5.2 Complex Numbers & Complex Arithmetic

5.2 Complex Numbers & Complex Arithmetic

In the real number system $\mathbb{R}$, quadratic equations such as $x^2 + 1 = 0$ or $x^2 + 16 = 0$ possess no real solutions because no real number multiplied by itself yields a negative quantity. To resolve this structural limitation, mathematics defines the complex number system (denoted by $\mathbb{C}$), which extends the 1D real continuum into a 2D algebraic field.

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The Imaginary Unit and Standard Form $a + bi$

The fundamental building block of complex numbers is the imaginary unit, denoted by $i$, defined as:

$$i = \sqrt{-1} \quad \implies \quad i^2 = -1$$

Standard Form Definition

A complex number $z$ is any expression written in standard form:

$$z = a + bi$$

where $a$ and $b$ are real numbers ($a, b \in \mathbb{R}$).

• $a = \text{Re}(z)$ represents the real part of $z$.
• $b = \text{Im}(z)$ represents the imaginary part of $z$ (note that $b$ itself is a real scalar coefficient).

Subsets of Complex Numbers

• If $b = 0$, $z = a + 0i = a$, which is a purely real number. Thus, $\mathbb{R} \subset \mathbb{C}$.
• If $a = 0$ and $b \neq 0$, $z = 0 + bi = bi$, which is a purely imaginary number (e.g., $5i$, $-i\sqrt{3}$).

Geometric Representation & Absolute Value (Modulus)

Complex numbers are plotted on the 2D Complex Plane (Argand Plane), where the horizontal axis represents $\text{Re}(z)$ and the vertical axis represents $\text{Im}(z)$. The modulus (or magnitude) $|z|$ represents the Euclidean distance from the origin $(0, 0)$ to $(a, b)$:

$$|z| = \sqrt{a^2 + b^2}$$

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Cyclical Powers of $i$

Because $i^2 = -1$, repeated multiplication by $i$ yields a period-4 repeating pattern. Computing sequential powers of $i$ reveals:

Exponent Reduction Formula

To simplify any high integer power $i^n$, divide the exponent $n$ by $4$ to find the integer remainder $r \in \{0, 1, 2, 3\}$ (where $n \equiv r \pmod 4$):

$$i^n = i^r$$

• If remainder $r = 0 \implies i^n = 1$
• If remainder $r = 1 \implies i^n = i$
• If remainder $r = 2 \implies i^n = -1$
• If remainder $r = 3 \implies i^n = -i$

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Complex Arithmetic Operations

Complex numbers obey standard algebraic field rules, treating $i$ like a variable while substituting $i^2 = -1$ whenever higher powers appear.

1. Complex Addition & Subtraction

Add or subtract corresponding real parts and imaginary parts:

$$(a + bi) + (c + di) = (a + c) + (b + d)i$$ $$(a + bi) - (c + di) = (a - c) + (b - d)i$$

2. Complex Multiplication

Expand the product using the FOIL method (First, Outer, Inner, Last) and substitute $i^2 = -1$:

$$(a + bi)(c + di) = ac + adi + bci + bdi^2 = ac + (ad + bc)i + bd(-1) = (ac - bd) + (ad + bc)i$$

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Complex Conjugates and Division

Definition of Complex Conjugate

For any complex number $z = a + bi$, its complex conjugate, denoted $\bar{z}$ (or $z^*$), is formed by negating the imaginary part:

$$\bar{z} = a - bi$$

Conjugate Product Identity

Multiplying a complex number by its conjugate produces a strictly non-negative real scalar equal to the sum of the squares of its components:

$$z \bar{z} = (a + bi)(a - bi) = a^2 - abi + abi - b^2 i^2 = a^2 - b^2(-1) = a^2 + b^2$$

This fundamental identity is used to eliminate imaginary terms from denominators.

Complex Division Procedure

To divide complex numbers $\frac{a + bi}{c + di}$, multiply both numerator and denominator by the complex conjugate of the denominator $(c - di)$:

$$\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} = \left(\frac{ac + bd}{c^2 + d^2}\right) + \left(\frac{bc - ad}{c^2 + d^2}\right)i$$

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Comprehensive Worked Examples

Worked Example 1: Simplifying Higher Powers of $i$

Problem: Fully simplify the complex expression:

$$E = 4 i^{23} - 3 i^{54} + 2 i^{101}$$

Step-by-Step Solution:

1. Evaluate exponent remainders modulo 4:
   • $23 \div 4 = 5$ with remainder $3 \implies i^{23} = i^3 = -i$.
   • $54 \div 4 = 13$ with remainder $2 \implies i^{54} = i^2 = -1$.
   • $101 \div 4 = 25$ with remainder $1 \implies i^{101} = i^1 = i$.
2. Substitute reduced powers back into the expression: $$E = 4(-i) - 3(-1) + 2(i)$$
3. Perform scalar multiplications: $$E = -4i + 3 + 2i$$
4. Combine real terms and imaginary terms into standard form $a + bi$: $$E = 3 + (-4i + 2i) = 3 - 2i$$

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Worked Example 2: Complex Multiplication and Division

Problem: Express $z = \frac{(4 - 2i)(3 + 5i)}{1 + 2i}$ in standard form $a + bi$.

Step-by-Step Solution:

1. Multiply numerator terms using FOIL: $$(4 - 2i)(3 + 5i) = 12 + 20i - 6i - 10i^2$$ Since $i^2 = -1$, $-10(-1) = 10$: $$\text{Numerator} = (12 + 10) + (20i - 6i) = 22 + 14i$$
2. Set up the division quotient: $$z = \frac{22 + 14i}{1 + 2i}$$
3. Multiply numerator and denominator by denominator conjugate $(1 - 2i)$: $$z = \frac{(22 + 14i)(1 - 2i)}{(1 + 2i)(1 - 2i)}$$
4. Expand numerator and denominator:
   • Numerator: $(22)(1) - 44i + 14i - 28i^2 = 22 - 30i + 28 = 50 - 30i$.
   • Denominator: $1^2 + 2^2 = 1 + 4 = 5$.
5. Separate into real and imaginary parts: $$z = \frac{50 - 30i}{5} = \frac{50}{5} - \frac{30}{5}i = 10 - 6i$$

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Worked Example 3: Modulus and Conjugate Algebraic Identity

Problem: Given $z = 3 - 4i$, calculate the conjugate product $z \bar{z}$, the square $z^2$, and evaluate $z \bar{z} + z^2$.

Step-by-Step Solution:

1. Find conjugate product: $z \bar{z} = (3 - 4i)(3 + 4i) = 3^2 + 4^2 = 9 + 16 = 25$.
2. Expand $z^2$: $(3 - 4i)^2 = 3^2 - 2(3)(4i) + (4i)^2 = 9 - 24i - 16 = -7 - 24i$.
3. Sum expressions: $25 + (-7 - 24i) = (25 - 7) - 24i = 18 - 24i$.