Solving The Quartic Equation Z^4 + 2z^2 Cos(α) + 1 = 0

Jul 11, 2025 by Jeany 55 views

Introduction

In this article, we delve into the fascinating world of complex numbers and polynomial equations, specifically focusing on solving the quartic equation z^4 + 2z^2 cos(α) + 1 = 0. This equation, while seemingly simple, unveils a rich interplay of algebraic techniques and trigonometric identities. Our goal is to provide a comprehensive and accessible solution, guiding you through each step with clarity and precision. Understanding how to solve such equations is crucial in various fields, including engineering, physics, and advanced mathematics. Before we dive into the solution, it's important to understand the fundamental concepts and methods we will be employing. This will allow you to not only follow the solution but also appreciate the elegance and power of the underlying mathematical principles. We will start by recapping some necessary concepts, such as complex numbers, trigonometric identities, and the methods for solving quadratic equations, before moving on to the core of solving our specific quartic equation. This will set a solid foundation for your understanding and enable you to tackle similar problems in the future.

Background: Complex Numbers and Quadratic Equations

To effectively tackle the quartic equation, we need a firm grasp of complex numbers and how to solve quadratic equations. Complex numbers, denoted in the form a + bi where a and b are real numbers and i is the imaginary unit (√-1), extend the real number system and allow us to find solutions to equations that have no real roots. Operations like addition, subtraction, multiplication, and division can be performed on complex numbers, and they follow specific rules. Understanding these operations is crucial for manipulating the equations we will encounter.

Quadratic equations, of the form ax^2 + bx + c = 0, are solved using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. This formula is a cornerstone of algebra and provides the roots of any quadratic equation, whether they are real or complex. The discriminant, b^2 - 4ac, determines the nature of the roots: if it's positive, there are two distinct real roots; if it's zero, there is one real root (a repeated root); and if it's negative, there are two complex conjugate roots. This background will be fundamental as we transform the quartic equation into a manageable quadratic form.

Understanding complex numbers involves visualizing them on the complex plane (also known as the Argand diagram), where the horizontal axis represents the real part and the vertical axis represents the imaginary part. A complex number a + bi can be represented as a point (a, b) on this plane. Moreover, complex numbers can also be expressed in polar form, which uses magnitude (or modulus) and argument (or angle). The magnitude r of a complex number z = a + bi is given by r = √(a^2 + b^2), and the argument θ is given by tan^(-1)(b/a). This polar representation is especially useful when dealing with complex number multiplication and powers, due to De Moivre's Theorem.

De Moivre's Theorem states that for any complex number z = r(cos θ + i sin θ) and any integer n, the power z^n is given by z^n = r^n(cos(nθ) + i sin(nθ)). This theorem is invaluable for finding powers and roots of complex numbers and will be critical in our solution to the quartic equation. Grasping this theorem's implications will significantly simplify the process of finding the roots of the equation.

Transforming the Quartic Equation

Our initial equation, z^4 + 2z^2 cos(α) + 1 = 0, is a quartic equation in z. However, we can recognize a crucial pattern: it's quadratic in z^2. This observation allows us to make a substitution, letting w = z^2. This substitution transforms the quartic equation into a quadratic equation in w: w^2 + 2w cos(α) + 1 = 0. This seemingly small step dramatically simplifies the problem, as we now have a quadratic equation that can be solved using the quadratic formula, a tool we are already familiar with. This technique of substitution is a powerful method in algebra for solving higher-degree equations by reducing them to simpler forms.

By making this substitution, we've effectively reduced the complexity of the problem. We've moved from dealing with a fourth-degree polynomial to a second-degree polynomial, a significant simplification. This kind of transformation is a common and effective strategy in mathematics: recognizing patterns and applying appropriate substitutions can make seemingly intractable problems much more manageable. This process of transformation highlights the importance of mathematical ingenuity and the ability to recognize underlying structures within complex equations.

Applying the quadratic formula to this transformed equation is the next logical step. Remember, the quadratic formula is a reliable tool for finding the roots of any quadratic equation, and in this case, it will provide us with the values of w. Once we have the values of w, we can then use the substitution w = z^2 to find the corresponding values of z, which are the solutions to our original quartic equation. This approach demonstrates the power of breaking down a complex problem into smaller, more manageable steps.

Solving the Quadratic Equation for w

Applying the quadratic formula to w^2 + 2w cos(α) + 1 = 0, we have a = 1, b = 2 cos(α), and c = 1. Plugging these values into the formula, we get:

w = (-2 cos(α) ± √((2 cos(α))^2 - 4(1)(1))) / (2(1))

Simplifying the expression under the square root, we have:

w = (-2 cos(α) ± √(4 cos^2(α) - 4)) / 2

Further simplification yields:

w = (-2 cos(α) ± 2√(cos^2(α) - 1)) / 2

w = -cos(α) ± √(cos^2(α) - 1)

Now, we use the trigonometric identity sin^2(α) + cos^2(α) = 1, which implies cos^2(α) - 1 = -sin^2(α). Substituting this into our equation, we get:

w = -cos(α) ± √(-sin^2(α))

Since the square root of -1 is i, the imaginary unit, we have:

w = -cos(α) ± i sin(α)

Thus, we have two solutions for w: w_1 = -cos(α) + i sin(α) and w_2 = -cos(α) - i sin(α). These solutions are complex conjugates, which is a common occurrence when solving quadratic equations with a negative discriminant. Recognizing and manipulating these solutions in their complex form is crucial for moving on to the next step, finding the solutions for z.

These solutions for w are expressed in rectangular form (a + bi), but it will be beneficial to express them in polar form for the next steps. Converting complex numbers into polar form allows us to utilize De Moivre's Theorem, which simplifies the process of finding roots of complex numbers. This involves finding the magnitude and argument of each complex number. The magnitude of a complex number a + bi is given by √(a^2 + b^2), and the argument is given by tan^(-1)(b/a). Understanding this conversion is fundamental to working with complex numbers and solving equations involving them.

Finding the Solutions for z

Recall that we made the substitution w = z^2. Now, we need to find z given the solutions for w. We have z^2 = w, so z = ±√w. We have two values for w: w_1 = -cos(α) + i sin(α) and w_2 = -cos(α) - i sin(α). To find the square roots, it's helpful to express w_1 and w_2 in polar form.

First, let's consider w_1 = -cos(α) + i sin(α). Its magnitude is √((-cos(α))^2 + (sin(α))^2) = √(cos^2(α) + sin^2(α)) = √1 = 1. Its argument is θ_1 such that cos(θ_1) = -cos(α) and sin(θ_1) = sin(α). This implies θ_1 = π - α. Therefore, w_1 = cos(π - α) + i sin(π - α).

Next, consider w_2 = -cos(α) - i sin(α). Its magnitude is also 1. Its argument is θ_2 such that cos(θ_2) = -cos(α) and sin(θ_2) = -sin(α). This implies θ_2 = π + α. Therefore, w_2 = cos(π + α) + i sin(π + α).

Now we can find the square roots. Using De Moivre's Theorem, the square roots of w_1 are given by:

z = ±√1(cos((π - α + 2πk)/2) + i sin((π - α + 2πk)/2)) for k = 0, 1

For k = 0, we have z_1 = cos((π - α)/2) + i sin((π - α)/2). For k = 1, we have z_2 = cos((3π - α)/2) + i sin((3π - α)/2).

Similarly, the square roots of w_2 are given by:

z = ±√1(cos((π + α + 2πk)/2) + i sin((π + α + 2πk)/2)) for k = 0, 1

For k = 0, we have z_3 = cos((π + α)/2) + i sin((π + α)/2). For k = 1, we have z_4 = cos((3π + α)/2) + i sin((3π + α)/2).

Thus, we have found the four roots of the quartic equation. Each root is a complex number, and their expressions involve trigonometric functions of α. This demonstrates the deep connection between algebra, trigonometry, and complex numbers. Understanding these connections is vital in advanced mathematical problem-solving.

Solutions and Conclusion

In conclusion, the solutions to the equation z^4 + 2z^2 cos(α) + 1 = 0 are:

z_1 = cos((π - α)/2) + i sin((π - α)/2)
z_2 = cos((3π - α)/2) + i sin((3π - α)/2)
z_3 = cos((π + α)/2) + i sin((π + α)/2)
z_4 = cos((3π + α)/2) + i sin((3π + α)/2)

These solutions showcase the power of combining algebraic manipulation with trigonometric identities and complex number theory. By transforming the quartic equation into a quadratic equation and utilizing De Moivre's Theorem, we were able to find the four complex roots. This approach provides a clear and structured method for solving similar quartic equations. Mastering this technique will enhance your problem-solving abilities in advanced mathematical contexts.

The process of solving this equation highlights several key mathematical concepts and techniques. The ability to recognize patterns and make appropriate substitutions, the use of the quadratic formula, the manipulation of complex numbers in both rectangular and polar forms, and the application of De Moivre's Theorem are all essential tools in the mathematician's toolkit. By understanding and applying these concepts, you can tackle a wide range of problems in algebra, trigonometry, and complex analysis.

This exploration into the solution of a specific quartic equation has broader implications for problem-solving in mathematics and beyond. The ability to break down a complex problem into smaller, more manageable parts, the importance of understanding fundamental concepts, and the power of combining different mathematical tools are all valuable lessons. As you continue your mathematical journey, keep these lessons in mind, and you will be well-equipped to tackle increasingly challenging problems.