Problem 33 - Advanced Engineering Mathematics

Question

In Problems 33-36, find all complex numbers for which the given statement is true. $$ z=z^{-1} $$

Step-by-Step Solution

Answer

The solutions are \( z = 1 \) and \( z = -1 \).

Step 1: Understand the Equation

The problem states that for a complex number \( z \), it is equal to its inverse. Mathematically, this means \( z = z^{-1} \). The inverse of \( z \) can be expressed as \( \frac{1}{z} \). Therefore, we can rewrite the equation as \( z = \frac{1}{z} \).

Step 2: Simplify the Equation

To eliminate the fraction \( \frac{1}{z} \), multiply both sides of the equation \( z = \frac{1}{z} \) by \( z \). This results in the quadratic equation \( z^2 = 1 \).

Step 3: Solve for \( z \)

Solving the equation \( z^2 = 1 \) involves finding the roots of the equation, which are the values of \( z \) that satisfy it. Solve \( z^2 - 1 = 0 \) by factoring it into \( (z - 1)(z + 1) = 0 \). The solutions to this equation are \( z = 1 \) and \( z = -1 \).

Step 4: Verify Complex Nature

The problem asks for complex numbers, so we check if \( z = 1 \) and \( z = -1 \) are complex. Both solutions are real, but since real numbers are a subset of complex numbers (where the imaginary part is zero), \( z = 1 \) and \( z = -1 \) are indeed complex numbers.

Key Concepts

Inverse of a Complex Number

The concept of an inverse of a complex number is akin to the idea of reciprocals in real numbers. If you have a complex number expressed as \( z \), its inverse, denoted as \( z^{-1} \), is given by \( \frac{1}{z} \).
For real numbers, this simply means flipping the number, such that the product of a number and its inverse is \( 1 \).For complex numbers, it works the same way, but there's an additional layer. Complex numbers are written in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The inverse of \( z \) is such that when \( z \) is multiplied by \( z^{-1} \), the result is \( 1 \), even in the complex plane.

• Example: If you have a complex number \( z = 1 \), then its inverse \( z^{-1} \) is also \( 1 \) because \( 1 \times 1 = 1 \).
• Important Note: Not all complex numbers have numeric inverses. Only non-zero complex numbers have an inverse because you can’t divide by zero.

It's crucial to grasp that even though we refer to them as complex numbers when they have zero imaginary parts, like \( z = 1 \), they still satisfy properties of complex numbers.

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). Solving these equations typically involves finding the values of \( x \) that satisfy the equation. In the context of complex numbers, this can mean solving equations like \( z^2 = 1 \), as seen in the original exercise.
The methods for solving quadratic equations can vary:

• Factoring: As demonstrated, \( z^2 - 1 \) can be factored into \( (z - 1)(z + 1) = 0 \). This means the solutions are \( z = 1 \) and \( z = -1 \).
• Quadratic Formula: Another method, although not needed here, is using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our specific case, there is an elegant symmetry because specific values are integers and straightforward, but often in more complex cases, these solutions can be in the realm of complex numbers where both real and imaginary parts play roles.

Real and Complex Numbers

Understanding the relationship between real and complex numbers enhances comprehension of concepts like those found in the exercise. Real numbers are a subset of complex numbers, where any real number can be seen as a complex number with zero imaginary part.
Complex numbers are typically denoted as \( z = a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit defined as \( i^2 = -1 \).

• Real Numbers: These include all the numbers on the traditional number line, like \( -2, 0, \text{or } 3.5 \).
• Complex Numbers: These include expressions like \( 5 + 2i \), where both components are explicit.
• Subset Relation: Any real number \( a \) can be written as \( a + 0i \), showing that real numbers are indeed complex numbers.

The two solutions \( z = 1 \) and \( z = -1 \) in the exercise illustrate real numbers, but because of their zero imaginary part, are also fit into the complex framework, underscoring the versatility of complex numbers.