this result by the $i$ we factored out at the beginning:

$$S = i 	imes (-1) = -i$$

Through careful manipulation and geometric insight, we have arrived at our answer. The final result is **$-i$**.

Animated Solution for Mathematics - Complex Numbers: The value of $\sum_{k = 1}^{10} (sin \frac{2 π k}{11} + i cos \frac{2 π k}{11})$ is

Select Answer:

Visualized Solution

Analyze the Summation

Factor out $i$

Simplify $\frac{1}{i}$

Apply Euler's Formula

Roots of Unity Concept

Sum of all $n^{t h}$ Roots

Write the Complete Sum

Split the Summation

Isolate the Required Sum

Final Substitution

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Algebraic Transformation

The Geometry of Unity

The Final Piece of the Puzzle

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ∑k=110​(sin112πk​+icos112πk​) is