Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3

Q: Why do we treat the die rolls as distinct? Shouldn't the order not matter?

In probability, when we calculate the sample space, we treat the trials as distinct events. Even if the set of values is the same, the sequence $(1, \omega, \omega^2)$ is a different outcome from $(\omega, 1, \omega^2)$ in the total sample space of $6^3$ outcomes. We must account for all permutations to get the correct probability.

Q: How do we know that $\omega^r$ only takes three values?

Because $\omega^3 = 1$, any power $\omega^r$ can be written as $\omega^{3k + m} = (\omega^3)^k \cdot \omega^m = 1^k \cdot \omega^m = \omega^m$, where $m$ is the remainder of $r$ divided by $3$. Since $m \in \{0, 1, 2\}$, the only possible values are $\omega^0=1, \omega^1=\omega, \omega^2=\omega^2$.

Q: Is it possible for the sum to be zero in any other way, like $1+1+\omega^2$?

No. Geometrically, $1, \omega, \omega^2$ are vectors pointing to the vertices of an equilateral triangle centered at the origin. Any other combination, like $1+1+\omega^2 = 2+\omega^2$, does not lie at the origin. The only way to balance the vectors to zero is to have one of each.

Question

Let $\omega$ be a complex cube root of unity with $\omega 
eq 1$. A fair die is thrown three times. If $r_1, r_2$ and $r_3$ are the numbers obtained on the die, then the probability that $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$ is

Options:

(A) 1/18

(B) 1/9

(C) 2/9

(D) 1/36

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a complex plane, looking at the unit circle. You have a fair die in your hand, and you are about to throw it three times. Each roll, $r_1, r_2, r_3$, acts as a key to a hidden symmetry.

We are looking for the probability that $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$, where $\omega$ is a complex cube root of unity. This is a beautiful puzzle of balance.

### The Cyclic Nature of $\omega$

First, let's demystify $\omega$. We know that $\omega^3 = 1$, which serves as the heartbeat of the problem.

Because the powers of $\omega$ are cyclic, we only care about $r \pmod 3$:
* If $r$ is a multiple of $3$, $\omega^r = 1$.
* If $r$ leaves a remainder of $1$, $\omega^r = \omega$.
* If $r$ leaves a remainder of $2$, $\omega^r = \omega^2$.

This reduces our possibilities down to the set $\{1, \omega, \omega^2\}$.

### Mapping the Die

Now, consider the faces of the die $\{1, 2, 3, 4, 5, 6\}$. We map these to our roots:
* If $r \in \{3, 6\}$, then $\omega^r = 1$.
* If $r \in \{1, 4\}$, then $\omega^r = \omega$.
* If $r \in \{2, 5\}$, then $\omega^r = \omega^2$.

Since the die is fair, the probability of rolling any specific root is:
$$P(\omega^r = 1) = P(\omega^r = \omega) = P(\omega^r = \omega^2) = \frac{2}{6} = \frac{1}{3}$$

We have a perfectly balanced system where each root is equally likely to appear.

### The Geometric Truth

Why must the sum be zero? Think of these roots as vectors on the complex plane.

Algebraically, the identity is:
$$1 + \omega + \omega^2 = 0$$

To get a sum of zero, you must have exactly one of each root. If you had two $1$s and one $\omega$, you would be at $2 + \omega$, which is not zero.

Thus, the condition $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$ is satisfied if and only if the set $\{\omega^{r_1}, \omega^{r_2}, \omega^{r_3}\}$ is a permutation of $\{1, \omega, \omega^2\}$.

### The Combinatorial Finale

We have three dice, and we need one of each root. The number of ways to arrange three distinct items is $3! = 6$.

For each of these $6$ sequences, the probability is:
$$\left( \frac{1}{3} ight) 	imes \left( \frac{1}{3} ight) 	imes \left( \frac{1}{3} ight) = \frac{1}{27}$$

Therefore, the total probability is:
$$6 	imes \frac{1}{27} = \frac{6}{27}$$

Simplifying this by dividing both the numerator and the denominator by $3$, we arrive at our final answer:
$$\mathbf{\frac{2}{9}}$$

Animated Solution for Mathematics - Complex Numbers: Let $ω$ be a complex cube root of unity with $ω \neq = 1$ . A fair die is thrown three times. If $r_{1}, r_{2}$ and $r_{3}$ are the numbers obtained on the die, then the probability that $ω^{r_{1}} + ω^{r_{2}} + ω^{r_{3}} = 0$ is

Select Answer:

Visualized Solution

Visualizing the Complex Plane

Properties of $ω$

Condition for Sum to be Zero

Grouping Die Faces

Probability of Each Root

Permutations of Roots

Final Probability Calculation

Simplifying the Result

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Cyclic Nature of $ω$

Mapping the Die

The Geometric Truth

The Combinatorial Finale

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Select Answer:

Visualized Solution

Visualizing the Complex Plane

Properties of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

Condition for Sum to be Zero

Grouping Die Faces

Probability of Each Root

Permutations of Roots

Final Probability Calculation

Simplifying the Result

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Cyclic Nature of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

Mapping the Die

The Geometric Truth

The Combinatorial Finale

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