Problem 59 - A Graphical Approach to College Algebra

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{3}=27$$

Step-by-Step Solution

Answer

The solutions are 3, \(-\frac{3}{2} + i\frac{3\sqrt{3}}{2}\), and \(-\frac{3}{2} - i\frac{3\sqrt{3}}{2}\).

Step 1: Understanding the Problem

We need to find the cube roots of 27. That means we need to find all complex numbers \(x\) such that \(x^3 = 27\).

Step 2: Express 27 in Polar Form

Since 27 is a positive real number, we can express it in polar form as \(27 = 27 e^{i0}\). Here, the magnitude (absolute value) is 27, and the argument (angle) is 0.

Step 3: Apply the Formula for Roots of Complex Numbers

According to De Moivre's Theorem, the \(n\)-th complex roots of a number \(re^{i heta}\) are given by \(x_k = r^{1/n}e^{i( heta + 2k\pi)/n}\) for \(k = 0, 1, ..., n-1\).

Step 4: Calculate the Magnitude of the Roots

Calculate the magnitude of the roots using \(r^{1/n} = 27^{1/3}\). So, the magnitude of each root is \(3\) because \(27^{1/3} = 3\).

Step 5: Find Each Root

Compute the roots using the formula: - For \(k = 0\): \(x_0 = 3 e^{i(0 + 2(0)\pi)/3} = 3 e^{i0} = 3\).- For \(k = 1\): \(x_1 = 3 e^{i(0 + 2(1)\pi)/3} = 3 e^{i2\pi/3}\).- For \(k = 2\): \(x_2 = 3 e^{i(0 + 2(2)\pi)/3} = 3 e^{i4\pi/3}\).

Step 6: Convert the Roots back to Cartesian Form

Transform each of the complex exponentials back into rectangular form: - \(x_0 = 3\)- \(x_1 = 3 \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)\), which simplifies to \(-\frac{3}{2} + i\frac{3\sqrt{3}}{2}\).- \(x_2 = 3 \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)\), which simplifies to \(-\frac{3}{2} - i\frac{3\sqrt{3}}{2}\).

Key Concepts

Cube Roots

When we talk about cube roots in the context of complex numbers, we are searching for all numbers, both real and complex, that can be multiplied by themselves twice to yield the original number. For the equation such as \(x^3 = 27\), we are essentially finding all the cube roots of 27. In the real number system, the cube root of 27 is straightforwardly 3 because \(3^3 = 27\).

However, in the complex number system, things become more interesting. Complex numbers have the unique property of having multiple roots for even something seemingly simple like cube roots. Here, we have more roots to consider in the complex plane.

The key is to find all values that satisfy the equation, which means looking into all possibilities in the complex number space.

Polar Form

Complex numbers can be expressed in polar form, which is very helpful when dealing with powers and roots. The polar form expresses complex numbers in terms of their magnitude and angle rather than the traditional rectangular form. When we say polar form, we are writing a complex number as \(re^{i\theta}\), where \(r\) is the magnitude and \(\theta\) is the angle or argument.

In our problem, 27 can be represented in polar form as \( 27e^{i0} \), where:

• The magnitude \(r\) is 27, which is simply the absolute value of the positive real number 27.
• The argument \(\theta\) is 0 since 27 is positive and lies on the positive x-axis in the complex plane.

This notation becomes powerful because now we can easily manipulate powers and roots.

De Moivre's Theorem

De Moivre's Theorem is a critical tool when it comes to finding the roots of complex numbers. The theorem simplifies the process of raising complex numbers to a power or extracting roots. Simply put, when a complex number is expressed in polar form \(re^{i\theta}\), De Moivre's Theorem helps us find the \(n\)-th roots using the formula:

• \(x_k = r^{1/n}e^{i(\theta + 2k\pi)/n}\) for \(k = 0, 1, ..., n-1\).

By applying this theorem, we calculate each root by adjusting the angle \(\theta\) while keeping \(r^{1/n}\) consistent.

For the example \(x^3 = 27\), using De Moivre's Theorem gives us three roots, each evenly spaced around the complex circle.

Cartesian Form

The Cartesian form of a complex number is the standard way we write complex numbers as \(a + bi\), where \(a\) is the real part, and \(b\) is the imaginary part. After solving for roots using polar form, De Moivre's Theorem, and getting results in the form \(re^{i\theta}\), we often need to convert back into Cartesian form for simplicity or clarity.

For example, converting \(x_1 = 3 e^{i2\pi/3}\) into Cartesian form involves:

• Calculating the real part as \(3 \cos(2\pi/3)\).
• Calculating the imaginary part as \(3 \sin(2\pi/3)\).

The final result is \(-\frac{3}{2} + i\frac{3\sqrt{3}}{2}\).

Doing this for all roots helps to solidify our understanding and gives clear insight into the positional layout of these complex numbers on the plane.