2.2 The Complex Number System and Operations

2.2 The Complex Number System and Operations

Quick Answer: The imaginary unit i satisfies i² = -1, meaning i = √(-1). Every complex number is expressed in standard form a + bi, where a is the real part and b is the imaginary part. Complex numbers add and subtract like algebraic polynomials, multiply using FOIL with the substitution i² = -1, and divide by multiplying the numerator and denominator by the complex conjugate of the denominator (a - bi).

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Defining the Imaginary Unit i and Standard Form (N-CN.1)

Throughout elementary algebra, students are taught that squaring any real number produces a non-negative result: $x^2 \ge 0$ for all $x \in \mathbb{R}$. Consequently, an equation as simple as $x^2 + 1 = 0$ (or $x^2 = -1$) has no real solutions. To resolve this limitation and provide solutions to all polynomial equations, mathematicians expanded the real number system $\mathbb{R}$ into the complex number system $\mathbb{C}$.

Formal Definition of the Imaginary Unit

The fundamental building block of complex numbers is the imaginary unit, designated by the symbol $i$:

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

Using this definition, the principal square root of any negative real number $-c$ (where $c > 0$) can be extracted algebraically:

$$\sqrt{-c} = \sqrt{c \cdot (-1)} = \sqrt{c} \cdot \sqrt{-1} = i\sqrt{c}$$

• Example 1: $\sqrt{-25} = i\sqrt{25} = 5i$
• Example 2: $\sqrt{-48} = \sqrt{16 \cdot 3 \cdot (-1)} = 4i\sqrt{3}$
• Example 3: $-\sqrt{-81} = -(9i) = -9i$

The Standard Form $a + bi$

A complex number is any number that can be written in canonical standard form:

$$z = a + bi$$

where:

• $a \in \mathbb{R}$ represents the real part, denoted $\text{Re}(z)$.
• $b \in \mathbb{R}$ represents the imaginary part, denoted $\text{Im}(z)$ (note that $b$ itself is a real number; it represents the real coefficient multiplying $i$).

Complex numbers encapsulate all familiar number sets:

• If $b = 0$, the number $z = a + 0i = a$ is a pure real number (e.g., $7$, $-\pi$, $\sqrt{2}$).
• If $a = 0$ and $b \neq 0$, the number $z = 0 + bi = bi$ is a pure imaginary number (e.g., $3i$, $-11i$).
• If $a \neq 0$ and $b \neq 0$, the number is a strictly non-real complex number (e.g., $4 - 7i$).

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The Cyclic Periodicity of Powers of i

A hallmark feature of the imaginary unit $i$ is that its successive positive integer powers cycle endlessly through four distinct values: $\{i, -1, -i, 1\}$.

Deriving the Four-Term Cycle

1. $i^1 = i$
2. $i^2 = -1$ (by definition)
3. $i^3 = i^2 \cdot i = (-1) \cdot i = -i$
4. $i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1$
5. $i^5 = i^4 \cdot i = 1 \cdot i = i$
6. $i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1$
7. $i^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i$
8. $i^8 = (i^4)^2 = 1^2 = 1$

Because $i^4 = 1$, multiplying by $i^4$ is equivalent to multiplying by 1. Therefore, any power $i^n$ depends entirely on the remainder when the exponent $n$ is divided by 4.

Algorithm for Evaluating Large Powers of $i$

To simplify $i^n$ for any positive integer $n$:

1. Divide the exponent $n$ by 4 using long division or mental arithmetic.
2. Identify the integer remainder $r \in \{0, 1, 2, 3\}$.
3. Replace $i^n$ with $i^r$ (recalling that $i^0 = 1$).

• Example 1: Simplify $i^{77}$. Divide $77$ by $4$: $77 = 4(19) + 1$. The remainder is 1, so $i^{77} = i^1 = i$.
• Example 2: Simplify $i^{242}$. Divide $242$ by $4$: $242 = 4(60) + 2$. The remainder is 2, so $i^{242} = i^2 = -1$.
• Example 3: Simplify $i^{2024}$. Divide $2024$ by $4$: $2024 = 4(506) + 0$. The remainder is 0, so $i^{2024} = 1$.
• Example 4: Negative powers: $i^{-3} = \frac{1}{i^3} = \frac{1}{-i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = i$. Alternatively, add multiples of 4 to the exponent until it becomes positive: $i^{-3} = i^{-3 + 4} = i^1 = i$.

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Addition and Subtraction of Complex Numbers (N-CN.2)

Adding and subtracting complex numbers operates identically to combining like terms in polynomial algebra: combine the real parts together, and combine the imaginary parts together.

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

Worked Problem 1: Multi-Term Complex Simplification

Problem: Simplify $(8 - 5i) - (3 - 9i) + (-4 + 2i)$ and write the result in standard $a + bi$ form.

• Step 1: Distribute the negative sign across the second binomial.

  $$(8 - 5i) - 3 + 9i + (-4 + 2i)$$

• Step 2: Group all real terms and all imaginary terms.

  $$\text{Real terms:} \quad 8 - 3 - 4 = 1$$ $$\text{Imaginary terms:} \quad -5i + 9i + 2i = 6i$$

• Step 3: Combine into standard $a + bi$ form.

  $$1 + 6i$$

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Multiplication of Complex Numbers

Multiplying complex numbers requires applying the distributive property (FOIL) exactly like binomial multiplication, followed by the mandatory substitution $i^2 = -1$.

$$(a + bi)(c + di) = ac + adi + bci + bdi^2$$

Because $i^2 = -1$, the term $bdi^2$ transforms into a real number: $bd(-1) = -bd$. Grouping real and imaginary parts yields:

$$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$$

Worked Problem 2: Multiplying Two Complex Numbers

Problem: Find the product of $(4 - 3i)$ and $(2 + 5i)$ in standard $a + bi$ form.

• Step 1: Expand using FOIL.

  $$\text{First:} \quad 4 \cdot 2 = 8$$ $$\text{Outer:} \quad 4 \cdot 5i = 20i$$ $$\text{Inner:} \quad (-3i) \cdot 2 = -6i$$ $$\text{Last:} \quad (-3i) \cdot (5i) = -15i^2$$

• Step 2: Sum the four partial products.

  $$8 + 20i - 6i - 15i^2$$

• Step 3: Substitute $i^2 = -1$.

  $$8 + 14i - 15(-1) = 8 + 14i + 15$$

• Step 4: Combine real terms to write in $a + bi$ form.

  $$(8 + 15) + 14i = 23 + 14i$$

Squaring Complex Binomials

When squaring a complex number, apply $(u + v)^2 = u^2 + 2uv + v^2$:

$$(3 - 5i)^2 = 3^2 - 2(3)(5i) + (5i)^2 = 9 - 30i + 25i^2 = 9 - 30i + 25(-1) = -16 - 30i$$

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Complex Conjugates and Division

Just as square root radicals in denominators are eliminated by rationalizing with conjugates, imaginary units in denominators are eliminated using complex conjugates.

The Complex Conjugate Definition and Product Theorem

For any complex number $z = a + bi$, its complex conjugate is denoted $\bar{z}$ and defined by reversing the sign of the imaginary part:

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

When a complex number is multiplied by its conjugate, the imaginary cross-terms cancel completely:

$$(a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 - b^2(-1) = a^2 + b^2$$

Fundamental Theorem: Because $a$ and $b$ are real numbers, $a^2 \ge 0$ and $b^2 \ge 0$. Therefore, the product of any complex number and its conjugate is ALWAYS a non-negative real number: $(a + bi)(a - bi) = a^2 + b^2$.

Complex Division Algorithm

To divide two complex numbers $\frac{a + bi}{c + di}$:

1. Identify the complex conjugate of the denominator: $\overline{c + di} = c - di$.
2. Multiply both numerator and denominator by this conjugate: $\frac{(a + bi)(c - di)}{(c + di)(c - di)}$.
3. Expand the denominator into the real number $c^2 + d^2$.
4. Expand the numerator using FOIL and replace $i^2$ with $-1$.
5. Split the resulting fraction into separate real and imaginary parts to express the answer in standard $a + bi$ form: $$\frac{ac + bd}{c^2 + d^2} + \left(\frac{bc - ad}{c^2 + d^2}\right)i$$

Worked Problem 3: Complex Division

Problem: Express the quotient $\frac{7 + i}{3 - 4i}$ in standard $a + bi$ form.

• Step 1: Identify the conjugate of the denominator. Denominator: $3 - 4i \implies$ Conjugate: $3 + 4i$.

• Step 2: Multiply numerator and denominator by $(3 + 4i)$.

  $$\frac{7 + i}{3 - 4i} \cdot \frac{3 + 4i}{3 + 4i} = \frac{(7 + i)(3 + 4i)}{(3 - 4i)(3 + 4i)}$$

• Step 3: Evaluate the denominator.

  $$(3 - 4i)(3 + 4i) = 3^2 + 4^2 = 9 + 16 = 25$$

• Step 4: Expand the numerator.

  $$(7 + i)(3 + 4i) = 21 + 28i + 3i + 4i^2 = 21 + 31i + 4(-1) = 17 + 31i$$

• Step 5: Divide both terms by 25 and write in standard form.

  $$\frac{17 + 31i}{25} = \frac{17}{25} + \frac{31}{25}i$$

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Common Regents Pitfalls and Exam Strategies

• Pitfall 1: The radical product fallacy. Students frequently multiply negative radicands under a single radical: $$\sqrt{-4} \cdot \sqrt{-9} = \sqrt{(-4)(-9)} = \sqrt{36} = 6 \quad \text{(WRONG!)}$$ The identity $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is valid only when at least one radicand is non-negative. To multiply radicals with negative radicands, you must convert to $i$-notation first: $$\sqrt{-4} \cdot \sqrt{-9} = (2i) \cdot (3i) = 6i^2 = 6(-1) = -6$$
• Pitfall 2: Leaving $i^2$ unsimplified. Leaving $i^2$ in a final answer (e.g., writing $21 - 4i^2$ instead of $25$) violates the "simplest form" requirement and routinely incurs a 1-credit deduction on Regents constructed-response items.
• Pitfall 3: Not splitting quotients into standard form. If a question instructs you to write an answer in $a + bi$ form, writing $\frac{17 + 31i}{25}$ may lose credit. Always split into $\frac{17}{25} + \frac{31}{25}i$.
• Pitfall 4: Subtraction distribution slips. In $(5 - 2i) - (8 - 6i)$, students often forget that the minus sign applies to the $-6i$, incorrectly writing $-2i - 6i = -8i$ instead of $-2i - (-6i) = +4i$.
• TI-84 Plus CE Calculator Check: Put your graphing calculator into complex mode: press [MODE], scroll to the line reading REAL a+bi re^(θi), and select a+bi. Press [2nd] [ . ] to type the imaginary unit $i$. You can directly evaluate expressions such as (7 + i)/(3 - 4i) and convert the result to exact fractions using [MATH] [1: >Frac].