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Chapter 1: Problem 37

Prove: For any complex numbers \(z, w \in \mathbf{C},|z+w| \leq|z|+|w|.\)

Short Answer

Expert verified
We prove the triangle inequality for complex numbers, \(|z+w| \leq |z| + |w|\), by calculating the square of the magnitude of the sum, \(|(z + w)|^2 = (a + c)^2 + (b + d)^2 = a^2 + 2ac + c^2 + b^2 + 2bd + d^2\). Then, we compare it to \((|z| + |w|)^2 = a^2 + b^2 + 2|z||w| + c^2 + d^2\). We show that \(ac + bd \leq |z||w|\) using the Cauchy-Schwarz inequality, \((ac + bd)^2 \leq (a^2 + b^2)(c^2 + d^2) = |z|^2|w|^2\). Thus, the Triangle Inequality holds for complex numbers.

Step by step solution

01

Write the given inequality in terms of complex numbers

We're given the inequality: \(|z + w| \leq |z| + |w|\), where \(z\) and \(w\) are arbitrary complex numbers. Our goal is to show that this inequality holds for all such complex numbers.
02

Understand the magnitude of a complex number

The magnitude of a complex number \(z = a + bi\) is defined as \(|z| = \sqrt{a^2 + b^2}\), the square root of the sum of the squares of its real and imaginary parts. We can express the magnitudes of \(z\) and \(w\) as follows: \(|z| = \sqrt{a^2 + b^2}\) and \(|w| = \sqrt{c^2 + d^2}\), where \(z = a + bi\) and \(w = c + di\).
03

Calculate the square of the magnitude of the sum

We're going to find the square of the magnitude of \(z + w\) first. That is, \(|(z + w)|^2\). Given \(z + w = (a + c) + (b + d)i\), using the definition of the magnitude, we get: \(|(z + w)|^2 = (a + c)^2 + (b + d)^2 = a^2 + 2ac + c^2 + b^2 + 2bd + d^2\)
04

Observe the terms in the inequality

To make it easier to prove the inequality, we're going to compare the squares of the magnitudes of both sides of the inequality: \(|(z + w)|^2 \leq (|z| + |w|)^2\) Notice that: \((|z| + |w|)^2 = (|z|)^2 + 2|z||w| + (|w|)^2 = a^2 + b^2 + 2|z||w| + c^2 + d^2\) Compare this with the expression we got for \(|(z + w)|^2\).
05

Prove the inequality using a key property

Now we need to show that: \(a^2 + 2ac + c^2 + b^2 + 2bd + d^2 \leq a^2 + b^2 + 2|z||w| + c^2 + d^2\) Which simplifies to: \(2ac + 2bd \leq 2|z||w|\) Divide by 2, we get: \(ac + bd \leq |z||w|\) Now, by Cauchy-Schwarz inequality, we know that: \((ac + bd)^2 \leq (a^2 + b^2)(c^2 + d^2) = |z|^2|w|^2\) Taking the square root of both sides, we get: \(ac + bd \leq |z||w|\), which means the inequality is true. Therefore, we have proved that for any complex numbers \(z\) and \(w\), the Triangle Inequality holds: \(|z + w| \leq |z| + |w|\).

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Key Concepts

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

Complex Numbers
Complex numbers are a cornerstone of higher mathematics, extending the familiar real number system to include solutions to equations that cannot be solved with real numbers alone. A complex number is written as \( z = a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit, satisfying \( i^2 = -1 \).

The real part of \( z \) is \( a \), and the imaginary part is \( b \). Complex numbers can be added, subtracted, multiplied, and divided, much like real numbers, but with special rules for the imaginary unit \( i \). The set of all complex numbers is denoted by \( \textbf{C} \) and comprises a two-dimensional plane, known as the complex plane, where the x-axis represents the real part and the y-axis represents the imaginary part of complex numbers.
Magnitude of Complex Numbers
The magnitude (or absolute value) of a complex number is a measure of its size. For a complex number \( z = a + bi \), the magnitude, denoted as \( |z| \), is calculated as \( |z| = \sqrt{a^2 + b^2} \).

This formula is derived from the Pythagorean theorem, treating \( a \) and \( b \) as the sides of a right triangle, with the magnitude being the length of the hypotenuse. The magnitude of a complex number is always non-negative and can offer insight into the
Cauchy-Schwarz Inequality
The Cauchy-Schwarz Inequality is a powerful tool in linear algebra and other mathematical disciplines. It provides a way to compare the dot product of two vectors with the product of their magnitudes.

In the context of complex numbers \( z = a + bi \) and \( w = c + di \), the inequality states that the square of the sum of the products of corresponding parts (here, \( ac + bd \)) is less than or equal to the product of the sums of the squares of the components of each complex number (\( (a^2 + b^2)(c^2 + d^2) \)).

<h4>Mathematical Representation of the Cauchy-Schwarz Inequality</h4>
Mathematically, the Cauchy-Schwarz Inequality can be expressed as follows: \[ (ac + bd)^2 \leq (a^2 + b^2)(c^2 + d^2) \], which can also be written as: \[ ac + bd \leq \sqrt{a^2 + b^2}\sqrt{c^2 + d^2} \], or, in terms of magnitudes of complex numbers: \[ ac + bd \leq |z||w| \].

This inequality is crucial in proving various results and in ensuring the mathematical validity of operations involving vectors or complex numbers.
Proof Techniques in Linear Algebra
Linear algebra provides a variety of proof techniques for demonstrating the validity of mathematical statements, one of which we encounter when handling the Triangle Inequality for complex numbers.

One common approach in linear algebra is to use known inequalities like the Cauchy-Schwarz Inequality to establish the bound of an expression. Proofs can also involve direct calculation, manipulation of equations, and the application of core principles such as vector and matrix operations or properties of determinants.

In the exercise provided, the proof required us to represent the Triangle Inequality in terms of the components of complex numbers, square both sides to eliminate the square roots, and rearrange terms to effectively apply the Cauchy-Schwarz Inequality, ultimately demonstrating the truth of the expression for all complex numbers.

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Most popular questions from this chapter

Let \(u=(5,4,1), v=(3,-4,1), w=(1,-2,3) .\) Which pair of vectors, if any, are perpendicular (orthogonal)?

Prove Theorem 1.2: For any \(u, v, w\) in \(\mathbf{R}^{n}\) and \(k\) in \(\mathbf{R}\) : (i) \((u+v) \cdot w=u \cdot w+v \cdot w\), (ii) \((k u) \cdot v=k(u \cdot v)\), (iii) \(u \cdot v=v \cdot u\), (iv) \(u \cdot u \geq 0\), and \(u \cdot u=0\) iff \(u=0\). Let \(u=\left(u_{1}, u_{2}, \ldots, u_{n}\right), v=\left(v_{1}, v_{2}, \ldots, v_{n}\right), w=\left(w_{1}, w_{2}, \ldots, w_{n}\right)\) (i) Because \(u+v=\left(u_{1}+v_{1}, u_{2}+v_{2}, \ldots, u_{n}+v_{n}\right)\), $$ \begin{aligned} (u+v) \cdot w &=\left(u_{1}+v_{1}\right) w_{1}+\left(u_{2}+v_{2}\right) w_{2}+\cdots+\left(u_{n}+v_{n}\right) w_{n} \\ &=u_{1} w_{1}+v_{1} w_{1}+u_{2} w_{2}+\cdots+u_{n} w_{n}+v_{n} w_{n} \\ &=\left(u_{1} w_{1}+u_{2} w_{2}+\cdots+u_{n} w_{n}\right)+\left(v_{1} w_{1}+v_{2} w_{2}+\cdots+v_{n} w_{n}\right) \\ &=u \cdot w+v \cdot w \end{aligned} $$ (ii) Because \(k u=\left(k u_{1}, k u_{2}, \ldots, k u_{n}\right)\), $$ (k u) \cdot v=k u_{1} v_{1}+k u_{2} v_{2}+\cdots+k u_{n} v_{n}=k\left(u_{1} v_{1}+u_{2} v_{2}+\cdots+u_{n} v_{n}\right)=k(u \cdot v) $$ (iii) \(u \cdot v=u_{1} v_{1}+u_{2} v_{2}+\cdots+u_{n} v_{n}=v_{1} u_{1}+v_{2} u_{2}+\cdots+v_{n} u_{n}=v \cdot u\) (iv) Because \(u_{i}^{2}\) is nonnegative for each \(i\), and because the sum of nonnegative real numbers is nonnegative, $$ u \cdot u=u_{1}^{2}+u_{2}^{2}+\cdots+u_{n}^{2} \geq 0 $$ Furthermore, \(u \cdot u=0\) iff \(u_{i}=0\) for each \(i\), that is, iff \(u=0\)

Prove: For any vectors \(u, v, w\) in \(\mathbf{C}^{n}\) : (a) \((u+v) \cdot w=u \cdot w+v \cdot w\) (b) \(w \cdot(u+v)=w \cdot u+w \cdot v\).

Find \(k\) so that \(u\) and \(v\) are orthogonal, where: (a) \(u=(1, k,-3)\) and \(v=(2,-5,4)\) (b) \(u=(2,3 k,-4,1,5)\) and \(v=(6,-1,3,7,2 k)\)

Normalize each vector: (a) \(u=(5,-7)\) (b) \(v=(1,2,-2,4)\) (c) \(w=\left(\frac{1}{2},-\frac{1}{3}, \frac{3}{4}\right)\)
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