Problem 26 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 23-30: Find the solutions of the equation. $$ x^{5}+1=0 $$

Step-by-Step Solution

Answer

The solutions are complex numbers, expressed in polar form with distinct angles.

Step 1: Set the Equation to Zero

The given equation is \(x^5 + 1 = 0\). To start solving it, we need to rewrite it as \(x^5 = -1\).

Step 2: Rewrite -1 as a Complex Number

The number \(-1\) can be expressed as a complex number in polar form: \(-1 = e^{i\pi}\), due to Euler's formula, since the cosine of \(\pi\) is -1 and the sine of \(\pi\) is 0.

Step 3: Identify the General Solution

Since we have \(x^5 = e^{i\pi}\), we apply De Moivre's Theorem, leading to the solutions \(x = e^{i(\pi+2k\pi)/5}\) for \(k = 0, 1, 2, 3, 4\). These five values for \(k\) generate the five distinct roots of the equation.

Step 4: Calculate the Roots

Compute each of the fifth roots by substituting \(k = 0, 1, 2, 3, 4\) into \(e^{i(\pi + 2k\pi)/5}\). Thus, the roots are:1. \(x_0 = e^{i\pi/5}\)2. \(x_1 = e^{i3\pi/5}\)3. \(x_2 = e^{i\pi}\)4. \(x_3 = e^{i7\pi/5}\)5. \(x_4 = e^{i9\pi/5}\).

Key Concepts

Polar Form

Complex numbers can be beautifully expressed in the polar form. The polar form allows us to represent a complex number in terms of a magnitude and an angle. For a complex number expressed as \( a + bi \), the polar form is expressed as \( r(\cos \theta + i\sin \theta) \), where \( r \) is the modulus (distance from the origin to the point in the complex plane) and \( \theta \) is the argument (the angle with the positive real axis).
To convert \( -1 \) into polar form, we recognize it has a modulus of 1, since its distance from the origin is 1, and an argument of \( \pi \), because it lies on the negative real axis. Therefore, the polar form of \( -1 \) is \( e^{i\pi} \).
This representation becomes crucial when solving polynomial equations, like \( x^5 + 1 = 0 \), as it allows us to find all the roots of the equation easily by rotating angles.

Euler's Formula

Euler's formula is an essential concept in complex numbers, creating a bridge between trigonometry and complex exponential functions. The formula is given by \( e^{ix} = \cos x + i\sin x \). This remarkable equation simplifies working with complex numbers, turning trigonometric expressions into exponential ones.
In our exercise, Euler's formula helps us transform the complex number \( -1 \) into \( e^{i\pi} \).
This transformation allows solutions of polynomials to be expressed neatly in exponential form, simplifying the process of calculating roots and powers.

De Moivre's Theorem

De Moivre's Theorem is a formula that connects complex numbers and roots of unity. The theorem states that for a complex number in polar form, \( (\cos \theta + i\sin \theta)^n = \cos(n\theta) + i\sin(n\theta) \).
De Moivre's theorem becomes particularly useful when dealing with powers of complex numbers. When solving \( x^5 + 1 = 0 \), once \( x^5 = e^{i\pi} \) is identified, De Moivre's theorem is utilized to find the roots \( x = e^{(i(\pi+2k\pi)/5)} \), with different integer values for \( k \).
This allows us to break down powers of complex numbers into repeated angles, revealing all possible roots.

Roots of Unity

Roots of unity are special solutions of the equation \( x^n = 1 \), but they extend to solutions like \( x^5 = -1 \) when combined with polar representations. These roots are spaced evenly on the complex plane, forming a symmetric pattern and making them easy to visualize.
In our exercise, the five roots of unity are derived from applying the exponential form of \(-1\) using different rotations: \( e^{i(\pi + 2k\pi)/5} \) for \( k = 0, 1, 2, 3, 4 \).
These roots form a regular polygon in the complex plane, showing the elegance and symmetry inherent in complex numbers.