Problem 56 - A Graphical Approach to Precalculus with Limits

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{2}=-4$$

Step-by-Step Solution

Answer

The solutions are \( x = 2i \) and \( x = -2i \).

Step 1: Identify the problem type

The equation given is of the form \( x^n = k \), where \( n = 2 \) and \( k = -4 \). This type of equation is known as a complex number equation.

Step 2: Express k in polar form

The complex number \( -4 \) can be represented in polar form. Since \( -4 \ _ 0 = 4 \ _ {180^\circ} \), it can be expressed as \( 4e^{i\pi} \) in polar form.

Step 3: Write the general expression for roots

According to De Moivre's Theorem, the solutions of \( x^n = k \) are of the form \( x = k^{1/n} \cdot e^{2\pi i m/n} \), where \( m \) ranges from 0 to \( n-1 \).

Step 4: Apply De Moivre's Theorem

For \( x^2 = -4 \), substitute \( k = -4 = 4e^{i\pi} \) and \( n = 2 \), giving us the expression \( x = 4^{1/2} e^{(i\pi + 2\pi m)/2} \).

Step 5: Calculate magnitude and argument

Calculate \( 4^{1/2} = 2 \). For the argument \( \frac{\pi + 2\pi m}{2} \):- When \( m = 0 \): Argument is \( \frac{\pi}{2} \), giving \( x = 2e^{i\frac{\pi}{2}} = 2i \).- When \( m = 1 \): Argument is \( \frac{3\pi}{2} \), giving \( x = 2e^{i\frac{3\pi}{2}} = -2i \).

Step 6: List the solutions

The solutions are \( x = 2i \) and \( x = -2i \).

Key Concepts

De Moivre's Theorem

To grasp De Moivre's Theorem, let's dive into its utility, especially when dealing with complex numbers. This powerful theorem is commonly used for finding roots of complex numbers and raising them to powers. It states that for a complex number in polar form written as \( r \left( \cos \theta + i \sin \theta \right) \), raising it to the power \( n \) results in:

• \[ r^n \left( \cos(n\theta) + i \sin(n\theta) \right) \]

In simpler terms, it helps by converting complex number multiplications into simpler calculations with magnitudes and angles, or arguments.
De Moivre's Theorem becomes a tool when you encounter complex numbers modeled as \( x^n = k \), translating to finding \( n \) roots of \( k \). By expressing \( k \) in polar form and applying the theorem, solutions emerge as elegant combinations of magnitudes and phased rotations. This breaks down potentially intimidating calculations into manageable steps for accurate solutions.

Polar Form

Understanding complex numbers in polar form is essential. A complex number, such as \( a + bi \), converts to polar form by identifying its magnitude (distance from the origin) and angle (direction from the positive real axis). Here's how the conversion goes:

• Magnitude \( r = \sqrt{a^2 + b^2} \)
• Angle \( \theta = \text{atan2}(b, a) \)

Thus, the polar form is represented as \( re^{i\theta} \).
In the example \( x^2 = -4 \), the complex number \(-4\) finds its polar form with magnitude 4 and angle \( \pi \) (since it lies along the negative real axis). This translates into \( 4e^{i\pi} \).
Transformation into polar form simplifies operations that involve multiplication and division, among other things, making calculations like powers or roots more straightforward and intuitive.

Complex Solutions

Finding complex solutions involves transforming equations into a form where De Moivre's theorem can be applied smoothly. From any equation such as \( x^n = k \), after expressing \( k \) in polar form \( re^{i\theta} \), solutions can be found using:

• \[ x = r^{1/n}e^{i(\theta + 2m\pi)/n} \]
• Where \( m \) ranges from 0 to \( n - 1 \)

By following this procedure in the context of \( x^2 = -4 \), the problem resolves to two solutions because \( n = 2 \). Solving as shown yields \( 2i \) and \( -2i \) as clear examples of how imaginary units can form our complex roots.
This approach allows us to grasp the solutions visually within the complex plane, providing assurance of their accuracy and completeness.