Problem 51 - Precalculus

Question

Find all complex solutions to the given equations. $$x^{4}-2 i=0$$

Step-by-Step Solution

Answer

The solutions are \( x = \sqrt[4]{2} e^{i \pi/8} \); \( x = \sqrt[4]{2} e^{i 5\pi/8} \); \( x = \sqrt[4]{2} e^{i 9\pi/8} \); \( x = \sqrt[4]{2} e^{i 13\pi/8} \).

Step 1: Rewrite the Equation

The given equation is \( x^4 - 2i = 0 \). We can rewrite this equation as \( x^4 = 2i \). Our objective is to express the right-hand side in a form that will allow us to find the fourth roots.

Step 2: Express in Polar Form

Given \( x^4 = 2i \), we express \( 2i \) in polar form. First, recall that the complex number \( 2i \) has a magnitude \( r = 2 \) and an argument \( \theta = \frac{\pi}{2} \), since it lies on the positive imaginary axis.

Step 3: Apply De Moivre's Theorem

To find the fourth roots, apply De Moivre's Theorem. The general formula for the \( n \)-th roots of a complex number is given by:\[ x_k = \sqrt[4]{r} \left( \cos\left( \frac{\theta + 2k\pi}{n} \right) + i \sin\left( \frac{\theta + 2k\pi}{n} \right) \right) \]where \( k = 0, 1, 2, ..., n-1 \). Here, \( r = 2 \), \( n = 4 \), and \( \theta = \frac{\pi}{2} \). Thus, \( \sqrt[4]{r} = \sqrt[4]{2} \).

Step 4: Calculate Each Solution

Compute the fourth roots by substituting \( k = 0, 1, 2, 3 \):- For \( k = 0 \), \[ x_0 = \sqrt[4]{2} \left( \cos\left( \frac{\pi/2}{4} + 0\cdot\frac{\pi}{2} \right) + i \sin\left( \frac{\pi/2}{4} + 0\cdot\frac{\pi}{2} \right) \right) \] \( = \sqrt[4]{2} \left( \cos\left( \frac{\pi}{8} \right) + i \sin\left( \frac{\pi}{8} \right) \right) \)- For \( k = 1 \), \[ x_1 = \sqrt[4]{2} \left( \cos\left( \frac{\pi/2 + 2\pi}{4} \right) + i \sin\left( \frac{\pi/2 + 2\pi}{4} \right) \right) \] \( = \sqrt[4]{2} \left( \cos\left( \frac{5\pi}{8} \right) + i \sin\left( \frac{5\pi}{8} \right) \right) \)- For \( k = 2 \), \[ x_2 = \sqrt[4]{2} \left( \cos\left( \frac{\pi/2 + 4\pi}{4} \right) + i \sin\left( \frac{\pi/2 + 4\pi}{4} \right) \right) \] \( = \sqrt[4]{2} \left( \cos\left( \frac{9\pi}{8} \right) + i \sin\left( \frac{9\pi}{8} \right) \right) \)- For \( k = 3 \), \[ x_3 = \sqrt[4]{2} \left( \cos\left( \frac{\pi/2 + 6\pi}{4} \right) + i \sin\left( \frac{\pi/2 + 6\pi}{4} \right) \right) \] \( = \sqrt[4]{2} \left( \cos\left( \frac{13\pi}{8} \right) + i \sin\left( \frac{13\pi}{8} \right) \right) \)

Key Concepts

De Moivre's Theorem

When dealing with complex numbers, De Moivre's Theorem is a powerful tool, especially for finding roots and powers. Imagine you have a complex number in polar form, written as \( z = r \left( \cos \theta + i \sin \theta \right) \). De Moivre's Theorem allows you to raise this number to any power or find its roots. It states: \[ z^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right) \] Here, \( n \) is an integer representing the power. This theorem makes computations straightforward since it breaks down the problem into multiplying magnitudes and adding angles. In the context of this exercise where you're finding the 4th roots, you reverse the process. Instead of multiplication, you're dividing the angle and taking the nth root of the magnitude.

Polar Form

To tackle complex numbers efficiently, understanding polar form is crucial.Polar form expresses complex numbers differently than the usual rectangular form, which uses \(a + bi\). Instead, it uses magnitude and angle. Imagine a point on a plane:

• Magnitude \(r\): This is the distance from the origin to the point, calculated as \( r = \sqrt{a^2 + b^2} \).
• Argument \(\theta\): The counterclockwise angle from the positive x-axis to the line connecting the origin with the point, \(\theta = \tan^{-1}(\frac{b}{a})\).

Thus, a complex number becomes \( r \left( \cos \theta + i \sin \theta \right) \). This form is incredibly handy for multiplication, division, and finding powers or roots, which can be cumbersome in rectangular form.

Nth Roots of Complex Numbers

Finding the nth roots of complex numbers is easier with De Moivre's Theorem and the polar form. Starting from a complex number in polar form \( z = r(\cos\theta + i\sin\theta) \), the nth roots are determined by using the formula: \[ x_k = \sqrt[n]{r} \left( \cos\left( \frac{\theta + 2k\pi}{n} \right) + i \sin\left( \frac{\theta + 2k\pi}{n} \right) \right) \] Here, \( k \) runs from 0 to \( n-1 \). Each value of \( k \) gives a unique root. This formula accounts for the rotational symmetry of complex numbers on the complex plane. Each root is equally spaced, making them visually easy to interpret. In our exercise, with \( x^4 = 2i \), the fourth roots are found by dividing the argument and adjusting by multiples of \( 2\pi \). This approach ensures all possible nth roots are covered, showing the beauty and completeness of complex number theory.