🎯⭐ INTERACTIVE LESSON

Complex Numbers and Operations

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Complex Numbers and Operations - Complete Interactive Lesson

Part 1: The Imaginary Unit & Powers of $i$

🌀 Complex Numbers and Operations

Part 1 of 7 — The Imaginary Unit & Powers of $i$

Topics in This Part

Section
Why We Need a New Number
Defining $i$ and Simplifying Square Roots
The Cycle of Powers of $i$

🔑 Key Concept: There is no real number whose square is negative. Mathematicians fixed this by inventing one — the imaginary unit $i$ , defined so that $i^2 = -1$ . That single idea unlocks the complex numbers, and once you master it, equations that "had no solution" suddenly do.

Why We Need a New Number

Try to solve $x^2 = -1$ . No real number works: a positive squared is positive, a negative squared is positive, and $0^2 = 0$ . So we define a brand-new number to plug the hole:

$i = \sqrt{}$

Concept Check 🎯

The Cycle of Powers of $i$

Watch what happens as we raise $i$ to higher powers:

$i^1 = i, \quad i^2 = -1, \quad i^3 = i^2\cdot i = -i, \quad i^4 = i^2\cdot i^2 = (-1)(-1) = 1$

Powers of $i$ 🧮

Use the remainder-on-division-by-4 trick. Enter each value as one of: 1, -1, i, or -i.

1) $i^{16} = \,?$ 2) $i^{30} = \,?$

Match Each Result 🔽

What You Can Now Do

You can simplify the square root of any negative number and evaluate any power of $i$ . These are the two building blocks of complex-number arithmetic.

In Part 2, we combine real and imaginary pieces into a single object — a complex number $a + bi$ — and learn to add, subtract, and multiply them.

Part 2: Adding, Subtracting & Multiplying

🌀 Complex Numbers and Operations

Part 2 of 7 — Adding, Subtracting & Multiplying

🔑 A complex number has the form $a + bi$ , where $a$ is the real part and $b$ is the imaginary part. Think of it as two numbers riding together: the real $a$ and the imaginary . Operations keep the two parts separate — until collapses an imaginary product back into a real number.

Part 3: Conjugates & Division

🌀 Complex Numbers and Operations

Part 3 of 7 — Conjugates & Division

🔑 The conjugate of $a + bi$ is $a - bi$ — same real part, flipped sign on the imaginary part. Multiplying a complex number by its conjugate always produces a real number, and that is exactly the trick that lets us divide.

The Complex Conjugate

The conjugate of $z = a + bi$ is written .

Part 4: The Complex Plane & Modulus

🌀 Complex Numbers and Operations

Part 4 of 7 — The Complex Plane & Modulus

🔑 Every complex number is a point. Plot $a + bi$ at coordinates $(a, b)$ in the complex plane: the horizontal axis is real, the vertical axis is imaginary. Its distance from the origin is the modulus $\lvert z\rvert$ .

Plotting in the Complex Plane

Part 5: Polar (Trigonometric) Form

🌀 Complex Numbers and Operations

Part 5 of 7 — Polar (Trigonometric) Form

🔑 A second address for every complex number. Instead of "go right $a$ , up $b$ ," polar form says "go out a distance $r$ at an angle $\theta$ ." This view makes multiplication and powers of complex numbers far easier.

Rectangular vs. Polar Form

A complex number $z = a + b$ can be described by its distance from the origin and the angle it makes with the positive real axis:

Part 6: Complex Roots of Quadratics

🌀 Complex Numbers and Operations

Part 6 of 7 — Complex Roots of Quadratics

🔑 Where complex numbers earn their keep. When the discriminant $b^2 - 4ac$ is negative, a quadratic has no real solutions — but it has two complex solutions, and they always come as a conjugate pair.

The Discriminant Tells You What to Expect

For $ax^2 + bx + c = 0$ , the quadratic formula is

Part 7: Mixed Mastery & Exit Quiz

🌀 Complex Numbers and Operations

Part 7 of 7 — Mixed Mastery & Exit Quiz

You can now (1) handle powers of $i$ , (2) add, subtract, multiply, and divide, (3) use conjugates and the modulus, (4) plot in the complex plane and convert to polar form, and (5) find complex roots of quadratics. Let's tie it all together.

Quick Reference

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

i = - 1, which means i^{2} = - 1

With $i$ in hand, we can take the square root of any negative number:

$\sqrt{-9} = \sqrt{9}\cdot\sqrt{-1} = 3i, \qquad \sqrt{-7} = \sqrt{7}\,i$

$\sqrt{-a} = i\sqrt{a} \quad (a > 0)$

Expression	Simplified
$\sqrt{-25}$	$5i$
$\sqrt{-16}$	$4i$
$\sqrt{-12}$	$\sqrt{4}\cdot\sqrt{3}\,i = 2i\sqrt{3}$
$\sqrt{-1}$	$i$

⚠️ Pull out the $i$ first. The rule $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ fails for negatives. For example $\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6$ , not $\sqrt{36} = 6$ . Always convert to $i$ before multiplying.

Then it repeats, because $i^5 = i^4\cdot i = 1\cdot i = i$ :

Power	$i^1$	$i^2$	$i^3$	$i^4$	$i^5$	$i^6$	$i^7$	$i^8$
Value	$i$	$-1$	$-i$	$1$	$i$	$-1$

🔑 The Period-4 Trick: $i^n$ depends only on the remainder when $n$ is divided by 4.

Remainder $0$ $1$ $2$ $3$
$i^n$ $1$ $i$ $-1$ $-i$

Example: $i^{27}$ . Divide $27 \div 4 = 6$ remainder $3$ . Remainder $3 \Rightarrow i^{27} = -i$ .

Addition & Subtraction

Combine like parts: real with real, imaginary with imaginary. It works exactly like combining like terms in algebra (treat $i$ as the "variable").

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

Worked Example

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

For subtraction, distribute the minus sign to both parts:

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

⚠️ Watch the sign on the imaginary part. Subtracting $-2i$ adds $2i$ , giving $5i + 2i = 7i$ , not $3i$ .

Add & Subtract 🧮

For each result $a + bi$ , enter the real part $a$ in the first box and the imaginary part $b$ in the second.

1) $(6 + 2i) + (1 + 9i)$ → $a = \,?,\ b = \,?$ 2) $(5 - 3i) - (2 + 4i)$ → $a = \,?,\ b = \,?$

Multiplication: FOIL, Then Replace $i^2$

Multiply two complex numbers just like two binomials (FOIL), then use $i^2 = -1$ to simplify.

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

$\underbrace{2\cdot 4}_{8} + \underbrace{2\cdot 5i}_{10i} + \underbrace{3i\cdot 4}_{12i} + \underbrace{3i\cdot 5i}_{15i^2}$

Combine the imaginary terms and replace $i^2 = -1$ :

$= 8 + 22i + 15i^2 = 8 + 22i + 15(-1) = 8 + 22i - 15 = -7 + 22i$

💡 The key step is $i^2 = -1$ . The last product $15i^2$ becomes the real number $-15$ , which merges with the .

Concept Check 🎯

Multiply It Out 🧮

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

1) $(1 + i)(1 - i)$ → $a = \,?,\ b = \,?$ 2) $(2 + 5i)(3 - i)$ → $a = \,?,\ b = \,?$

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

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

The middle terms cancel and $i^2 = -1$ turns the subtraction into addition. The answer $a^2 + b^2$ is real and non-negative — no $i$ left.

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

💡 This is a difference of squares in disguise: $(a+bi)(a-bi) = a^2 + b^2$ . Memorize it — it is the engine of complex division.

Concept Check 🎯

Dividing Complex Numbers

A fraction with $i$ in the denominator isn't fully simplified. Multiply the top and bottom by the conjugate of the denominator — this makes the denominator real.

Worked Example: $\dfrac{4 + 7i}{2 + i}$

Multiply by $\dfrac{2 - i}{2 - i}$ (the conjugate of $2 + i$ ):

$\frac{4 + 7i}{2 + i}\cdot\frac{2 - i}{2 - i} = \frac{(4 + 7i)(2 - i)}{(2 + i)(2 - i)}$

Denominator: $\;(2)^2 + (1)^2 = 4 + 1 = 5$ .

Numerator: $\;(4 + 7i)(2 - i) = 8 - 4i + 14i - 7i^2 = 8 + 10i + 7 = 15 + 10i$ .

$= \frac{15 + 10i}{5} = 3 + 2i$

🔑 Standard form is $a + bi$ . To get there, split the fraction over the real denominator: $\dfrac{15 + 10i}{5} = \dfrac{15}{5} + \dfrac{10}{5}i = 3 + 2i$ .

Walk Through the Division 🔽

You are simplifying $\dfrac{1 + 3i}{1 - i}$ . Choose what happens at each stage.

Divide & Simplify 🧮

Write each quotient as $a + bi$ . Enter $a$ , then $b$ .

1) $\dfrac{5}{i}$ → $a = \,?,\ b = \,?$ (hint: conjugate of $i$ is $-i$ ) 2) $\dfrac{10 + 5i}{1 + 2i}$ → $a = \,?,\ b = \,?$

The complex plane (also called the Argand diagram) replaces the usual $x$ - and $y$ -axes:

Real axis (horizontal) ← the real part $a$
Imaginary axis (vertical) ← the imaginary part $b$

So $z = a + bi$ lives at the point $(a, b)$ .

Complex Number	Point $(a, b)$	Location
$3 + 2i$	$(3, 2)$	Quadrant I
$-4 + i$	$(-4, 1)$	Quadrant II
$-2 - 3i$	$(-2, -3)$	Quadrant III
$5$	$(5, 0)$	on the real axis
$-6i$	$(0, -6)$	on the imaginary axis

💡 A real number sits on the horizontal axis ( $b = 0$ ); a pure imaginary number sits on the vertical axis ( $a = 0$ ).

Locate Each Number 🔽

The Modulus (Absolute Value)

The modulus $\lvert z\rvert$ is the distance from the origin to the point $(a, b)$ — straight from the Pythagorean theorem:

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

Worked Example: $z = 3 + 4i$

$\lvert z\rvert = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Worked Example: $z = -5 + 12i$

$\lvert z\rvert = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

⚠️ Square first, so signs disappear. $(-5)^2 = 25$ , not $-25$ . The modulus is always $\ge 0$ .

Concept Check 🎯

Compute the Modulus 🧮

Each of these gives a whole-number answer. Enter the modulus.

1) $\lvert 9 + 12i\rvert = \,?$ 2) $\lvert -7 + 24i\rvert = \,?$ 3) $\lvert 6 - 8i\rvert = \,?$

$r = \lvert z\rvert = \sqrt{a^2 + b^2}, \qquad a = r\cos\theta, \qquad b = r\sin\theta$

Substituting gives the trigonometric (polar) form:

$z = r(\cos\theta + i\sin\theta)$

Here $r$ is the modulus and $\theta$ is the argument (the angle).

$\tan\theta = \frac{b}{a}$

…but you must check the quadrant of $(a, b)$ to pick the correct angle, because $\tan$ repeats every $180^\circ$ .

💡 The reference angle from $\tan^{-1}\left\lvert\frac{b}{a}\right\rvert$ tells you the size; the quadrant tells you the true direction.

Worked Example: Convert $z = 1 + i\sqrt{3}$ to Polar Form

Step 1 — modulus: $r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

Step 2 — argument. The point $(1, \sqrt{3})$ is in Quadrant I: $\tan\theta = \frac{\sqrt{3}}{1} = \sqrt{3} \;\Rightarrow\; \theta = 60^\circ$

Step 3 — write it: $z = 2\left(\cos 60^\circ + i\sin 60^\circ\right)$

✅ Check by going back: $2\cos 60^\circ = 2\cdot\tfrac{1}{2} = 1$ and ✓

Concept Check 🎯

Reminder: $\cos 90^\circ = 0,\ \sin 90^\circ = 1,\ \cos 180^\circ = -1,\ \sin 180^\circ = 0$ , and $\cos 45^\circ = \sin 45^\circ = \dfrac{\sqrt{2}}{2}$ .

Why Polar Form Is Powerful

To multiply in polar form: multiply the moduli, add the angles.

$r_1(\cos\theta_1 + i\sin\theta_1)\cdot r_2(\cos\theta_2 + i\sin\theta_2) = r_1 r_2\big(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big)$

Example

$\big[2(\cos 30^\circ + i\sin 30^\circ)\big]\cdot\big[3(\cos 40^\circ + i\sin 40^\circ)\big] = 6\big(\cos 70^\circ + i\sin 70^\circ\big)$

Moduli $2\cdot 3 = 6$ ; angles $30^\circ + 40^\circ = 70^\circ$ .

🔑 This pattern is why polar form matters: repeated multiplication (powers) becomes repeated angle addition, which is the heart of De Moivre's Theorem in later courses.

Polar Operations 🔽

Polar Form Practice 🧮

1) The modulus $r$ of $z = -4 + 3i$ is $\,?$ (whole number) 2) Multiply $3(\cos 25^\circ + i\sin 25^\circ)\cdot 4(\cos 35^\circ + i\sin 35^\circ)$ . The new modulus is $\,?$ 3) …and the new angle (in degrees, without the degree symbol) is $\,?$

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The discriminant $\Delta = b^2 - 4ac$ decides the type of roots:

Discriminant $\Delta$	Roots
$\Delta > 0$	two distinct real roots
$\Delta = 0$	one repeated real root
$\Delta < 0$	two complex roots (a conjugate pair)

💡 When $\Delta < 0$ , the $\sqrt{\;}$ contains a negative number — and that is precisely where $i$ steps in: $\sqrt{-k} = i\sqrt{k}$ .

Concept Check 🎯

Worked Example: Solve $x^2 + 4x + 13 = 0$

Step 1 — identify $a = 1,\ b = 4,\ c = 13$ .

Step 2 — discriminant: $\Delta = 4^2 - 4(1)(13) = 16 - 52 = -36$

Negative, so expect a complex conjugate pair.

Step 3 — quadratic formula: $x = \frac{-4 \pm \sqrt{-36}}{2(1)} = \frac{-4 \pm 6i}{2}$

Step 4 — simplify by dividing each term by $2$ : $x = -2 \pm 3i$

So the roots are $-2 + 3i$ and $-2 - 3i$ — conjugates, as promised.

⚠️ Divide BOTH terms by $2a$ . Writing $\frac{-4 \pm 6i}{2}$ as $-4 \pm 3i$ is wrong; the must also be divided, giving .

Solve the Quadratic 🧮

Solve $x^2 - 6x + 25 = 0$ . The roots are $p \pm qi$ .

1) Discriminant $b^2 - 4ac = \,?$ 2) Real part $p = \,?$ 3) Imaginary coefficient $q = \,?$ (the positive value)

Build the Equation 🎯

⚠️ Top mistakes to avoid: dropping the $i$ from a negative square root; forgetting $i^2 = -1$ in a product; and not dividing both terms by $2a$ in the quadratic formula.

Mixed Practice 🔽

Mixed Computation 🧮

1) Write $(2 + i)(3 + 4i)$ as $a + bi$ : $a = \,?,\ b = \,?$ 2) Write $\dfrac{8 + i}{2 - i}$ as $a + bi$ : $a = \,?,\ b = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.

(2+i)(2−i)(4+7i)(2−i)

2\sin 60^\circ = 2\cdot\tfrac{\sqrt{3}}{2} = \sqrt{3}

Complex Numbers and Operations

The General Rule

Addition & Subtraction

Worked Example

Multiplication: FOIL, Then Replace $i^2$

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

The Magic Property

Worked Example

Dividing Complex Numbers

Worked Example: $\dfrac{4 + 7i}{2 + i}$

The Modulus (Absolute Value)

Worked Example: $z = 3 + 4i$

Worked Example: $z = -5 + 12i$

Finding $\theta$

Worked Example: Convert $z = 1 + i\sqrt{3}$ to Polar Form

Why Polar Form Is Powerful

Example

Worked Example: Solve $x^2 + 4x + 13 = 0$

Complex Numbers and Operations

Complex Numbers and Operations - Complete Interactive Lesson

Part 1: The Imaginary Unit & Powers of $i$

🌀 Complex Numbers and Operations

Topics in This Part

Why We Need a New Number

The Cycle of Powers of iii

What You Can Now Do

Part 2: Adding, Subtracting & Multiplying

🌀 Complex Numbers and Operations

Part 3: Conjugates & Division

🌀 Complex Numbers and Operations

The Complex Conjugate

Part 4: The Complex Plane & Modulus

🌀 Complex Numbers and Operations

Plotting in the Complex Plane

Part 5: Polar (Trigonometric) Form

🌀 Complex Numbers and Operations

Rectangular vs. Polar Form

Part 6: Complex Roots of Quadratics

🌀 Complex Numbers and Operations

The Discriminant Tells You What to Expect

Part 7: Mixed Mastery & Exit Quiz

🌀 Complex Numbers and Operations

Quick Reference

The General Rule

Addition & Subtraction

Worked Example

Multiplication: FOIL, Then Replace i2i^2i2

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

The Magic Property

Worked Example

Dividing Complex Numbers

Worked Example: 4+7i2+i\dfrac{4 + 7i}{2 + i}2+i4+7i​

The Modulus (Absolute Value)

Worked Example: z=3+4iz = 3 + 4iz=3+4i

Worked Example: z=−5+12iz = -5 + 12iz=−5+12i

Finding θ\thetaθ

Worked Example: Convert z=1+i3z = 1 + i\sqrt{3}z=1+i3​ to Polar Form

Why Polar Form Is Powerful

Example

Worked Example: Solve x2+4x+13=0x^2 + 4x + 13 = 0x2+4x+13=0

The Cycle of Powers of $i$

Multiplication: FOIL, Then Replace $i^2$

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

Worked Example: $\dfrac{4 + 7i}{2 + i}$

Worked Example: $z = 3 + 4i$

Worked Example: $z = -5 + 12i$

Finding $\theta$

Worked Example: Convert $z = 1 + i\sqrt{3}$ to Polar Form

Worked Example: Solve $x^2 + 4x + 13 = 0$