🎯⭐ INTERACTIVE LESSON

The Complex Number System

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The Complex Number System - Complete Interactive Lesson

Part 1: The Imaginary Unit i

🌀 The Complex Number System

Part 1 of 5 — The Imaginary Unit $i$

Topics in This Part

Section
Why We Invented $i$
Simplifying $\sqrt{-n}$
The Powers of $i$

🔑 Key Concept: No real number squares to a negative. So mathematicians defined a new number, $i$ , with the single property $i^2 = -1$ . That one definition unlocks an entire number system.

Why We Invented $i$

Try to solve $x^2 = -1$ . On the real number line there is no answer — any real number squared is zero or positive. To fill that gap, we define the imaginary unit:

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

Concept Check 🎯

Simplify the Radical 🧮

Write each as a real coefficient times $i$ . Enter just the coefficient of $i$ (the number in front).

1) $\sqrt{-36} = \,?\,i$

The Powers of $i$ Cycle

Watch what happens when we keep multiplying by $i$ :

$i^1 = i \qquad i^2 = -1 \qquad i^3 = i^2\cdot i = -i \qquad i^4 = i^2\cdot i^2 = 1$

Cycle Through the Powers 🔽

Use the remainder-after-dividing-by- $4$ shortcut.

Part 1 Recap

You now own the two facts the whole system rests on:

$i^2 = -1$ — the definition. Use it to simplify $\sqrt{-n} = \sqrt{n}\,i$ .

Part 2: Standard Form, the Complex Plane & Adding

🌀 The Complex Number System

Part 2 of 5 — Standard Form, the Complex Plane & Adding

🔑 The Idea: A complex number glues a real part and an imaginary part together: $a + bi$ . Real numbers and imaginary numbers are both just special cases of this one form.

Standard Form: $a + bi$

Every complex number can be written as

$a + bi \qquad\text{where } a \text{ and } b \text{ are real numbers.}$

Part 3: Multiplying Complex Numbers

🌀 The Complex Number System

Part 3 of 5 — Multiplying Complex Numbers

🔑 Why it works: Multiply complex numbers exactly like binomials (FOIL), then replace every $i^2$ with $-1$ and recombine. That single substitution is the entire trick.

FOIL, Then Replace $i^2$

Part 4: Conjugates, Division & Modulus

🌀 The Complex Number System

Part 4 of 5 — Conjugates, Division & Modulus

🔑 Big Payoff: You can't leave an $i$ in a denominator. Multiplying top and bottom by the conjugate clears it, turning any complex quotient into clean $a + bi$ form.

The Complex Conjugate

The conjugate of $a + bi$ is — same real part, opposite sign on the imaginary part. We write it .

Part 5: Mixed Practice & Mastery Check

🌀 The Complex Number System

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) simplify roots and powers of $i$ , (2) add and subtract, (3) multiply, and (4) take conjugates, divide, and find the modulus. Let's put it all together.

Quick Reference

Goal	Key move
Simplify $\sqrt{-n}$

That's the whole idea. Everything else follows from $i^2 = -1$ .

Square Roots of Negatives

For any positive number $n$ :

$\sqrt{-n} = \sqrt{n}\cdot\sqrt{-1} = i\sqrt{n}$

Expression	Rewrite	Simplified
$\sqrt{-9}$	$\sqrt{9}\cdot\sqrt{-1}$	$3i$
$\sqrt{-25}$	$\sqrt{25}\cdot\sqrt{-1}$
$\sqrt{-12}$	$\sqrt{4}\cdot\sqrt{3}\cdot\sqrt{-1}$

⚠️ Pull out $i$ first. The rule $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ fails for negatives: $\sqrt{-4}\cdot\sqrt{-9}\ne\sqrt{36}$ . Convert each to $i$ -form before multiplying.

\sqrt{-100} = \,?\,i

\sqrt{-1} = \,?\,i

Then it repeats — $i^5 = i^4\cdot i = i$ , and so on. The powers cycle through four values:

Power	$i^1$	$i^2$	$i^3$	$i^4$	$i^5$	$i^6$
Value	$i$	$-1$	$-i$	$1$	$i$	$-1$

🔑 The Shortcut: Divide the exponent by $4$ and use the remainder:

remainder $1 \to i$

remainder $2 \to -1$

remainder $3 \to -i$

remainder $0 \to 1$

Example: $i^{23}$ . Since $23 \div 4 = 5$ remainder $3$ , we get $i^{23} = -i$ .

Powers of $i$ cycle every 4 — use the remainder of the exponent over

4

In Part 2 we combine a real part and an imaginary part into a single complex number and learn to add and subtract them.

a + bi where a and b are real numbers.

$a$ is the real part, written $\text{Re}(z)$ .
$b$ is the imaginary part, written $\text{Im}(z)$ — note $b$ is the real coefficient, not $bi$ .

Number	Real part $a$	Imaginary part $b$
$3 + 5i$	$3$	$5$
$-2 - 7i$	$-2$	$-7$
$4i$	$0$	$4$
$6$	$6$	$0$

💡 A real number is just a complex number with $b = 0$ , and a pure imaginary number is one with $a = 0$ . Complex numbers contain the reals.

We graph $a + bi$ as the point $(a, b)$ : the horizontal axis is the real axis and the vertical axis is the imaginary axis. So $3 + 5i$ lives at the point $(3, 5)$ .

Concept Check 🎯

Adding & Subtracting

Combine like parts — reals with reals, imaginaries with imaginaries — just like combining like terms:

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

Worked Example: Add

$(3 + 5i) + (4 - 2i) = (3+4) + (5-2)i = 7 + 3i$

Worked Example: Subtract

Distribute the minus sign to both parts of the second number:

$(6 - i) - (2 + 4i) = (6-2) + (-1-4)i = 4 - 5i$

⚠️ The classic slip: forgetting to subtract the imaginary part too. $-(2 + 4i) = -2 - 4i$ , so the $i$ -terms become $-1 - 4 = -5$ , not .

Add & Subtract 🧮

Each answer is a complex number $a + bi$ . Enter $a$ in the first box and $b$ in the second.

1) $(2 + 3i) + (5 + 4i) = a + bi$ 2) $(7 - 2i) - (3 + 6i) = a + bi$

Classify & Combine 🔽

Part 2 Recap

Standard form is $a + bi$ : real part $a$ , imaginary part $b$ .
Plot it at $(a, b)$ in the complex plane.
Add/subtract by combining real with real and imaginary with imaginary — and distribute the minus sign carefully.

Next up: multiplying complex numbers, where $i^2 = -1$ does the heavy lifting.

To multiply $(a + bi)(c + di)$ , distribute as usual, then use $i^2 = -1$ .

Worked Example: $(2 + 3i)(4 + 5i)$

\begin{aligned} (2 + 3i)(4 + 5i) &= 8 + 10i + 12i + 15i^2 \\ &= 8 + 22i + 15(-1) \\ &= 8 + 22i - 15 \\ &= -7 + 22i \end{aligned}

⚠️ Don't forget $i^2 = -1$ . The $15i^2$ term becomes the real number $-15$ , which changes the real part. Leaving it as $+15$ is the #1 multiplication error.

Multiplying by $i$ Alone

Distribute, then apply $i^2 = -1$ :

$i(3 - 4i) = 3i - 4i^2 = 3i - 4(-1) = 4 + 3i$

A Useful Special Product

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

The cross terms cancel and $-b^2 i^2 = +b^2$ , so the result is the real number $a^2 + b^2$ . We'll use this everywhere in Part 4.

Product	Result
$(3 + 2i)(3 - 2i)$	$3^2 + 2^2 = 13$

Concept Check 🎯

Multiply It Out 🧮

Write each product as $a + bi$ . Enter $a$ then $b$ .

1) $(2 + i)(1 + 3i) = a + bi$ 2) $i(2 - 5i) = a + bi$

Spot the Real Result 🔽

Each conjugate product collapses to a real number $a^2 + b^2$ .

Part 3 Recap

Multiply by FOIL/distribution, then replace each $i^2$ with $-1$ .
$i$ times a number rotates real ↔ imaginary parts (and flips a sign).
The conjugate product $(a + bi)(a - bi) = a^2 + b^2$ is always real.

That last fact is the key to dividing complex numbers — our Part 4 topic.

$z$	$\overline{z}$
$3 + 5i$	$3 - 5i$
$-2 - 7i$	$-2 + 7i$
$4i$	$-4i$
$6$	$6$

The magic property, from Part 3:

$z\cdot\overline{z} = (a + bi)(a - bi) = a^2 + b^2 \quad(\text{always a non-negative real number})$

Find the Conjugate 🔽

Flip the sign of the imaginary part only.

Dividing Complex Numbers

To divide, multiply numerator and denominator by the conjugate of the denominator. This makes the denominator real.

Worked Example: $\dfrac{2 + 3i}{1 - i}$

The denominator's conjugate is $1 + i$ :

$\frac{2 + 3i}{1 - i}\cdot\frac{1 + i}{1 + i} = \frac{(2 + 3i)(1 + i)}{(1 - i)(1 + i)}$

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

Denominator: $(1 - i)(1 + i) = 1^2 + 1^2 = 2$

$= \frac{-1 + 5i}{2} = -\frac{1}{2} + \frac{5}{2}i$

💡 Split the final fraction into real and imaginary pieces: $\dfrac{-1 + 5i}{2} = \dfrac{-1}{2} + \dfrac{5}{2}i$ . Standard form always separates the two parts.

Divide & Simplify 🧮

Simplify $\dfrac{5}{2 + i}$ to the form $a + bi$ .

1) Multiply top and bottom by the conjugate $2 - i$ . The new denominator is $\,?$ 2) The real part $a = \,?$ (decimal ok) 3) The imaginary part $b = \,?$ (decimal ok)

Modulus (Absolute Value)

The modulus $|z|$ is the distance from $a + bi$ to the origin in the complex plane — a straight Pythagorean distance:

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

Examples

$|3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

$|{-5} + 12i| = \sqrt{(-5)^2 + 12^2} = \sqrt{169} = 13$

💡 Notice $|z| = \sqrt{z\cdot\overline{z}}$ , since . The conjugate and the modulus are two sides of the same coin.

Concept Check 🎯

Part 4 Recap

The conjugate of $a + bi$ is $a - bi$ ; their product $a^2 + b^2$ is real.
Divide by multiplying top and bottom by the denominator's conjugate, then split into $a + bi$ .
Modulus $|a + bi| = \sqrt{a^2 + b^2}$ is the distance to the origin.

Part 5 puts all four skills together — then an Exit Quiz.

⚠️ The two errors that cost the most points: dropping $i^2 = -1$ when multiplying, and not distributing the minus sign to the imaginary part when subtracting.

Mixed Skills 🔽

One quick check from each part.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.

(1−i)(1+i)(2+3i)(1+i)

z\cdot\overline{z} = a^2 + b^2

The Complex Number System

The Complex Number System - Complete Interactive Lesson

Part 1: The Imaginary Unit i

🌀 The Complex Number System

Topics in This Part

Why We Invented iii

The Powers of iii Cycle

Part 1 Recap

Part 2: Standard Form, the Complex Plane & Adding

🌀 The Complex Number System

Standard Form: a+bia + bia+bi

Part 3: Multiplying Complex Numbers

🌀 The Complex Number System

FOIL, Then Replace i2i^2i2

Part 4: Conjugates, Division & Modulus

🌀 The Complex Number System

The Complex Conjugate

Part 5: Mixed Practice & Mastery Check

🌀 The Complex Number System

Quick Reference

Square Roots of Negatives

The Complex Plane

Adding & Subtracting

Worked Example: Add

Worked Example: Subtract

Part 2 Recap

Worked Example: (2+3i)(4+5i)(2 + 3i)(4 + 5i)(2+3i)(4+5i)

Multiplying by iii Alone

A Useful Special Product

Part 3 Recap

Dividing Complex Numbers

Worked Example: 2+3i1−i\dfrac{2 + 3i}{1 - i}1−i2+3i​

Modulus (Absolute Value)

Examples

Part 4 Recap

Why We Invented $i$

The Powers of $i$ Cycle

Standard Form: $a + bi$

FOIL, Then Replace $i^2$

Worked Example: $(2 + 3i)(4 + 5i)$

Multiplying by $i$ Alone

Worked Example: $\dfrac{2 + 3i}{1 - i}$