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

**URL:** <https://sorumatik.co/t/6th-roots-of-unity-explain-everything-corresponding-to-it-including-it-s-equation-1793157226853/698554>\
**Category:** Question Answer\
**Created:** [Ekim 28, 2026, 3:13öö UTC](https://sorumatik.co/t/6th-roots-of-unity-explain-everything-corresponding-to-it-including-it-s-equation-1793157226853/698554 "2026-10-28T03:13:47Z")\
**Posts on this page:** 2\
**Page:** 1

<div class="post-metadata">

**Author:** ![Dhiya\_S\_Khan](https://sorumatik.co/user_avatar/sorumatik.co/dhiya_s_khan/32/238873_2.png) [@Dhiya\_S\_Khan](https://sorumatik.co/u/Dhiya_S_Khan)\
**Post date:** [Ekim 28, 2026, 3:13öö UTC](https://sorumatik.co/t/6th-roots-of-unity-explain-everything-corresponding-to-it-including-it-s-equation-1793157226853/698554/1 "2026-10-28T03:13:47Z")

</div>

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

---

<div class="post-metadata">

**Author:** ![Akademik\_Rehber](https://sorumatik.co/user_avatar/sorumatik.co/akademik_rehber/32/235405_2.png) [@Akademik\_Rehber](https://sorumatik.co/u/Akademik_Rehber)\
**Post date:** [Ekim 28, 2026, 3:14öö UTC](https://sorumatik.co/t/6th-roots-of-unity-explain-everything-corresponding-to-it-including-it-s-equation-1793157226853/698554/2 "2026-10-28T03:14:05Z")

</div>

## The 6th Roots of Unity

💡 **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)

* * *

🧠 **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:**

- **For k = 1:**

- **For k = 2:**

- **For k = 3:**

- **For k = 4:**

- **For k = 5:**

* * *

**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:**

2. **Product of Roots:**

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

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

━━━━━━━━━━━━━━━━━━━━━━━━━━━  
✅ **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}\}.  
━━━━━━━━━━━━━━━━━━━━━━━━━━━

🎯 **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.

⚠ **COMMON MISTAKES:**

- **❌ Incorrect Approach:** Assuming z^6 = 1 yields only the real solutions z = \pm 1.
- **✅ 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?