The value of $\sum_{k=1}^6 \left(\sin \frac{2\pi k}{7} - i \cos \frac{2\pi k}{7}\right)$ is Options: (A) -1 (B) 0 (C) -i (D) i (E) None

### Analyzing the Setup The problem asks us to evaluate the sum: $$\sum_{k=1}^{6} \left( \sin \frac{2\pi k}{7} - i \cos \frac{2\pi k}{7} \right)$$ At first glance, this expression deviates from the standard polar form of a complex number, $z = \cos \theta + i \sin \theta$. The real and imaginary parts are swapped, and the signs are unconventional. ### Unmasking the Form To simplify, we factor out $-i$ from the expression: $$\sin \theta - i \cos \theta = -i \left( \cos \theta + \frac{1}{-i} \sin \theta \right)$$ Since $\frac{1}{-i} = i$, the expression simplifies beautifully to: $$-i (\cos \theta + i \sin \theta) = -i e^{i \theta}$$ Here, $\theta = \frac{2\pi k}{7}$. Thus, our summation transforms into: $$-i \sum_{k=1}^{6} e^{i \frac{2\pi k}{7}}$$ ### The Power of Euler Let us define $\omega = e^{i \frac{2\pi}{7}}$. This $\omega$ represents the fundamental $7^{th}$ root of unity. The summation now takes the elegant form: $$-i \sum_{k=1}^{6} \omega^k$$ ### The Geometry of the Heptagon The $7^{th}$ roots of unity are $1, \omega, \omega^2, \omega^3, \omega^4, \omega^5, \omega^6$. These points form the vertices of a regular heptagon inscribed in the unit circle. Due to the perfect rotational symmetry of these roots, their sum is zero: $$1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0$$ Our target summation, $\sum_{k=1}^{6} \omega^k$, represents the sum of all these roots excluding the root $1$. Therefore: $$1 + \sum_{k=1}^{6} \omega^k = 0 \implies \sum_{k=1}^{6} \omega^k = -1$$ ### Final Calculation We now multiply the result of the summation by the constant $-i$ that we factored out earlier. The final result is: $$-i \times (-1) = \mathbf{i}$$ By recognizing the symmetry of the roots of unity, we have bypassed tedious trigonometric expansion to arrive at the solution **$i$**.

JEE Advanced 1987

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: The value of $\sum_{k = 1}^{6} (sin \frac{2 π k}{7} - i cos \frac{2 π k}{7})$ is

Q: Wait, why do we factor out $-i$ instead of just $i$?

Great observation! We factor out $-i$ specifically to force the expression into the standard Euler form, which is $\cos \theta + i \sin \theta$. If you factor out $i$, you get $i(-i \sin \theta - \cos \theta)$, which doesn't immediately look like $e^{i\theta}$. By factoring out $-i$, the real part becomes $\cos \theta$ and the imaginary part becomes $i \sin \theta$, which is exactly what we need to use Euler's formula.

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

Think of the $n^{th}$ roots of unity as vectors pointing from the origin to the vertices of a regular $n$-sided polygon inscribed in the unit circle. Because this polygon is perfectly symmetric about the origin, the vector sum of all these vertices must cancel out to zero. It is a beautiful consequence of rotational symmetry in the complex plane!

Select Answer:

Visualized Solution

Analyzing the Given Series

We need to evaluate: $\sum_{k = 1}^{6} (sin \frac{2 π k}{7} - i cos \frac{2 π k}{7})$
Notice the terms inside the bracket. They look like a complex number in polar form, but the real part is sine and the imaginary part is cosine.
Standard polar form is $cos θ + i sin θ$ .

Factoring out $- i$

Let $θ = \frac{2 π k}{7}$ . The term is $sin θ - i cos θ$ .
We can factor out $- i$ to rearrange the terms.
$- i (cos θ + \frac{1}{- i} sin θ)$
Since $\frac{1}{- i} = i$ , this becomes $- i (cos θ + i sin θ)$ .

Applying Euler's Formula

Using Euler's formula: $e^{i θ} = cos θ + i sin θ$
Substitute $θ = \frac{2 π k}{7}$ :
The term becomes $- i e^{i \frac{2 π k}{7}}$
The summation is now: $- i \sum_{k = 1}^{6} e^{i \frac{2 π k}{7}}$

Introducing Roots of Unity

Let $ω = e^{i \frac{2 π}{7}}$ . This is the fundamental $7^{t h}$ root of unity.
The terms in our sum are $ω^{k}$ for $k = 1, 2, \dots, 6$ .
Geometrically, these represent $6$ points on the unit circle in the complex plane.

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

The sum of all $n^{t h}$ roots of unity is always zero.
For $n = 7$ , the roots are $1, ω, ω^{2}, \dots, ω^{6}$ .
Therefore, $1 + ω + ω^{2} + ω^{3} + ω^{4} + ω^{5} + ω^{6} = 0$ .

Evaluating the Summation

We need the sum $S = \sum_{k = 1}^{6} ω^{k} = ω + ω^{2} + \dots + ω^{6}$ .
From the property: $1 + S = 0$ .
Rearranging this gives: $S = - 1$ .

Final Calculation

Our original expression was $- i \sum_{k = 1}^{6} ω^{k}$ .
Substitute the value of the sum we just found: $S = - 1$ .
Final Value $= - i \times (- 1) = i$ .
The correct option is $i$ .

00:00 / 00:00

The Sigma Insight: Cube Roots and nth Roots of Unity

Solution Diagram

Analyzing the Setup

The problem asks us to evaluate the sum:

k = 1 \sum 6 (sin \frac{2 π k}{7} - i cos \frac{2 π k}{7})

At first glance, this expression deviates from the standard polar form of a complex number,

z = cos θ + i sin θ

. The real and imaginary parts are swapped, and the signs are unconventional.

Unmasking the Form

To simplify, we factor out

- i

from the expression:

sin θ - i cos θ = - i (cos θ + \frac{1}{- i} sin θ)

Since

\frac{1}{- i} = i

, the expression simplifies beautifully to:

- i (cos θ + i sin θ) = - i e^{i θ}

Here,

θ = \frac{2 π k}{7}

. Thus, our summation transforms into:

- i k = 1 \sum 6 e^{i \frac{2 π k}{7}}

The Power of Euler

Let us define

ω = e^{i \frac{2 π}{7}}

. This

ω

represents the fundamental

7^{t h}

root of unity.

The summation now takes the elegant form:

- i k = 1 \sum 6 ω^{k}

The Geometry of the Heptagon

The

7^{t h}

roots of unity are

1, ω, ω^{2}, ω^{3}, ω^{4}, ω^{5}, ω^{6}

. These points form the vertices of a regular heptagon inscribed in the unit circle.

Due to the perfect rotational symmetry of these roots, their sum is zero:

1 + ω + ω^{2} + ω^{3} + ω^{4} + ω^{5} + ω^{6} = 0

Our target summation,

\sum_{k = 1}^{6} ω^{k}

, represents the sum of all these roots excluding the root

1

. Therefore:

1 + k = 1 \sum 6 ω^{k} = 0 ⟹ k = 1 \sum 6 ω^{k} = - 1

Final Calculation

We now multiply the result of the summation by the constant

- i

that we factored out earlier.

The final result is:

- i \times (- 1) = i

By recognizing the symmetry of the roots of unity, we have bypassed tedious trigonometric expansion to arrive at the solution $i$ .

Common Traps & FAQs

Wait, why do we factor out

- i

instead of just

i

Great observation! We factor out

- i

specifically to force the expression into the standard Euler form, which is

cos θ + i sin θ

. If you factor out

i

, you get

i (- i sin θ - cos θ)

, which doesn't immediately look like

e^{i θ}

. By factoring out

- i

, the real part becomes

cos θ

and the imaginary part becomes

i sin θ

, which is exactly what we need to use Euler's formula.

Why is the sum of all

n^{t h}

roots of unity always zero?

Think of the

n^{t h}

roots of unity as vectors pointing from the origin to the vertices of a regular

n

-sided polygon inscribed in the unit circle. Because this polygon is perfectly symmetric about the origin, the vector sum of all these vertices must cancel out to zero. It is a beautiful consequence of rotational symmetry in the complex plane!

Can I just use the sum of a geometric progression formula instead of the roots of unity property?

Absolutely! You can use the formula for the sum of a geometric progression:

S = a \frac{r ^{n} - 1}{r - 1}

. Here,

a = ω

and

r = ω

. The sum is

ω \frac{ω ^{6} - 1}{ω - 1}

. Since

ω^{7} = 1

, this simplifies to

ω \frac{ω ^{- 1} - 1}{ω - 1}

, which also leads to

- 1

. However, recognizing the roots of unity property is much faster and demonstrates a deeper geometric intuition, which is exactly what JEE examiners love to see.

Animated Solution for Mathematics - Complex Numbers: The value of $\sum_{k = 1}^{6} (sin \frac{2 π k}{7} - i cos \frac{2 π k}{7})$ is

Select Answer:

Visualized Solution

Analyzing the Given Series

Factoring out $- i$

Applying Euler's Formula

Introducing Roots of Unity

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

Evaluating the Summation

Final Calculation

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

Unmasking the Form

The Power of Euler

The Geometry of the Heptagon

Final Calculation

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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=16​(sin72πk​−icos72πk​) is

Select Answer:

Visualized Solution

Analyzing the Given Series

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Applying Euler's Formula

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Evaluating the Summation

Final Calculation

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

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The Power of Euler

The Geometry of the Heptagon

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Animated Solution for Mathematics - Complex Numbers: The value of $\sum_{k = 1}^{6} (sin \frac{2 π k}{7} - i cos \frac{2 π k}{7})$ is

Factoring out $- i$

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

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