Solving The Complex Equation Z³ - 1 + J√3 = 0 In X + Jy Form
Introduction
In this article, we will delve into the fascinating world of complex numbers and explore how to solve a specific cubic equation involving them. Our main focus will be on finding the solutions to the equation z³ - 1 + j√3 = 0, where 'z' represents a complex number and 'j' is the imaginary unit (√-1). We will express our answers in the standard rectangular form of complex numbers, x + jy, where 'x' and 'y' are real numbers representing the real and imaginary parts, respectively. This exploration will not only enhance our understanding of complex number algebra but also showcase the elegance and power of mathematical techniques in solving seemingly intricate problems.
Understanding Complex Numbers
Before diving into the solution, it's crucial to have a solid grasp of complex numbers. A complex number is a number that can be expressed in the form a + bj, where 'a' and 'b' are real numbers, and 'j' is the imaginary unit, defined as j² = -1. The real part of the complex number is 'a', and the imaginary part is 'b'. Complex numbers extend the real number system by including an imaginary component, allowing us to work with the square roots of negative numbers and solve equations that have no real solutions. Complex numbers can be visualized on a complex plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part. This geometric representation provides a powerful tool for understanding the operations and properties of complex numbers.
Polar Form of Complex Numbers
An alternative way to represent complex numbers is the polar form. Instead of using the real and imaginary components, we describe a complex number by its magnitude (or modulus) 'r' and its argument (or angle) 'θ'. The magnitude 'r' is the distance from the origin to the point representing the complex number on the complex plane, and the argument 'θ' is the angle between the positive real axis and the line connecting the origin to the point. The polar form of a complex number z = x + jy is given by z = r(cos θ + j sin θ), where r = √(x² + y²) and θ = arctan(y/x). The polar form is particularly useful when dealing with multiplication, division, and powers of complex numbers, as these operations become much simpler in this representation.
Euler's Formula
A fundamental connection between exponential and trigonometric functions is provided by Euler's formula, which states that e^(jθ) = cos θ + j sin θ. This remarkable formula allows us to express the polar form of a complex number in a compact exponential form: z = re^(jθ). Euler's formula is a cornerstone of complex analysis and has wide-ranging applications in mathematics, physics, and engineering. It elegantly bridges the gap between the seemingly disparate worlds of exponential functions and trigonometric functions, revealing their deep interrelationship.
Transforming the Equation
Now, let's return to our original equation: z³ - 1 + j√3 = 0. To solve it, we first isolate the z³ term:
z³ = 1 - j√3
Our next step is to express the complex number on the right-hand side, 1 - j√3, in its polar form. This will make it easier to find the cube roots.
Finding the Magnitude
The magnitude 'r' of 1 - j√3 is calculated as follows:
r = √(1² + (-√3)²) = √(1 + 3) = √4 = 2
Finding the Argument
The argument 'θ' is the angle whose tangent is the ratio of the imaginary part to the real part. In this case:
θ = arctan(-√3 / 1) = -π/3
However, we need to consider that the arctangent function only gives us the principal value, which lies in the range (-π/2, π/2). Since the complex number 1 - j√3 lies in the fourth quadrant, the argument is indeed -π/3. We can also express this angle as 5π/3, which is coterminal with -π/3.
Polar Form Representation
Therefore, the polar form of 1 - j√3 is:
2(cos(-π/3) + j sin(-π/3)) or 2e^(-jπ/3)
Finding the Cube Roots
Now that we have z³ = 2e^(-jπ/3), we need to find the cube roots of this complex number. In general, a complex number has 'n' distinct 'n'th roots. To find these roots, we use De Moivre's theorem, which states that for any complex number z = re^(jθ) and any integer 'n', the 'n'th roots of 'z' are given by:
z^(1/n) = r^(1/n) * e^(j(θ + 2πk)/n), where k = 0, 1, 2, ..., n-1
In our case, we want to find the cube roots (n = 3) of 2e^(-jπ/3). So, we have:
z = (2e(-jπ/3))(1/3) = 2^(1/3) * e^(j(-π/3 + 2πk)/3), where k = 0, 1, 2
Let's calculate the three roots for k = 0, 1, and 2:
Root 1 (k = 0)
z₁ = 2^(1/3) * e^(j(-π/3)/3) = 2^(1/3) * e^(-jπ/9)
Root 2 (k = 1)
z₂ = 2^(1/3) * e^(j(-π/3 + 2π)/3) = 2^(1/3) * e^(j(5π/3)/3) = 2^(1/3) * e^(j5π/9)
Root 3 (k = 2)
z₃ = 2^(1/3) * e^(j(-π/3 + 4π)/3) = 2^(1/3) * e^(j(11π/3)/3) = 2^(1/3) * e^(j11π/9)
Converting Back to Rectangular Form
We have found the three cube roots in polar form. Now, we need to convert them back to the rectangular form x + jy. We will use Euler's formula to do this:
e^(jθ) = cos θ + j sin θ
Root 1 (z₁)
z₁ = 2^(1/3) * e^(-jπ/9) = 2^(1/3) * (cos(-π/9) + j sin(-π/9))
Approximating the trigonometric values:
z₁ ≈ 2^(1/3) * (0.9397 - j0.3420) ≈ 1.1835 - j0.4306
Root 2 (z₂)
z₂ = 2^(1/3) * e^(j5π/9) = 2^(1/3) * (cos(5π/9) + j sin(5π/9))
Approximating the trigonometric values:
z₂ ≈ 2^(1/3) * (-0.1736 + j0.9848) ≈ -0.2185 + j1.2351
Root 3 (z₃)
z₃ = 2^(1/3) * e^(j11π/9) = 2^(1/3) * (cos(11π/9) + j sin(11π/9))
Approximating the trigonometric values:
z₃ ≈ 2^(1/3) * (-0.7660 - j0.6428) ≈ -0.9650 - j0.8045
Solutions in Rectangular Form
Therefore, the three solutions to the equation z³ - 1 + j√3 = 0 in the form x + jy are approximately:
- z₁ ≈ 1.1835 - j0.4306
- z₂ ≈ -0.2185 + j1.2351
- z₃ ≈ -0.9650 - j0.8045
Conclusion
In this article, we have successfully solved the complex equation z³ - 1 + j√3 = 0 and found its three roots in the form x + jy. We utilized the polar form of complex numbers, De Moivre's theorem, and Euler's formula to navigate the intricacies of complex number algebra. This process highlights the power and elegance of mathematical tools in tackling complex problems. The solutions we obtained showcase the rich structure of complex numbers and their ability to provide solutions where real numbers fall short. Understanding and manipulating complex numbers is crucial in various fields, including mathematics, physics, and engineering, where they play a vital role in modeling and solving real-world problems.
