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Complex Numbers: Algebraic Approach

The system of real numbers, while vast, is incomplete. A simple quadratic equation such as \(x^2 = -1\) has no solution within the real numbers. To address this, we extend the real number line into a two-dimensional plane, introducing a new number, the imaginary unit \(i\). This extension forms the set of complex numbers, \(\C\), a system where not only does \(x^2 = -1\) have a solution, but every non-constant polynomial with real (or complex) coefficients has a complete set of solutions in \(\C\) (Fundamental Theorem of Algebra). This chapter introduces the algebraic foundations of complex numbers, their operations, and their power in solving equations.

The Number \(i\) and the Set of Complex Numbers

Definition The Imaginary Unit \(i\)
The imaginary unit, denoted by \(i\), is defined as a number such that:$$\textcolor{colordef}{i^2 = -1}$$
Definition Complex Number
A complex number is a number of the form \(z = a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. The set of all complex numbers is denoted by \(\C\).
Note
  • \(i\) cannot be a real number, since no real number squared is equal to \(-1\).
  • A real number \(a\) is a complex number where \(b=0\), written as \(a = a+0i\). Thus \(\R \subset \C\).
  • A number where \(a=0\), such as \(bi\), is called a purely imaginary number.
Definition Real and Imaginary Parts
For a complex number \(z = a + bi\) (with \(a,b \in \R\)):
  • \(a\) is the real part of \(z\), denoted \(\Re(z)\).
  • \(b\) is the imaginary part of \(z\), denoted \(\Im(z)\).
By definition, \(\Re(z)\) and \(\Im(z)\) are real numbers.
Example
For \(z = 2+3i\):
  • \(\Re(2+3i)=2\)
  • \(\Im(2+3i)=3\)

Operations with Complex Numbers

The arithmetic of complex numbers follows the standard rules of algebra, together with the key relation \(i^2=-1\).
Definition Algebraic Operations
Let \(z = a+bi\) and \(w = c+di\), where \(a,b,c,d \in \R\).
  • Addition:$$\begin{aligned}[t]z + w &= (a+bi) + (c+di) \\ &= (a+c)+(b+d)i.\end{aligned}$$
  • Subtraction:$$\begin{aligned}[t]z - w &= (a+bi) - (c+di) \\ &= (a-c)+(b-d)i.\end{aligned}$$
  • Multiplication:$$\begin{aligned}[t]z \times w &= (a+bi)(c+di) \\ &= ac + adi + bci + bdi^2 \\ &= ac + adi + bci - bd \quad (i^2 = -1)\\ &= (ac-bd)+ (ad+bc)i.\end{aligned}$$
  • Division (with \(w \neq 0\)):$$\begin{aligned}[t]\dfrac{z}{w}&= \frac{a+bi}{c+di} = \left(\frac{a+bi}{c+di}\right)\left(\frac{c-di}{c-di}\right) \\ &= \frac{(a+bi)(c-di)}{(c+di)(c-di)} \\ &= \frac{ac-adi+bci-bdi^2}{c^2-(di)^2} \\ &= \frac{(ac+bd) + (bc-ad)i}{c^2+d^2}.\end{aligned}$$

Equality of Complex Numbers

Proposition Equality of Complex Numbers
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal:$$ a+bi=c+di \;\Leftrightarrow\; a=c \text{ and } b=d. $$

  • (\(\Rightarrow\)) Suppose \(a+bi=c+di\). Then$$(a-c) + (b-d)i = 0.$$This implies \(a-c = i(d-b)\). If \(d \neq b\), then$$i = \frac{a-c}{d-b}.$$Since \(a,b,c,d\) are real, the right-hand side is a real number, while \(i\) is not real (because \(i^2=-1\) has no real solution). This is a contradiction. Therefore, the assumption \(d \neq b\) must be false, which means \(d=b\). Substituting this back into \(a-c = i(d-b)\) gives \(a-c = 0\), so \(a=c\).
  • (\(\Leftarrow\)) Suppose \(a=c\) and \(b=d\). Then \(bi=di\). Adding the equalities \(a=c\) and \(bi=di\) gives$$a+bi = c+di.$$

Complex Conjugate

Definition Complex Conjugate
The complex conjugate of \(z = a + bi\) (with \(a,b\in\R\)) is \(\conjugate{z} = a - bi\). The complex conjugate of \(z\) is often denoted as \(\conjugate{z}\) or \(z^{*}\).
Example
The complex conjugate of \(2+3i\) is \(\conjugate{2+3i} = 2-3i\).
Proposition Properties of the Conjugate
Given two complex numbers \(z\) and \(w\):$$\begin{aligned}[t]\conjugate{\conjugate{z}} &= z, \\ \conjugate{z + w} &= \conjugate{z} + \conjugate{w}, \\ \conjugate{z - w} &= \conjugate{z} - \conjugate{w}, \\ \conjugate{zw} &= \conjugate{z} \; \conjugate{w}, \\ \conjugate{\left(\frac{z}{w}\right)} &= \frac{\conjugate{z}}{\conjugate{w}},\quad \text{if } w \neq 0.\end{aligned}$$

Let \(z=a+bi\) and \(w=c+di\).For example,$$\begin{aligned}\conjugate{z + w}&=\conjugate{(a+c)+(b+d)i}\\ &=(a+c)-(b+d)i \\ &= a-bi+c-di \\ &= \conjugate{z}+\conjugate{w}.\end{aligned}$$The other properties are proven similarly, by expanding and using \(i^2=-1\).