Problem 55 - A Graphical Approach to College Algebra

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{2}=-1$$

Step-by-Step Solution

Answer

The solutions are \(i\) and \(-i\).

Step 1: Understand the Equation

The equation given is \(x^2 = -1\). This means we have to find all the complex numbers \(x\) such that when squared, they result in \(-1\).

Step 2: Recognize the Roots Formula

The equation \(x^n = k\), where \(k\) is a complex number, has \(n\) distinct complex solutions given by \[x = k^{1/n} \cdot e^{(2\pi i k)/n}\] for \(k = 0, 1, 2, \ldots, n-1\). For our equation, \(k = -1\) and \(n = 2\).

Step 3: Convert \\(k\\) to Polar Form

First, write \(k = -1\) in polar form. The modulus \(|-1| = 1\), and the argument is \(\pi\) (as -1 is on the negative x-axis). Thus, \(-1\) can be written as \( e^{\pi i} \).

Step 4: Calculate the Solutions

Use the roots formula: \[x_k = (-1)^{1/2} \cdot e^{(2\pi i k)/2}\]. For \(k = 0\), \(x_0 = e^{(\pi i)/2}\), which equals \(i\). For \(k = 1\), \(x_1 = e^{(\pi i + 2\pi i)/2} = e^{(3\pi i)/2}\), which equals \(-i\).

Step 5: List All Solutions

The solutions to the equation \(x^2 = -1\) are \(i\) and \(-i\). These are the points on the unit circle at angles \(\pi/2\) and \(3\pi/2\), respectively.

Key Concepts

Polar Form

When dealing with complex numbers, converting them to polar form can often simplify calculations. Polar form is a way of expressing a complex number using a modulus (or magnitude) and an angle (also known as the argument or phase). The modulus is the distance of the complex number from the origin in the complex plane, found using the formula \(|z| = \sqrt{x^2 + y^2}\) where \(z = x + yi\).
The argument is the angle formed with the positive real axis, usually measured in radians.
To write a complex number in polar form, we use:

• Modulus \(r = |z|\)
• Argument \(\theta\), where \(z = r(\cos\theta + i\sin\theta) = r\cdot e^{i\theta}\)

In the exercise, the number \(-1\) was converted to polar form. Since \(-1\) lies on the negative x-axis, its modulus is 1 and its argument is \(\pi\). Thus, in polar form, \(-1\) is represented as \(e^{\pi i}\). This conversion helps in easily finding complex roots by leveraging properties of exponents.

Roots of Unity

Roots of unity are special solutions to polynomial equations of the form \(x^n = 1\). They are fundamental in understanding the solutions to similar equations like \(x^{n} = k\) for complex numbers. Essentially, if we find the \(n\)th roots of any complex number, we use the concept of rotating around the unit circle in the complex plane.
For a given \(n\), the roots of unity are equally spaced points around the unit circle.
Consider the complex number \(-1\) in the equation \(x^2 = -1\). To find the square roots, we apply the idea that these roots are placed at angles that divide the circle's circumference equally. For the roots of unity related to \(-1\), there are \(n = 2\) such points, separated by exactly \(\pi\) radians (180 degrees) on the unit circle.

• \(i\) is at angle \(\frac{\pi}{2}\).
• \(-i\) is at angle \(\frac{3\pi}{2}\).

These solutions are essentially rotations by \(90\) degrees from each other, helping in solving our initial equation efficiently.

Complex Solutions

Complex solutions often involve finding the roots of an equation in the complex plane. These solutions can be visualized using the unit circle concept, where each point represents a possible solution to the equation. In our case, the equation \(x^2 = -1\) is looking for numbers that, when squared, return \(-1\).
To solve such an equation, the polar form and roots of unity approach are practical tools.
The steps are broken down as follows:

• Convert the number to polar form, as done for \(-1\) becoming \(e^{\pi i}\).
• Apply the roots formula, \(x = (-1)^{1/2} \cdot e^{(2\pi i \cdot k)/2}\).
• Identify dividers of the full circle, leading to solutions like \(i\) and \(-i\).

These complex solutions result from understanding both the geometry of the unit circle and the algebra inherent in solving polynomial equations in the complex number system.