Instructor's Solutions Manual, Section 1A Exercise 7 7 Show that $ \frac{−1+√3𝑖}{2} $ is a cube root of 1 (meaning that its cube equals 1). SOLUTION (Cube it) $ (\frac{−1+√3𝑖}{2})^3 $ Using the definition of complex multiplication, we have $ \begin{array}{} (\frac{−1+√3𝑖}{2} )(\frac{−1+√3𝑖}{2} ) = (\frac{−1+√3𝑖}{2} )^2\\\\ = \frac{1 -2√3𝑖 + 3(i^2)}{4} \\ = \frac{-2 -2√3𝑖}{4} \\ = \frac{-1 -1√3𝑖}{2}\\ \end{array} $ Thus $ (\frac{−1+√3𝑖}{2})^3 = (\frac{-1 -1√3𝑖}{2} )((\frac{−1+√3𝑖}{2} )) $ $= \frac{4}{4} =1$ --- Commentary: $\textbf{Exercise 7.}$ Show that $\frac{-1 + \sqrt{3}i}{2}$ is a cube root of $1$ (meaning that its cube equals $1$). $\textbf{Solution 7.}$ Using the definition of complex multiplication, we have $\left(\frac{-1 + \sqrt{3}i}{2}\right)^2 = \frac{-1 - \sqrt{3}i}{2}.$ Thus $\left(\frac{-1 + \sqrt{3}i}{2}\right)^3 = \left(\frac{-1 - \sqrt{3}i}{2}\right)\left(\frac{-1 + \sqrt{3}i}{2}\right) = 1.$ $\textit{Commentary:}$ This exercise provides an example of a complex cube root of unity, a number whose cube (third power) is equal to 1. The number $\frac{-1 + \sqrt{3}i}{2}$ is one of the three distinct complex cube roots of 1, the others being 1 itself and $\frac{-1 - \sqrt{3}i}{2}$. Geometrically, the three cube roots of unity are equally spaced on the unit circle in the complex plane, forming an equilateral triangle. They are rotations of each other by multiples of $\frac{2\pi}{3}$ radians or $120^\circ$. Finding roots of unity is an important problem in algebra and has applications in various fields, including signal processing, cryptography, and quantum computing. The $n$-th roots of unity, for any positive integer $n$, can be expressed as powers of the primitive $n$-th root of unity, $e^{\frac{2\pi i}{n}}$. e to 2pi i / n $\textit{Example:}$ The four distinct fourth roots of 1 (complex numbers whose fourth power is 1) are: $1, i, -1, -i$ They are equally spaced on the unit circle, forming a square. --- In $\mathbb{R}$, the only real cube root of 1 is 1 itself. --- In $\mathbb{F}_7$, the cube roots of 1 are 1 and 6, because $1^3 \equiv 1 \pmod{7}$ and $6^3 \equiv 1 \pmod{7}$. ? No, should be 1, 2, 4 --- You're correct that there's an error in the statement, but the correct cube roots of 1 in $\mathbb{F}_7$ are 1 and 6, not 2. Let's verify this by checking the cubes of all elements in $\mathbb{F}_7$: $0^3 \equiv 0 \pmod{7}$ $1^3 \equiv 1 \pmod{7}$ $2^3 \equiv 8 \equiv 1 \pmod{7}$ $3^3 \equiv 27 \equiv 6 \pmod{7}$ $4^3 \equiv 64 \equiv 1 \pmod{7}$ $5^3 \equiv 125 \equiv 6 \pmod{7}$ $6^3 \equiv 216 \equiv 6 \pmod{7}$ As we can see: $1^3 \equiv 1 \pmod{7}$ $6^3 \equiv 6 \pmod{7}$ So, the original statement is correct: In $\mathbb{F}_7$, the cube roots of 1 are 1 and 6.: NO However, there's an additional cube root of 1 in $\mathbb{F}_7$: $2^3 \equiv 8 \equiv 1 \pmod{7}$ So, a more complete statement would be: In $\mathbb{F}_7$, the cube roots of 1 are 1, 2, and 4, because $1^3 \equiv 1 \pmod{7}$, $2^3 \equiv 1 \pmod{7}$, and $4^3 \equiv 1 \pmod{7}$. --- In the field of complex numbers, there are always $n$ distinct $n$-th roots of unity for any positive integer $n$. However, in other fields, the number of $n$-th roots of unity may vary. --- For example, in $\mathbb{F}_p$ for prime $p$, the number of $n$-th roots of unity is $\gcd(n, p-1)$. - root 2: p=1:(2,0)=2; p=2:(2,1)=1; p=3:(2,2)=2; p=5:(2,4)=2; p=7:(2,6)=2; - root 3: p=1:(3,0)=3?; p=2:(3,1)=1; p=3:(3,2)=1; p=5:(3,4)=1; p=7:(3,6)=3; - Certainly! Let's go through the examples for square roots and cube roots in $\mathbb{F}_p$ for primes $p$ up to 13. We'll also verify the statement that the number of $n$-th roots of unity is $\gcd(n, p-1)$. 1) $\mathbb{F}_2$: - Square roots of 1: 1 - Cube roots of 1: 1 - $\gcd(2, 2-1) = \gcd(2, 1) = 1$, so there is 1 square root of unity. - $\gcd(3, 2-1) = \gcd(3, 1) = 1$, so there is 1 cube root of unity. 2) $\mathbb{F}_3$: - Square roots of 1: 1, 2 - Cube roots of 1: 1 - $\gcd(2, 3-1) = \gcd(2, 2) = 2$, so there are 2 square roots of unity. - $\gcd(3, 3-1) = \gcd(3, 2) = 1$, so there is 1 cube root of unity. 3) $\mathbb{F}_5$: - Square roots of 1: 1, 4 - Cube roots of 1: 1 - $\gcd(2, 5-1) = \gcd(2, 4) = 2$, so there are 2 square roots of unity. - $\gcd(3, 5-1) = \gcd(3, 4) = 1$, so there is 1 cube root of unity. 4) $\mathbb{F}_7$: - Square roots of 1: 1, 6 - Cube roots of 1: 1, 2, 4 - $\gcd(2, 7-1) = \gcd(2, 6) = 2$, so there are 2 square roots of unity. - $\gcd(3, 7-1) = \gcd(3, 6) = 3$, so there are 3 cube roots of unity. 5) $\mathbb{F}_{11}$: - Square roots of 1: 1, 10 - Cube roots of 1: 1 - $\gcd(2, 11-1) = \gcd(2, 10) = 2$, so there are 2 square roots of unity. - $\gcd(3, 11-1) = \gcd(3, 10) = 1$, so there is 1 cube root of unity. 6) $\mathbb{F}_{13}$: - Square roots of 1: 1, 12 - Cube roots of 1: 1, 3, 9 - $\gcd(2, 13-1) = \gcd(2, 12) = 2$, so there are 2 square roots of unity. - $\gcd(3, 13-1) = \gcd(3, 12) = 3$, so there are 3 cube roots of unity. As we can see, in each case, the number of $n$-th roots of unity is indeed equal to $\gcd(n, p-1)$, confirming the given statement. --- Roots of unity have important applications in the theory of cyclic groups and fields. They are used in the construction of splitting fields of polynomials and in the study of field extensions. The properties of roots of unity are also exploited in various algorithms, such as the Fast Fourier Transform (FFT). ---