Math Problem Statement

چهار ریشه مختلط معادله z^4 + z^3 + z^2 + z + 1 = 0 را بیابید.

Solution

The problem here is to find the four complex roots of the polynomial equation:

z4+z3+z2+z+1=0z^4 + z^3 + z^2 + z + 1 = 0

This is a specific type of equation known as a "cyclotomic equation." The given equation is the 5th cyclotomic polynomial, which can be solved by finding the 5th roots of unity, excluding the root z=1z = 1.

Steps to Solve

  1. Recognize the Polynomial: The polynomial z4+z3+z2+z+1=0z^4 + z^3 + z^2 + z + 1 = 0 is equivalent to solving for the roots of unity, specifically the non-trivial roots of z5=1z^5 = 1 (excluding z=1z = 1).

  2. Find the Roots of Unity: The solutions to z5=1z^5 = 1 are the complex numbers zk=e2πik/5z_k = e^{2\pi i k / 5} for k=0,1,2,3,4k = 0, 1, 2, 3, 4.

  3. Exclude z=1z = 1: We exclude z=e2πi0/5=1z = e^{2\pi i \cdot 0 / 5} = 1, leaving us with four roots: z1=e2πi/5,z2=e4πi/5,z3=e6πi/5,z4=e8πi/5.z_1 = e^{2\pi i / 5}, \quad z_2 = e^{4\pi i / 5}, \quad z_3 = e^{6\pi i / 5}, \quad z_4 = e^{8\pi i / 5}.

Roots in Terms of Cosine and Sine

The roots can also be expressed in terms of cosine and sine as follows:

  • z1=cos(2π5)+isin(2π5)z_1 = \cos\left(\frac{2\pi}{5}\right) + i \sin\left(\frac{2\pi}{5}\right)
  • z2=cos(4π5)+isin(4π5)z_2 = \cos\left(\frac{4\pi}{5}\right) + i \sin\left(\frac{4\pi}{5}\right)
  • z3=cos(6π5)+isin(6π5)z_3 = \cos\left(\frac{6\pi}{5}\right) + i \sin\left(\frac{6\pi}{5}\right)
  • z4=cos(8π5)+isin(8π5)z_4 = \cos\left(\frac{8\pi}{5}\right) + i \sin\left(\frac{8\pi}{5}\right)

These are the four complex roots of the equation.

Would you like more details on how to calculate the exact values of these roots or any further explanation?


Here are 5 follow-up questions to deepen understanding:

  1. How can you represent each root in polar form and Cartesian form?
  2. What is the significance of cyclotomic polynomials in algebra and number theory?
  3. How does Euler's formula relate to solving equations with complex roots?
  4. How can you verify that each of these roots satisfies the original polynomial equation?
  5. How would the approach differ if the polynomial was z5+z4+z3+z2+z+1=0z^5 + z^4 + z^3 + z^2 + z + 1 = 0?

Tip: For solving polynomial equations with complex roots, Euler's formula eiθ=cosθ+isinθe^{i\theta} = \cos \theta + i \sin \theta is a powerful tool. It simplifies the process of finding roots on the complex plane.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Cyclotomic Polynomials
Roots of Unity

Formulas

z^n = 1 for roots of unity
Euler's formula: e^(iθ) = cos(θ) + i sin(θ)

Theorems

Roots of Unity Theorem
Cyclotomic Polynomial Theorem

Suitable Grade Level

Grades 11-12