Complex Numbers | Math Learning Guide

📋 Prerequisites

When the discriminant is negative, you used to stop and say "no real solutions." Complex numbers let you keep going. They show up everywhere in advanced math — polynomial roots, electrical engineering, quantum mechanics. For competition math, you mainly need the basics: arithmetic, modulus, and the conjugate root theorem.

📚 Key Concepts

What Is a Complex Number?

The imaginary unit: $ i = \sqrt{-1} $, so $ i^2 = -1 $

A complex number has the form $ a + bi $, where:

$a$ = real part (Re)
$b$ = imaginary part (Im)

Real numbers are complex numbers with $b = 0$. Pure imaginary numbers have $a = 0$.

The powers of i cycle every 4:

$ i^1 = i $
$ i^2 = -1 $
$ i^3 = -i $
$ i^4 = 1 $
Then it repeats: $ i^5 = i $, $ i^6 = -1 $, etc.

Trick: $ i^n = i^{n \mod 4} $. For $ i^{37} $: 37 mod 4 = 1, so $ i^{37} = i^1 = i $.

Adding & Subtracting Complex Numbers

Combine real parts and imaginary parts separately:

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

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

Example: $ (3 + 2i) + (1 - 4i) = 4 - 2i $

Multiplying Complex Numbers

FOIL like binomials, then simplify $i^2 = -1$:

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

Example: $ (2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 2 + i + 3 = 5 + i $

Complex Conjugate

The conjugate of $ a + bi $ is $ a - bi $ (flip the sign of the imaginary part).

Notation: $ \overline{a + bi} = a - bi $

Key property: Multiplying a complex number by its conjugate gives a real number:

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

This is the difference of squares pattern, and it's how you divide complex numbers.

Dividing Complex Numbers

Multiply numerator and denominator by the conjugate of the denominator (like rationalizing):

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

Result: real denominator, standard $a + bi$ form.

Modulus (Absolute Value)

The modulus (or magnitude) of $ z = a + bi $ is its distance from the origin in the complex plane:

\[ |z| = |a + bi| = \sqrt{a^2 + b^2} \]

Properties:

$ |z \cdot w| = |z| \cdot |w| $
$ |\frac{z}{w}| = \frac{|z|}{|w|} $
$ z \cdot \overline{z} = |z|^2 $

Complex Conjugate Root Theorem

If a polynomial has real coefficients and $ a + bi $ is a root, then its conjugate $ a - bi $ is also a root.

Complex roots come in conjugate pairs!

Useful for factoring: if you know one complex root, you automatically know another. Multiply $(x - (a+bi))(x - (a-bi)) = x^2 - 2ax + (a^2 + b^2)$ to get a real quadratic factor.

⚠️ Common Mistake: $ \sqrt{-a} = i\sqrt{a} $, not $ \sqrt{-a} = \sqrt{-1}\sqrt{a} = i\sqrt{a} $ (which is correct). But be careful: $ \sqrt{a} \times \sqrt{b} = \sqrt{ab} $ only works when both $a, b \ge 0$. $ \sqrt{-1} \times \sqrt{-1} \neq \sqrt{1} $ — that would give 1 instead of -1. Always convert to $i$ form first.

💡 Key Insight: For competition problems, the conjugate root theorem is the most powerful tool. If a real-coefficient polynomial has a complex root, its twin is automatically there too. This means odd-degree real polynomials must have at least one real root (complex roots come in pairs, can't account for all of them). Also: when dividing complex numbers, always multiply by the conjugate — it's the same move as rationalizing radicals.

✏️ Example Problems

📝 Example 1 (Power of i)

What is the value of $ i^{42} $?
Express your answer as a simplified complex number $a + bi$. What is $a + b$?

Pattern: Powers of i cycle every 4.

$ i^1 = i $, $ i^2 = -1 $, $ i^3 = -i $, $ i^4 = 1 $

Step 1: Find 42 mod 4.

$ 42 \div 4 = 10 $ remainder $ 2 $

$ 42 \mod 4 = 2 $

Step 2: $ i^{42} = i^2 = -1 $

In $a + bi$ form: $ -1 + 0i $

So $ a = -1 $, $ b = 0 $

$ a + b = -1 + 0 = -1 $

Answer: -1

📝 Example 2 (Multiply complex numbers)

Compute: $ (2 + i)(3 + 4i) $
Express as $a + bi$. What is $a + b$?

Step 1: FOIL (First, Outer, Inner, Last).

\[ (2 + i)(3 + 4i) = 2(3) + 2(4i) + i(3) + i(4i) \]

Step 2: Simplify each term.

$ = 6 + 8i + 3i + 4i^2 $

Step 3: Replace $i^2$ with $-1$ and combine like terms.

$ = 6 + 11i + 4(-1) $

$ = 6 + 11i - 4 $

$ = 2 + 11i $

Step 4: $ a = 2 $, $ b = 11 $

$ a + b = 2 + 11 = 13 $

Answer: 13

📝 Example 3 (Modulus & conjugate)

What is the modulus of $ 3 + 4i $?

Modulus formula: $ |a + bi| = \sqrt{a^2 + b^2} $

Step 1: Identify $a = 3$, $b = 4$.

Step 2: Apply formula.

\[ |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Intuition: This is the 3-4-5 right triangle! In the complex plane, the point (3, 4) is 5 units away from the origin.

Check with conjugate: $ (3+4i)(3-4i) = 9 + 16 = 25 = 5^2 = |z|^2 $. ✓

Answer: 5