01 · 0–4 minutes

Answers on a line

We have always put real solutions on a number line. Today we will give some other answers a place.

For example, x² = 4 has two real solutions: −2 and 2. We can mark both.

Our route today

Move from the real line to an Argand plane; name points as complex numbers; connect the unit circle, multiplication and rotation.

02 · 4–10 minutes

An equation that does not fit

Let us stay with the number line and try x² + 1 = 0.

For every real x, x² ≥ 0. So the real line has no solution to show us.

There is a familiar movement that gives us a clue. Turn a plane vector 90° twice, and it ends as its negative. We can watch those two turns slowly.

R₉₀² = −I

Here I is the identity transformation. R90 is a matrix acting on vectors; this movement suggests how we might extend our number system.

Start with v = (2, 1).

03 · 10–15 minutes

Widen the number system

We can give these missing real solutions a place by widening our number system.

A big-picture result, the Fundamental Theorem of Algebra, says a nonconstant polynomial of degree n has n complex roots when we count repeated roots. We will use its promise, without proving it today.

Introduce a number i with the rule

i² = −1.

With this rule, i and −i both solve x² + 1 = 0. We can show them in a plane.

Keep the rule i² = −1 beside you. We will build its meaning one step at a time.

04 · 15–21 minutes

One point, two descriptions

The point stays where it is. We are giving it a second name.

An ordinary coordinate grid calls it (a, b). An Argand diagram calls the same point a + bi. The horizontal coordinate is the real part; the vertical coordinate is the imaginary part.

Worked example(−2, 3) ↔ −2 + 3i

Move 2 left and 3 up in either description.

05 · 21–27 minutes

Modulus is vector length

Think of z = a + bi as the vector (a, b). Its modulus, |z|, is the distance from zero.

|a + bi| = √(a² + b²)

Worked example: z = 3 + 2i

1. Read the coordinates: a = 3, b = 2.

2. Use the vector length: |z| = √(3² + 2²).

3. |z| = √(9 + 4) = √13.

Your turn: what is |5 − 2i|? Write it before checking.

|5 − 2i| = √(5² + (−2)²) = √29.

06 · 27–33 minutes

Argument is direction

The argument of a nonzero complex number is its angle from the positive real axis. We count anticlockwise as positive.

For today, use the principal interval (−π, π]. Check the quadrant before interpreting an arctangent.

Worked examples: check the quadrant first

3 + 2i is (3, 2), in quadrant I. The reference angle is arctan(2/3), so

arg(3 + 2i) = arctan(2/3).

5 − 2i is (5, −2), in quadrant IV. Its reference angle is arctan(2/5); the principal argument is negative:

arg(5 − 2i) = −arctan(2/5).

Zero has modulus 0, but it has no direction, so arg(0) is undefined.

07 · 33–41 minutes

The unit circle, newly named

You already know this point. Last lesson we called it (cos θ, sin θ).

On the Argand plane, the same point is cos θ + i sin θ. We shorten that expression to cis θ:

cis θ = cos θ + i sin θ

At θ = π/3, the coordinates are (1/2, √3/2), so

cis(π/3) = 1/2 + (√3/2)i.

Nothing happened to the point. The new notation helps us see the connection.

Angle

Use the sliders or their arrow keys. The exact worked values are in the explanation.

08 · 41–48 minutes

Keep direction; grow radius

On a bigger circle, a point on the same ray has the same direction and scaled coordinates.

z = r(cos θ + i sin θ) = r cis θ

Here r = |z| is the radius. For a nonzero point, θ is an argument.

Worked example

Double the radius at θ = π/3:

2cis(π/3) = 1 + √3 i.

Its modulus is 2 and its argument is π/3.

Convert 3 + 2i. First r = √(3² + 2²) = √13. Next, (3, 2) is in quadrant I, so θ = arctan(2/3).

3 + 2i = √13 cis(arctan(2/3)).

Check: r cosθ = 3 and r sinθ = 2.

Angle Radius

Use the sliders or their arrow keys. The exact worked values are in the explanation.

09 · 48–55 minutes

Multiply: a worked method

Before we investigate lengths and angles, let us make one product by algebra we trust.

Worked example(2 + i)(1 − 2i)

= 2(1 − 2i) + i(1 − 2i)

= 2 − 4i + i − 2i²

= 2 − 3i + 2, since i² = −1

= 4 − 3i.

We distribute as usual. The i² term becomes real. Keep this method beside you for the table.

Follow the terms

Real × real: 2

Two imaginary terms: −4i + i = −3i

The i² term: −2i² = +2

4 − 3i

10 · 55–66 minutes

Investigation: products and lengths

Start row A together. Then try B and C with a partner, keeping your own calculations. What pattern do the three lengths suggest?

First step for A(3 + 2i)² = 9 + 6i + 6i + 4i²

Now use i² = −1, collect terms, and find the three moduli.

Pair	a	b	\|a\|, \|b\|, \|ab\|
A	3 + 2i	3 + 2i	\|a\|\|b\|\|ab\|
B	5 − 2i	3 + 2i	\|a\|\|b\|\|ab\|
C	√3/2 − i/2	1/√2 + i/√2	\|a\|\|b\|\|ab\|

Type your working values; sqrt(13) and pi/12 are fine. Entries stay as you move between slides. Checking keeps your entries.

Pair	ab	\|a\|	\|b\|	\|ab\|
A	5 + 12i	√13	√13	13
B	19 + 4i	√29	√13	√377
C	(√6+√2)/4 + ((√6−√2)/4)i	1	1	1

Each time, |ab| = |a||b|.

11 · 66–75 minutes

Investigation: directions

Return to the same pairs. Find each argument in (−π, π]. Plot first, then calculate; inverse tangent needs a quadrant check.

A a = b = 3 + 2iB a = 5 − 2i; b = 3 + 2iC a = √3/2 − i/2; b = 1/√2 + i/√2

Pair	a	b
A	3 + 2i	3 + 2i
B	5 − 2i	3 + 2i
C	√3/2 − i/2	1/√2 + i/√2

For C, look for unit-circle angles before reaching for your calculator.

Type your working values; sqrt(13) and pi/12 are fine. Entries stay as you move between slides. Checking keeps your entries.

Pair	arg(a)	arg(b)	arg(ab)
A	arctan(2/3)	arctan(2/3)	arctan(12/5)
B	−arctan(2/5)	arctan(2/3)	arctan(4/19)
C	−π/6	π/4	π/12

In every row, the product's argument is the sum of the factor arguments, understood modulo a full turn.

12 · 75–77 minutes

The pattern we found

Our table shows a relationship between a product, lengths and directions. For nonzero complex numbers a and b:

|ab| = |a||b|arg(ab) ≡ arg(a) + arg(b) (mod 2π)

Adding a full turn changes the angle we write, but not the direction of the point.

In words

Multiply the distances from zero.

Add the turns from the positive real axis.

Then check that the result makes sense on Argand axes.

13 · 77–81 minutes

Multiplying by i

Since i = cis(π/2), multiplication by i keeps the modulus and adds a quarter-turn anticlockwise.

i(a + bi) = −b + ai(a, b) ↦ (−b, a)

This matches the action of R₉₀ on a vector. Number multiplication and a rotation matrix give two descriptions of the same movement.

Worked check: i(2 − 3i) = 3 + 2i.

14 · 81–85 minutes

A unit factor and a matrix

Choose a complex number c = a + bi with modulus 1. To find its matrix, watch what multiplying by c does to the two basis vectors.

Start with the horizontal basis vector1 = (1, 0) → c = a + bi = (a, b)

So the first matrix column is (a, b).

Then the vertical basis vectori = (0, 1) → ci = −b + ai = (−b, a)

So the second column is (−b, a). Here we used i² = −1.

a−bba

For a unit number, a = cosθ and b = sinθ. Substituting gives the familiar rotation matrix:

cosθ−sinθsinθcosθ

For example, c = i sends 1 to i and i to −1: both basis vectors turn 90° anticlockwise.

15 · 85–90 minutes

Show me the connection

On your own, put z = 2cis(π/6) on an Argand diagram.

Where does zi land? Give its cis form, its Cartesian form and one sentence explaining the movement.

Take your time to make a prediction. Your explanation helps me see which connection we should revisit next lesson.

z = √3 + i. Multiplying by i preserves modulus 2 and adds π/2: zi = 2cis(2π/3) = −1 + √3 i. The point turns 90° anticlockwise.