Problem 34 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

\( z \) lies on the unit circle: \( z = e^{i\theta} \), \( \theta \in \mathbb{R} \).

Step 1: Understanding Complex Conjugates

The problem statement involves the complex conjugate \( \bar{z} \). Recall that if \( z = a + bi \) (where \( a \) and \( b \) are real numbers), then the complex conjugate \( \bar{z} = a - bi \). The problem states \( \bar{z} = \frac{1}{z} \).

Step 2: Expressing \( z \) in Standard Form

Let \( z = a + bi \). We need to find such \( z \) that \( \bar{z} = \frac{1}{z} \). With \( \bar{z} = a - bi \), we equate this to the reciprocal of \( z \).

Step 3: Finding the Reciprocal of \( z \)

The reciprocal of \( z \) is given as \( \frac{1}{z} = \frac{1}{a+bi} \). Expand this by multiplying the numerator and denominator by the complex conjugate of the denominator, so \( \frac{1}{z} = \frac{a-bi}{a^2 + b^2} \).

Step 4: Equating Conjugate and Reciprocal

Based on the equality \( a - bi = \frac{a - bi}{a^2 + b^2} \), since the numerators are equal, we can equate the magnitude, resulting in \( 1 = \frac{1}{a^2 + b^2} \). Simplifying gives \( a^2 + b^2 = 1 \).

Step 5: Solving for \( z \)

The equation \( a^2 + b^2 = 1 \) represents a circle of radius 1 in the complex plane. Thus, \( z \) must lie on this circle. This gives us all complex numbers of the form \( z = \cos \theta + i \sin \theta \) for \( 0 \leq \theta < 2\pi \).

Step 6: Conclusion

The complex numbers \( z \) that satisfy the given condition are those lying on the unit circle, expressed as \( z = e^{i\theta} \) for any real number \( \theta \).

Key Concepts

Complex Conjugate

A complex number is typically expressed in the form \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The complex conjugate of this number, denoted as \( \bar{z} \), is formed by changing the sign of the imaginary part. Therefore, \( \bar{z} = a - bi \). Complex conjugates have some interesting properties that make them useful in simplifying complex expressions, especially when dealing with division and complex fractions.
For example, when multiplying a complex number by its conjugate, \( z \cdot \bar{z} \), the result is always a real number \( a^2 + b^2 \). This feature is particularly beneficial when finding the reciprocal of a complex number, as it allows the denominator to become a real number, simplifying calculations. In the context of the given problem, the solution involves equating the complex conjugate \( \bar{z} \) to \( \frac{1}{z} \). This equation paves the way for understanding that \( z \) and its conjugate exhibit symmetrical properties about the real axis.

Unit Circle

The unit circle is a fundamental concept in mathematics, especially in complex numbers and trigonometry. It is a circle with a radius of 1, centered at the origin \((0,0)\) of the complex plane. Every point \( z \) on the unit circle can be expressed using the formula \( z = \cos \theta + i \sin \theta \), where \( \theta \) is the angle formed with the positive real axis.
The relation \( a^2 + b^2 = 1 \) is well-known as the equation of the unit circle. In the context of complex numbers, this expression means that all complex numbers satisfying this equation have magnitudes (or moduli) of 1. A complex number \( z \) lying on the unit circle has a special property: its magnitude is 1, making it easy to equate \( z \) with its reciprocal, as shown in the problem solution. This principle helps us visualize that complex numbers, when plotted, form a circle of radius 1 centered at the origin.

Reciprocal of Complex Numbers

The reciprocal of a complex number \( z = a + bi \) is calculated using its conjugate. To find \( \frac{1}{z} \), you first multiply both the numerator and the denominator by the complex conjugate of \( z \). Thus, \[ \frac{1}{z} = \frac{1}{a+bi} = \frac{a - bi}{a^2 + b^2} \].
Multiplying by the conjugate clears the imaginary unit \( i \) from the denominator, making calculations more straightforward. As per the solution given, equating \( \bar{z} \) to \( \frac{1}{z} \) demonstrates a key feature of reciprocal relationships in complex numbers. Particularly, when the moduli of \( z \) is 1 (true for any \( z \) on the unit circle), the reciprocal \( \frac{1}{z} \) has the same structure as \( \bar{z} \).
This equality implies that for complex numbers of magnitude 1, their reciprocals are simply their conjugates, leading to the elegant result that all such numbers lie on the unit circle.