6th roots of unity explain everything corresponding to it including it’s equation | 1793157226853

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6th roots of unity explain everything corresponding to it including it’s equation

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The 6th Roots of Unity

:light_bulb: FORMULAS & RULES USED:

  • Fundamental Equation: z^6 = 1
  • Euler’s Formula & Polar Form: z_k = e^{i \frac{2k\pi}{n}} = \cos\left(\frac{2k\pi}{n}\right) + i \sin\left(\frac{2k\pi}{n}\right), where n = 6 and k \in \{0, 1, 2, 3, 4, 5\}
  • Sum of n-th Roots: \sum_{k=0}^{n-1} z_k = 0
  • Factorization: z^6 - 1 = (z^3 - 1)(z^3 + 1) = (z - 1)(z^2 + z + 1)(z + 1)(z^2 - z + 1)

:brain: STEP-BY-STEP SOLUTION:

Step 1 — Defining the Governing Equation and Polar Representation
To find the 6th roots of unity, we seek all complex numbers z satisfying the equation:

z^6 = 1

We express the number 1 in complex exponential form using its modulus r = 1 and argument \theta = 0 + 2k\pi:

1 = e^{i 2k\pi}, \quad k \in \mathbb{Z}

Applying De Moivre’s Theorem, the roots are obtained by taking the 6\text{th} root of both sides:

z_k = e^{i \frac{2k\pi}{6}} = e^{i \frac{k\pi}{3}}, \quad \text{for } k = 0, 1, 2, 3, 4, 5

Step 2 — Calculating All Six Distinct Roots
Evaluating z_k = \cos\left(\frac{k\pi}{3}\right) + i \sin\left(\frac{k\pi}{3}\right) for each value of k:

  • For k = 0:

    z_0 = e^{i \cdot 0} = \cos(0) + i \sin(0) = 1

  • For k = 1:

    z_1 = e^{i \frac{\pi}{3}} = \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + i\frac{\sqrt{3}}{2}

  • For k = 2:

    z_2 = e^{i \frac{2\pi}{3}} = \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}

  • For k = 3:

    z_3 = e^{i \pi} = \cos(\pi) + i \sin(\pi) = -1

  • For k = 4:

    z_4 = e^{i \frac{4\pi}{3}} = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}

  • For k = 5:

    z_5 = e^{i \frac{5\pi}{3}} = \cos\left(\frac{5\pi}{3}\right) + i \sin\left(\frac{5\pi}{3}\right) = \frac{1}{2} - i\frac{\sqrt{3}}{2}

Step 3 — Geometric Representation on the Complex Plane
In the complex plane (Argand diagram):

  • All roots have an absolute value of |z_k| = 1, meaning they lie on the unit circle.
  • The angular separation between consecutive roots is \Delta\theta = \frac{2\pi}{6} = \frac{\pi}{3} (60^\circ).
  • Connecting the points z_0, z_1, z_2, z_3, z_4, z_5 sequentially forms a regular hexagon inscribed in the unit circle, with one vertex fixed at (1, 0).

Step 4 — Algebraic Properties and Factorization

  1. Sum of Roots:

    z_0 + z_1 + z_2 + z_3 + z_4 + z_5 = 0

    (Symmetry dictates that the center of mass of the vertices is at the origin).

  2. Product of Roots:

    \prod_{k=0}^{5} z_k = (-1)^{6-1} \cdot (-1) = -1

  3. Primitive Roots:
    A root \omega is primitive if its powers generate all six roots. This occurs when \gcd(k, 6) = 1:

    \text{Primitive roots: } z_1 = e^{i\frac{\pi}{3}} \quad \text{and} \quad z_5 = e^{i\frac{5\pi}{3}}

  4. Polynomial Factorization:
    Factoring z^6 - 1 completely into real irreducible polynomials:

    z^6 - 1 = (z - 1)(z + 1)(z^2 + z + 1)(z^2 - z + 1)

    The roots of z^2 - z + 1 = 0 are precisely the primitive 6th roots (z_1 and z_5), corresponding to the 6th cyclotomic polynomial \Phi_6(z) = z^2 - z + 1.

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:white_check_mark: FINAL ANSWER:
The equation is z^6 = 1 (or z^6 - 1 = 0), and its six roots are:

\left\{ 1, \; \frac{1}{2} + i\frac{\sqrt{3}}{2}, \; -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \; -1, \; -\frac{1}{2} - i\frac{\sqrt{3}}{2}, \; \frac{1}{2} - i\frac{\sqrt{3}}{2} \right\}

In compact form: z \in \{\pm 1, \; \pm\frac{1}{2} \pm i\frac{\sqrt{3}}{2}\}.
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:bullseye: KEY CONCEPTS:

  • Cyclic Group: The set of roots forms a cyclic group of order 6 under complex multiplication, isomorphic to \mathbb{Z}_6.
  • Cyclotomic Polynomial (\Phi_6(z)): The minimal polynomial over \mathbb{Q} containing only the primitive 6th roots is \Phi_6(z) = z^2 - z + 1.

:warning: COMMON MISTAKES:

  • :cross_mark: Incorrect Approach: Assuming z^6 = 1 yields only the real solutions z = \pm 1.
  • :white_check_mark: Correct Approach: By the Fundamental Theorem of Algebra, a degree-6 polynomial must have exactly 6 roots in \mathbb{C} (2 real and 4 strictly complex conjugate pairs).

Would you like to explore how these roots relate to the roots of unity of lower orders, such as the 2nd and 3rd roots?