Imaginary Number Rules

Why This Matters

Imaginary numbers aren't just a mathematical curiosity—they're the gateway to the entire complex number system that powers everything from electrical engineering to quantum physics. In Honors Algebra II, you're being tested on your ability to manipulate these numbers fluently, recognize patterns in powers of $i$, and perform operations that combine real and imaginary components. The skills you build here directly connect to polynomial equations with no real solutions, the Fundamental Theorem of Algebra, and graphical representations in the complex plane.

Here's the key insight: complex numbers follow predictable rules that mirror what you already know about real number operations—with one crucial twist involving $i^2 = -1$. Don't just memorize formulas—understand why multiplying by a conjugate eliminates imaginary denominators, how the powers of $i$ cycle, and what the geometric meaning of these operations looks like on the complex plane. That conceptual understanding is what separates students who struggle from those who ace the exam.

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The Foundation: Defining $i$ and Its Powers

Before you can work with complex numbers, you need to internalize what $i$ actually represents and how it behaves when raised to powers. The key mechanism is that $i$ is defined to make $i^2 = -1$, which creates a predictable four-step cycle.

The Definition of $i$

• $i = \sqrt{-1}$—this definition extends our number system beyond the reals to include solutions to equations like $x^2 + 1 = 0$
• Complex numbers take the form $a + bi$, where $a$ is the real part and $b$ is the imaginary part
• Applications span STEM fields—AC circuit analysis, signal processing, and fluid dynamics all rely on complex number operations

Powers of $i$

• $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$—then the cycle repeats every four powers
• To simplify $i^n$, divide $n$ by 4 and use the remainder: remainder 0 → 1, remainder 1 → $i$, remainder 2 → $-1$, remainder 3 → $-i$
• This pattern appears constantly on exams—expect questions asking you to simplify expressions like $i^{47}$ or $i^{102}$

Simplifying Square Roots of Negative Numbers

• $\sqrt{-a} = i\sqrt{a}$ for any positive real number $a$—always extract the $i$ first
• Write answers in standard form $a + bi$—this is required for most exam responses
• Common error to avoid: never write $\sqrt{-4} \cdot \sqrt{-9} = \sqrt{36}$; instead, convert to $2i \cdot 3i = 6i^2 = -6$

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Operations: Adding, Subtracting, and Multiplying

Complex number arithmetic follows familiar rules with one essential modification: whenever $i^2$ appears, replace it with $-1$. This is where your algebra fundamentals meet the new number system.

Adding and Subtracting Complex Numbers

• Combine like terms separately—add real parts together and imaginary parts together: $(a + bi) + (c + di) = (a + c) + (b + d)i$
• Subtraction works identically—distribute the negative sign, then combine: $(a + bi) - (c + di) = (a - c) + (b - d)i$
• The result stays in standard form—no simplification of $i$ needed since you're not multiplying imaginary terms

Multiplying Complex Numbers

• Use FOIL or distribution on $(a + bi)(c + di)$ to get $ac + adi + bci + bdi^2$
• Replace $i^2$ with $-1$ and combine: final answer is $(ac - bd) + (ad + bc)i$
• Watch for special products—multiplying conjugates $(a + bi)(a - bi) = a^2 + b^2$ always yields a real number

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Division and Conjugates

Division of complex numbers requires a clever technique: multiplying by a form of 1 that eliminates the imaginary part from the denominator. The conjugate is your essential tool here.

The Complex Conjugate

• The conjugate of $a + bi$ is $a - bi$—same real part, opposite sign on the imaginary part
• Key property: $(a + bi)(a - bi) = a^2 + b^2$, which is always a real, non-negative number
• Conjugates appear in polynomial theory—complex roots of polynomials with real coefficients always come in conjugate pairs

Dividing Complex Numbers

• Multiply numerator and denominator by the conjugate of the denominator: $\frac{a + bi}{c + di} \cdot \frac{c - di}{c - di}$
• The denominator becomes real: $c^2 + d^2$, eliminating all $i$ terms below the fraction bar
• Simplify to standard form—separate into $\frac{\text{real part}}{c^2 + d^2} + \frac{\text{imaginary part}}{c^2 + d^2}i$

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Geometric Representations

Complex numbers aren't just algebraic objects—they have a rich geometric interpretation. The complex plane lets you visualize operations as transformations, and polar form reveals the rotational nature of multiplication.

Graphing on the Complex Plane

• Horizontal axis = real part, vertical axis = imaginary part—the complex number $a + bi$ corresponds to point $(a, b)$
• Addition becomes vector addition—geometrically, you're placing arrows tip-to-tail
• This visualization helps with modulus and argument—distance from origin and angle from the positive real axis

Absolute Value (Modulus)

• $|a + bi| = \sqrt{a^2 + b^2}$—this is the distance from the origin to the point $(a, b)$
• Connects to the Pythagorean theorem—the modulus is the hypotenuse of a right triangle with legs $a$ and $b$
• Modulus of a product: $|z_1 \cdot z_2| = |z_1| \cdot |z_2|$—multiplying complex numbers multiplies their distances from the origin

Polar Form

• Standard form: $r(\cos\theta + i\sin\theta)$ where $r$ is the modulus and $\theta$ is the argument (angle)
• Multiplication in polar form is elegant: multiply moduli and add arguments—$(r_1 \text{cis } \theta_1)(r_2 \text{cis } \theta_2) = r_1 r_2 \text{cis}(\theta_1 + \theta_2)$
• Euler's formula connects to exponentials: $re^{i\theta} = r(\cos\theta + i\sin\theta)$—this bridge to calculus appears in advanced courses

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Quick Reference Table

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Self-Check Questions

1. What is $i^{87}$? Explain how you used the four-step cycle to find your answer without computing all 87 powers.

2. Which two operations—addition or multiplication—require you to use the property $i^2 = -1$, and why doesn't the other operation need it?

3. Compare and contrast: How does multiplying a complex number by its conjugate differ from multiplying two arbitrary complex numbers? What's special about the result?

4. If you need to divide $\frac{3 + 2i}{1 - 4i}$, explain each step of the process and why multiplying by the conjugate works.

5. A complex number has modulus 5 and argument $\frac{\pi}{3}$. Write it in both polar form and standard form $a + bi$. Which form would you prefer for multiplying this number by another complex number, and why?