Chapter 1: Problem 75
Solve each equation in the complex number system. $$ x^{4}=16 $$
Short Answer
Expert verified
The solutions are \( x = 2 \), \( x = 2i \), \( x = -2 \), and \( x = -2i \).
Step by step solution
01
Rewrite the Equation
Rewrite the given equation in the form \[x^4 = 16\]. Recognize that this implies \[x^4 = 16\text{, or }16 = 2^4\].
02
Apply Root Extraction
Taking the fourth root on both sides of the equation gives us \[\text{Fourth root of } x^4 = \text{Fourth root of } 16\], which simplifies to \[ \text{x} = \root 4 \/ \text{16} \].
03
Introduce Complex Roots
Consider the complex roots of unity involved since \[\text{x} = 2 e^{i(2k\frac{\text{Ï€}}{4})}\] for \(k=0,1,2,3\) (the principal values of the fourth roots of unity).
04
Compute the Values
Substitute \(k=0,1,2,3\) to find all possible solutions:
05
List All Solutions
The solutions to the equation \( x^4 = 16 \) in the complex number system are \( x = 2, 2i, -2, \text{and} -2i \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Roots of Unity
To solve the equation \(\text{x}^4 = 16\) in the complex number system, we need to consider the concept of complex roots of unity. Complex roots of unity are the solutions to the equation \(x^n = 1\) where \(n\) is a positive integer. These roots are equally spaced on the unit circle in the complex plane.
For any \(\text{n-th}\) root of unity, we use the formula \(e^{i \frac{2k\text{Ï€}}{n}}\) where \(k\) ranges from 0 to \(n-1\). In our problem, we need the fourth roots of unity. The fourth roots of unity are therefore given by \(\text{e}^{i \frac{2k\text{Ï€}}{4}}\) for \(k = 0, 1, 2 \text{ and } 3\).
In simpler terms, these represent angles of \(0, \frac{\text{Ï€}}{2}, \text{Ï€}, \text{ and } \frac{3\text{Ï€}}{2} \) radians on the unit circle:
For any \(\text{n-th}\) root of unity, we use the formula \(e^{i \frac{2k\text{Ï€}}{n}}\) where \(k\) ranges from 0 to \(n-1\). In our problem, we need the fourth roots of unity. The fourth roots of unity are therefore given by \(\text{e}^{i \frac{2k\text{Ï€}}{4}}\) for \(k = 0, 1, 2 \text{ and } 3\).
In simpler terms, these represent angles of \(0, \frac{\text{Ï€}}{2}, \text{Ï€}, \text{ and } \frac{3\text{Ï€}}{2} \) radians on the unit circle:
- When \(k = 0\): \(e^{i \times 0} = 1\)
- When \(k = 1\): \(e^{i \frac{\text{Ï€}}{2}} = i \)
- When \(k = 2\): \(e^{i \text{Ï€}} = -1 \)
- When \(k = 3\): \(e^{i \frac{3\text{Ï€}}{2}} = -i \)
Fourth Roots
A fourth root of a complex number is one of the four numbers that, when raised to the power of four, yield the initial number.
In our equation \(x^4 = 16\), solving for \(x\) means finding all fourth roots of 16. To do this, we first express 16 in the form of \(2^4\), then use the fourth roots of unity:
\[\begin{equation} x = 2 e^{i\frac{2k\text{Ï€}}{4}} \end{equation}\]. We follow these steps:
::The fourth roots of 16 are therefore: \(+2, 2i, -2, \text{ and } -2i\).
In our equation \(x^4 = 16\), solving for \(x\) means finding all fourth roots of 16. To do this, we first express 16 in the form of \(2^4\), then use the fourth roots of unity:
\[\begin{equation} x = 2 e^{i\frac{2k\text{Ï€}}{4}} \end{equation}\]. We follow these steps:
- First, take the principal root: \(x = 2 \)
- Then, we apply the roots of unity: multiply by \(e^{i\frac{2k\text{Ï€}}{4}}\) for \(k=0, 1, 2, \text{ and } 3\)
::The fourth roots of 16 are therefore: \(+2, 2i, -2, \text{ and } -2i\).
Complex Number System
Solving equations in the complex number system involves considering both real and imaginary parts. Complex numbers are of the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.
In this system, a number like \(16\) can take on several roots, both real and imaginary.
In this system, a number like \(16\) can take on several roots, both real and imaginary.
- The equations can often be rewritten or reinterpreted using Euler's formula: \(e^{i\theta} = \text{cos}(\theta) + i \text{sin}(\theta)\), which is critical for understanding rotations in the complex plane.
- When raising to fractions of \(\text{Ï€}\), such as \(e^{i\frac{\text{Ï€}}{2}}\), the imaginary unit \(i\), an undefined square root appears.