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Chapter 1: Problem 45

Suppose \(w=\bar{z} / z\). Without doing any calculations, explain why \(|w|=1\).

Short Answer

Expert verified
The modulus of a complex number and its conjugate are equal, hence \(|w|=1\).

Step by step solution

01

Understanding the Problem

We need to determine why the modulus of the complex number \(w = \frac{\bar{z}}{z}\) is equal to 1. The modulus of a complex number \(w\) is given by \(|w| = \sqrt{a^2 + b^2}\) if \(w = a + bi\), but here we should focus on properties of the division of complex numbers.
02

Identify the Moduli of Components

Recall that for any complex number \(z = a + bi\), the modulus is \(|z| = \sqrt{a^2 + b^2}\). Moreover, the modulus of the conjugate \(\bar{z}\) is the same as the modulus of \(z\): \(|\bar{z}| = |z|\).
03

Using the Modulus Property of Division

For complex numbers, the property \(|\frac{\bar{z}}{z}| = \frac{|\bar{z}|}{|z|}\) applies. Since \(|\bar{z}| = |z|\), it follows that \(|\frac{\bar{z}}{z}| = \frac{|z|}{|z|} = 1\). Thus, \(|w| = 1\) without needing further calculation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Modulus of a Complex Number
The modulus of a complex number is a fundamental concept that helps in understanding the size or length of the number on the complex plane. When given a complex number in the form \(z = a + bi\), where \(a\) and \(b\) are real numbers, the modulus, denoted as \(|z|\), is calculated as \(\sqrt{a^2 + b^2}\). This formula gives you the distance from the origin to the point \((a, b)\) in the complex plane.
  • The modulus is always a non-negative real number.
  • If both \(a\) and \(b\) are zero, the modulus is zero, representing the origin.
  • The modulus provides insights into the amplitude or "magnitude" of oscillations in physics and engineering.
Understanding this concept is vital because it allows you to grasp the "size" of complex numbers and is a key component when performing operations like multiplication and division with complex numbers.
Complex Conjugate
In the world of complex numbers, each number has a twin known as the complex conjugate. For a complex number \(z = a + bi\), its conjugate is denoted as \(\bar{z} = a - bi\). The conjugate changes the sign of the imaginary part, flipping it over the real axis in the complex plane.
  • The modulus of a complex number \(z\) and its conjugate \(\bar{z}\) are identical, i.e., \(|z| = |\bar{z}|\).
  • Conjugates are leveraged to simplify division and other operations involving complex numbers.
  • In the field of signal processing, conjugates help in analyzing waveforms and functions.
This fundamental symmetry between a complex number and its conjugate helps maintain balance when dividing complex numbers. It's why the modulus of \(\frac{\bar{z}}{z}\) results in 1, as illustrated in the problem.
Properties of Division of Complex Numbers
Dividing complex numbers might initially seem complex, but it's simplified through understanding certain properties. When dividing two complex numbers, you multiply both the numerator and the denominator by the complex conjugate of the denominator. This eliminates the imaginary part in the denominator.
  • When dividing \(\frac{\bar{z}}{z}\), the modulus property states that \(|\frac{\bar{z}}{z}| = \frac{|\bar{z}|}{|z|}\).
  • Since \(|\bar{z}| = |z|\) for any complex number \(z\), this simplifies to \(\frac{|z|}{|z|} = 1\).
  • The outcome implies the modulus of the resulting quotient is 1, meaning it鈥檚 unit modulus.
This property proves highly beneficial, especially in simplifying expressions and solving equations involving complex numbers. By understanding these core principles, operations with complex numbers become much more intuitive and manageable.

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Most popular questions from this chapter

Use complex notation and inequalities in parts (a) and (b). (a) Make up a list of five sets in the complex plane that are connected. (b) Make up a list of five sets in the complex plane that are not connected.

Use a CAS as an aid in factoring the given quadratic polynomial. $$ z^{2}-3 i z-2 $$

Use (4) to compute all roots. Give the principal \(n\) the root in each case. Sketch the roots \(w_{0}, w_{1}, \ldots, w_{n-1}\) on an appropriate circle centered at the origin. $$ (8)^{1 / 3} $$

Use a CAS as an aid in factoring the given quadratic polynomial. $$ z^{2}-\sqrt{3} z-i $$

Solve the given quadratic equation using the quadratic formula. Then use (5) to factor the polynomial. $$ z^{2}-(1+9 i) z-20+5 i=0 $$
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