Linear-Algebra-Done-Right · Unit 1 · Video 1 · Interactive Practice

Complex Numbers as Pairs of Real Numbers

IKey Formulas

Formula	Name	What it says
$\mathbf{C} = \{a + bi : a, b \in \mathbf{R}\}$	Complex numbers	$a + bi$ is the ordered pair $(a, b)$ ; equal exactly when $a = c$ and $b = d$
$(a + bi) + (c + di) = (a + c) + (b + d)i$	Addition	Part by part
$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$	Multiplication	Rederive it: distribute, then $i^2 = -1$
$\beta - \alpha = \beta + (-\alpha)$ , $\;\beta / \alpha = \beta (1/\alpha)$ for $\alpha \neq 0$	Subtraction, division	Built from the unique inverses

Key Insight: The definition assumes nothing about $i$ . Writing $i = 0 + 1i$ , the product formula proves $i \cdot i = (0 \cdot 0 - 1 \cdot 1) + (0 \cdot 1 + 1 \cdot 0)i = -1$ . Throughout the course $\mathbf{F}$ stands for $\mathbf{R}$ or $\mathbf{C}$ , and its elements are called scalars.

IIVisualization 1 — Arithmetic on Ordered Pairs

A complex number is an ordered pair of reals, so sums and differences are computed part by part.

Operation

IIIVisualization 2 — Where the Product Formula Comes From

Distributing and replacing $i^2$ by $-1$ rederives the product formula $(ac - bd) + (ad + bc)i$ .

a 2 b 3 c 4 d 5

Step 1

IVVisualization 3 — Hunting for the Multiplicative Inverse

Each nonzero $\alpha$ has exactly one $\beta$ with $\alpha\beta = 1$ ; zero has none.

💡 Challenge: find $1/\alpha$ for every nonzero choice of $\alpha$ ; the theorem guarantees exactly one point works each time, and $\beta/\alpha$ is then just $\beta$ times that point.

VQuiz Questions

Problem 1 · Multiply by Distributing

Given: $\alpha = 3 - 2i$ and $\beta = 1 + 4i$ — find $\alpha\beta$ .

✅ Correct! The four terms are

3 + 12i - 2i - 8i^2

, and

-8i^2 = +8

, so

\alpha\beta = 11 + 10i

❌ Close, but… you treated

i^2

+1

. With

i^2 = -1

the term

-8i^2

+8

, not

-8

❌ Not quite. Multiplication is not part by part:

(3)(1) + (-2)(4)i

drops the cross terms

12i

and

-2i

and the

i^2

term. Only addition works part by part.

❌ Close, but check the sign. The cross term

(-2i)(1)

-2i

, so the

i

-part is

12 - 2 = 10

, not

12 + 2

❌ Not quite. Distribute into four terms, then replace

i^2

-1

Show solution

Distribute, using the commutative property to move real factors past $i$ :

(3 - 2i)(1 + 4i) = 3 \cdot 1 + 3 \cdot 4i + (-2i) \cdot 1 + (-2i)(4i) = 3 + 12i - 2i - 8i^2.

Replace $i^2$ by $-1$ , so $-8i^2 = 8$ , and collect real and $i$ terms:

(3 + 8) + (12 - 2)i = 11 + 10i.

Check with the formula ( $a = 3$ , $b = -2$ , $c = 1$ , $d = 4$ ): $(ac - bd) + (ad + bc)i = (3 - (-8)) + (12 + (-2))i = 11 + 10i$ ✓

Problem 2 · Equality Means Both Parts Agree

Given: real numbers $x$ and $y$ with $(x + 2y) + (x - y)i = 5 + 2i$ — find $x$ and $y$ .

What is x?

What is y?

✅ Correct! Equal complex numbers have equal first parts and equal second parts, giving

x + 2y = 5

and

x - y = 2

, so

x = 3

y = 1

❌ Not quite.

x = 5

would need

y = 0

in the real part, but then the

i

-part is

5 - 0 = 5 \neq 2

. Both parts must agree at once.

❌ Close, but check the subtraction.

(x + 2y) - (x - y) = 3y

, and

5 - 2 = 3

, so

3y = 3

and

y = +1

❌ Check x. Equate the real parts and the

i

-parts separately, then solve the two equations together.

❌ Check y. Equate the real parts and the

i

-parts separately, then solve the two equations together.

Show solution

A complex number is an ordered pair, so $a + bi = c + di$ exactly when $a = c$ and $b = d$ . Since $x$ and $y$ are real, $x + 2y$ and $x - y$ are the two parts:

x + 2y = 5, \qquad x - y = 2.

Subtracting the second equation from the first gives $3y = 3$ , so $y = 1$ , and then $x = 2 + y = 3$ .

Check: $(3 + 2) + (3 - 1)i = 5 + 2i$ ✓

Problem 3 · Inverse, Then Division

Given: $\alpha = 1 + 2i$ and $\beta = 3 + i$ — find $1/\alpha$ , then $\beta/\alpha = \beta(1/\alpha)$ .

What is 1/α?

What is β/α?

✅ Correct!

(1 + 2i)\left(\tfrac{1}{5} - \tfrac{2}{5}i\right) = \tfrac{1}{5}(1 + 4) = 1

, and

\beta/\alpha = (3 + i)\left(\tfrac{1}{5} - \tfrac{2}{5}i\right) = 1 - i

❌ Not quite. Inverting each part is the part-by-part trap:

(1 + 2i)(1 + \tfrac{1}{2}i) = 1 + \tfrac{1}{2}i + 2i + i^2 = \tfrac{5}{2}i

, not

1

❌ Close, but…

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

, not

1

. Divide by

5

❌ Close, but check the sign.

(1 + 2i)(1 - 2i) = 1 - 4i^2 = 1 + 4 = 5

, not

1 - 4 = -3

❌ Check 1/α. Multiply your answer by

1 + 2i

: it must give exactly

1

❌ Not quite. Division is not part by part:

\beta/\alpha

is defined as

\beta

times

1/\alpha

, and

(1 + 2i)(3 + \tfrac{1}{2}i) = 2 + \tfrac{13}{2}i \neq 3 + i

❌ Close, but… that is

\beta\alpha/5 = (3 + i)(1 + 2i)/5

. Multiply

\beta

1/\alpha = \tfrac{1}{5} - \tfrac{2}{5}i

, not by

\alpha

❌ Close, but… you multiplied by

1 - 2i

instead of

1/\alpha = \tfrac{1}{5}(1 - 2i)

. Divide by

5

❌ Check β/α. Compute

(3 + i)\left(\tfrac{1}{5} - \tfrac{2}{5}i\right)

by distributing.

Show solution

Step 1: the inverse. Since $(1 + 2i)(1 - 2i) = 1 - 2i + 2i - 4i^2 = 1 + 4 = 5$ , dividing $1 - 2i$ by the real number $5$ gives a number whose product with $\alpha$ is $1$ :

\frac{1}{\alpha} = \frac{1}{5} - \frac{2}{5}i, \qquad (1 + 2i)\Big(\frac{1}{5} - \frac{2}{5}i\Big) = \frac{1}{5}(1 + 2i)(1 - 2i) = \frac{5}{5} = 1.

The theorem says this inverse is unique, so $1/\alpha$ names exactly this number.

Step 2: division. By definition $\beta/\alpha = \beta(1/\alpha)$ :

(3 + i)\Big(\frac{1}{5} - \frac{2}{5}i\Big) = \frac{1}{5}(3 + i)(1 - 2i) = \frac{1}{5}(3 - 6i + i - 2i^2) = \frac{1}{5}(5 - 5i) = 1 - i.

Check: $\alpha(1 - i) = (1 + 2i)(1 - i) = 1 - i + 2i - 2i^2 = 3 + i = \beta$ ✓

Problem 4 · Powers of a Scalar

Given: the scalar $\alpha = 1 + i \in \mathbf{C}$ — find $\alpha^4$ , using $(\alpha^m)^n = \alpha^{mn}$ .

✅ Correct!

\alpha^2 = 2i

, so

\alpha^4 = (\alpha^2)^2 = (2i)^2 = 4i^2 = -4

❌ Not quite.

(1 + i)^4

is not

1^4 + i^4

: a power of a sum is not the sum of the powers. Only

(\alpha\beta)^m = \alpha^m\beta^m

holds, for products.

❌ Close, but…

(2i)^2 = 2^2 \cdot i^2 = 4 \cdot (-1)

(\alpha\beta)^m = \alpha^m \beta^m

. The

i^2

contributes a factor

-1

❌ Close, but…

(2i)^2

squares both factors:

(2i)^2 = 2^2 i^2

, not

2^2 i

❌ Not quite. Compute

\alpha^2

first, then square it.

Show solution

Step 1: $\alpha^2 = (1 + i)(1 + i) = 1 + i + i + i^2 = 1 + 2i - 1 = 2i$ .

Step 2: with $m = 2$ , $n = 2$ , the rule $(\alpha^m)^n = \alpha^{mn}$ gives $\alpha^4 = (\alpha^2)^2$ , and $(\alpha\beta)^2 = \alpha^2\beta^2$ gives

\alpha^4 = (2i)^2 = 2^2 \, i^2 = 4 \cdot (-1) = -4.

A nonreal scalar can have a negative real fourth power: in $\mathbf{C}$ , $x^4 = -4$ has the solution $x = 1 + i$ , while in $\mathbf{R}$ it has none.

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