Problem 87 - Algebra and Trigonometry

Question

Solve the equation. $$ z^{4}+1=0 $$

Step-by-Step Solution

Answer

The roots are \( \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \), \( -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \), \( -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \), and \( \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \).

Step 1: Recognize the Equation Type

The given equation is a polynomial of the form \( z^4 +1 = 0 \). It can be rewritten as \( z^4 = -1 \). This is an example of finding complex roots because \(-1 = e^{i\pi} \).

Step 2: Represent \\( -1 \\\) Using Euler's Formula

We can express \( -1 \) in terms of polar coordinates using Euler's formula: \(-1 = e^{i (\pi + 2 k \pi)}\) for any integer \(k\).

Step 3: Apply De Moivre's Theorem

Using De Moivre's Theorem, we find the fourth roots by setting \( z = e^{i(\pi + 2 k \pi)/4} \). We will consider the roots for \( k = 0, 1, 2, 3 \) to cover a full cycle of roots.

Step 4: Calculate Roots for Each k

- For \( k = 0 \): \( z = e^{i\pi/4} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \)- For \( k = 1 \): \( z = e^{i3\pi/4} = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \)- For \( k = 2 \): \( z = e^{i5\pi/4} = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \)- For \( k = 3 \): \( z = e^{i7\pi/4} = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \)

Step 5: Verify Periodicity and Correctness

The roots repeat with a periodicity of \(2\pi\), which means by using \( k = 0, 1, 2, 3 \), we have found all distinct roots within one full rotation of the unit circle. As such, these four values are indeed the fourth roots of \(-1\).

Key Concepts

Understanding Polynomial Equations

The given equation, \( z^4 + 1 = 0 \), is a type of polynomial equation. Polynomial equations are algebraic expressions that involve terms raised to a power or exponent. In this case, \( z^4 \) represents the fourth power of a complex number \( z \). A good first step in handling polynomial equations is to recognize their general form, which in this case allows us to write \( z^4 = -1 \). Solving polynomial equations often involves finding all values for \( z \) that make the equation true. Here, where we work with complex numbers, we will explore how these equations can yield what are known as complex roots.

Exploring Complex Roots

When dealing with polynomial equations, such as \( z^4 = -1 \), we often encounter not just real numbers but complex roots as well. A complex number is typically written in the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit satisfying \( i^2 = -1 \).In our example, we seek the roots of the equation, meaning values of \( z \) that satisfy \( z^4 = -1 \). These roots are complex numbers because \( -1 \) can be represented as \( e^{i\pi} \) using Euler's formula. By using complex numbers, we can solve polynomial equations more broadly, opening up solutions that may not appear in real numbers alone.

Using De Moivre's Theorem

De Moivre's Theorem is a powerful tool for finding roots of complex numbers. The theorem states that for any real number \( \theta \) and positive integer \( n \), \( (\cos \theta + i\sin \theta)^n = \cos(n\theta) + i\sin(n\theta) \). In the context of our example, this theorem helps us find the fourth roots of \(-1\) by expressing \( -1 \) in exponential form as \( e^{i\pi} \). Thus, for finding all the roots, we set \( z = e^{i(\pi + 2k\pi)/4} \).De Moivre's Theorem simplifies the process, allowing us to directly compute the roots for each integer \( k \) within a complete cycle:

• \( k = 0 \): \( z = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \)
• \( k = 1 \): \( z = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \)
• \( k = 2 \): \( z = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \)
• \( k = 3 \): \( z = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \)

These are the unique fourth roots of \(-1\).

Explaining Euler's Formula

Euler's formula is a key concept that links complex analysis to trigonometry. It states that for any real number \( x \), \( e^{ix} = \cos x + i\sin x \). This formula allows us to express complex exponentials as points on the unit circle in the complex plane.In solving \( z^4 + 1 = 0 \), Euler's formula is essential for expressing \( -1 \) as \( e^{i\pi} \) and extending it to \( e^{i(\pi + 2k\pi)} \), representing the periodicity in the complex plane.Understanding this formula is crucial because it shows how exponentials relate to rotations in the plane, which is why complex numbers are so naturally expressed in polar form. This insight is the bridge that allows De Moivre's Theorem to be used effectively in solving polynomial equations with complex roots.