Let z_k = cos(2kπ/10) + i sin(2kπ/10); k = 1, 2, ..., 9. Match the statements in List-I with those i

Q: Wait, why is the sum of all roots of unity zero?

Think of the roots as vectors pointing from the origin to the vertices of a regular polygon. By symmetry, the net force or vector sum must be zero. Algebraically, the sum of the roots of the polynomial $z^n - 1 = 0$ is the coefficient of the $z^{n-1}$ term, which is zero.

Q: How do we know the product of $|1-z_k|$ is 10?

By evaluating the polynomial identity $\frac{z^{10}-1}{z-1} = \prod_{k=1}^{9} (z-z_k)$ at $z=1$. The left side becomes the sum of a geometric series $1 + z + \dots + z^9$, which equals 10 when $z=1$. Thus, the product of the distances is 10.

Q: Is there a trick for the cosine sum, or do I have to calculate each one?

Calculating each one would be a nightmare! The trick is to use the property that the sum of all roots is zero. The real part of this sum is $1 + \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right) = 0$. It is much faster and more elegant than summing the series manually.

Question

Let $z_k = \cos\left(\frac{2k\pi}{10}\right) + i\sin\left(\frac{2k\pi}{10}\right); k = 1, 2, \dots, 9$. Match the statements in List-I with those in List-II.

List-I:

P. For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$

Q. There exists a $k \in \{1, 2, \dots, 9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers

R. $\frac{|1 - z_1||1 - z_2| \dots |1 - z_9|}{10}$ equals

S. $1 - \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right)$ equals

List-II:

1. True

2. False

3. 1

4. 2

Sigma Percentile · Accepted Answer

### The Geometry of Unity: A Journey into the Complex Plane

Welcome, fellow traveler of the mathematical realm. Today, we are not just solving a problem; we are exploring the elegant, symmetrical world of the $10^{th}$ roots of unity.

When you see $z_k = \cos\left(\frac{2k\pi}{10}\right) + i\sin\left(\frac{2k\pi}{10}\right)$, do not just see a formula. See a decagon. Imagine a circle of radius $1$ centered at the origin of the complex plane.

These ten numbers are the vertices of a perfect, ten-sided polygon inscribed within that circle. They are equally spaced, starting from $1$ (which is $z_0$) and rotating around the circle. This visualization is your most powerful tool.

### Phase 1: The Symmetry of Inverses

Statement P asks us about the existence of inverses. It asks if for every $z_k$, there exists a $z_j$ such that $z_k \cdot z_j = 1$.

In the world of complex numbers, multiplying by a number is like rotating and scaling. If the magnitude of $z_k$ is $1$, then its reciprocal is simply its complex conjugate, $\bar{z}_k$.

Because our roots are symmetrically distributed, the conjugate of any root is also a root in our set. For example, the conjugate of $z_1 = e^{i\frac{2\pi}{10}}$ is $e^{-i\frac{2\pi}{10}}$, which is $z_9$. Thus, every root has a partner, making Statement P undeniably **True**.

### Phase 2: The Illusion of Unsolvability

Statement Q suggests that the equation $z_1 \cdot z = z_k$ might have no solution. But let us pause and think. We are working in the field of complex numbers.

Division is perfectly defined as long as we are not dividing by zero. Since $z_1 = e^{i\frac{2\pi}{10}}$ is a point on the unit circle, it is clearly not zero.

Therefore, $z = \frac{z_k}{z_1}$ is always a valid complex number. The claim that no solution exists is a trap designed to test your confidence in the field properties of complex numbers. Statement Q is **False**.

### Phase 3: The Polynomial Masterpiece

Now, we reach the heart of the problem: Statement R. We need to evaluate the product $\frac{|1 - z_1||1 - z_2| \dots |1 - z_9|}{10}$.

This looks intimidating, but it is a classic application of polynomial theory. The $10^{th}$ roots of unity are the roots of the equation $z^{10} - 1 = 0$. We can factorize this as:

$$z^{10} - 1 = (z - 1)(z - z_1)(z - z_2) \dots (z - z_9)$$

If we divide both sides by $(z - 1)$, we get the identity:

$$\frac{z^{10} - 1}{z - 1} = (z - z_1)(z - z_2) \dots (z - z_9)$$

The left side is a geometric series: $1 + z + z^2 + \dots + z^9$. Now, here is the magic: substitute $z = 1$.

The left side becomes $1 + 1 + \dots + 1 = 10$. The right side becomes $(1 - z_1)(1 - z_2) \dots (1 - z_9)$.

Taking the absolute value, we find that the product of these chord lengths is exactly $10$. Dividing by $10$, we get $1$. The elegance of this cancellation is why we love mathematics.

### Phase 4: The Sum of Cosines

Finally, Statement S asks for $1 - \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right)$. We know that the sum of all $10^{th}$ roots of unity is zero:

$$1 + z_1 + z_2 + \dots + z_9 = 0$$

By taking the real part of this equation, we get:

$$1 + \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right) = 0$$

This implies that the sum of the cosines is $-1$. Substituting this into our expression, we get $1 - (-1) = 2$.

We have navigated the geometry, the algebra, and the symmetry of these roots. Each step was not just a calculation, but a revelation of the underlying structure. Keep this perspective, and no problem will ever be too complex.

Animated Solution for Mathematics - Complex Numbers: Let $z_{k} = cos (\frac{2 k π}{10}) + i sin (\frac{2 k π}{10}); k = 1, 2, \dots, 9$ . Match the statements in List-I with those in List-II.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Visualizing the $1 0^{t h}$ Roots of Unity

Statement P: Existence of Inverses

Statement P: Conjugate Pairs

Statement Q: Solutions to $z_{1} \cdot z = z_{k}$

Statement Q: Evaluating the Solution

Statement R: Roots of the Polynomial

Statement R: Polynomial Division

Statement R: Substituting $z = 1$

Statement R: Product of Chord Lengths

Statement S: Sum of Roots

Statement S: Sum of Cosines

Statement S: Final Evaluation

The Sigma Insight: Cube Roots and nth Roots of Unity

The Geometry of Unity

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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

Let a complex number $α, α \neq = 1$ , be a root of the equation $z^{p + q} - z^{p} - z^{q} + 1 = 0$ , where $p, q$ are distinct primes. Show that either $1 + α + α^{2} + \dots + α^{p - 1} = 0$ or $1 + α + α^{2} + \dots + α^{q - 1} = 0$ , but not both together.

Let $ω$ be a complex number such that $2 ω + 1 = z$ where $z = - 3$ . If $111 1 - ω^{2} - 1 ω^{2} 1 ω^{2} ω^{7} = 3 k$ , then k is equal to:

The value of $\sum_{k = 1}^{6} (sin \frac{2 π k}{7} - i cos \frac{2 π k}{7})$ is

If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^{3} + 2 z^{2} + 2 z + 1 = 0$ , is equal to :

If $z^{2} + z + 1 = 0, z \in C$ , then $\sum_{n = 1}^{15} (z^{n} + (- 1)^{n} \frac{1}{z ^{n}})^{2}$ is equal to \_\_\_\_\_.

Prove that $\sum_{k = 1}^{n - 1} (n - k) cos \frac{2 k π}{n} = - n /2$ , where $n \geq 3$ is an integer.

Let $z = a + ib, b \neq = 0$ be complex numbers satisfying $z^{2} = \overset{z}{ˉ} \cdot 2^{1 - ∣ z ∣}$ . Then the least value of $n \in N$ , such that $z^{n} = (z + 1)^{n}$ , is equal to

If $z^{2} + z + 1 = 0$ , where $z$ is complex number, then the value of $(z + 1/ z)^{2} + (z^{2} + 1/ z^{2})^{2} + (z^{3} + 1/ z^{3})^{2} + \dots + (z^{6} + 1/ z^{6})^{2}$ is

Let $z_{1}$ and $z_{2}$ be $n^{t h}$ roots of unity which subtend a right angle at the origin. Then $n$ must be of the form

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let zk​=cos(102kπ​)+isin(102kπ​);k=1,2,…,9. Match the statements in List-I with those in List-II.