Complex Numbers and Their Algebra

Try this

Choose “ $z + w$ ”. The green sum is exactly the fourth vertex of the parallelogram with sides $z$ and $w$ . Drag $z$ : how does the parallelogram change?
Choose “ $z - w$ ”. How is the green point related to the arrow that goes from $w$ to $z$ ?
Choose “ $z \times w$ ” and drag $w$ to $i$ (the point $(0, 1)$ ). Through what angle has the product turned compared with $z$ ? Then drag $w$ to $- 1$ .
Choose “conjugate $\overset{z}{ˉ}$ ”. Why are the two readings $z + \overset{z}{ˉ}$ and $z \overset{z}{ˉ}$ always real?

Add one new number $i$ to the real numbers, with the rule $i^{2} = - 1$ , and allow it to be added, subtracted, multiplied and divided with real numbers. The result is the complex numbers $z = a + bi$ ( $a, b \in R$ ). $a$ is the real part $Re z$ and $b$ is the imaginary part $Im z$ .

$b = 0$ : a real number; $b \neq = 0$ : an imaginary number; $a = 0$ and $b \neq = 0$ : a purely imaginary number.
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. One complex equation is two real equations: this is called “turning the complex into the real”.
Calculate by the usual rules for polynomials, and at the end replace $i^{2}$ by $- 1$ :

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i .

Each complex number $a + bi$ corresponds to the point $(a, b)$ of the plane, and also to the vector from the origin to that point. This plane is the complex plane: the horizontal axis is the real axis and the vertical axis is the imaginary axis.

Predict first

What is $(1 + i)^{2}$ ?

2

2 i

0

2 + 2 i

Choose an answer first; then you can see the result.

Why

Multiplication is just “the distributive law + $i^{2} = - 1$ ”. Division needs one small trick: multiply the denominator by its conjugate $\overset{w}{ˉ}$ . Since $w \overset{w}{ˉ} = c^{2} + d^{2}$ is a positive real number, the denominator becomes real.

One product, one quotient: (1+2i)(2−i) and (4+3i)÷(1+2i)

Step 1 / 5

Set up the two numbers
$z = 1 + 2 i$ , $w = 2 - i$ . First guess: which quadrant will the product land in?

$(1 + 2 i) (2 - i) = ?$
Expand like a polynomial
Replace i² by −1
Now divide back: make the denominator real
Back where we started

The conjugate has a few more handy properties: $\overline{z_{1} \pm z_{2}} = \overset{z}{ˉ}_{1} \pm \overset{z}{ˉ}_{2}$ , $\overline{z_{1} z_{2}} = \overset{z}{ˉ}_{1} \overset{z}{ˉ}_{2}$ , $z + \overset{z}{ˉ} = 2 Re z$ , $z \overset{z}{ˉ} = a^{2} + b^{2}$ . In particular, $z$ is real $⟺ \overset{z}{ˉ} = z$ , and $z$ is purely imaginary $⟺ \overset{z}{ˉ} = - z$ and $z \neq = 0$ .

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $a + bi = 0$ implies $a = b = 0$

When $a, b$ are real, $a + bi$ is the point $(a, b)$ of the complex plane. If it equals $0$ it must be the origin, so $a = 0$ and $b = 0$ .

This is exactly why “turning the complex into the real” works: for example, from $(x + 2 y) + (x - y) i = 3$ (with $x, y$ real) we immediately get $x + 2 y = 3$ and $x - y = 0$ , so $x = y = 1$ .

Practice

Exercise 1Basic

Compute $(3 + 2 i) (1 - 4 i)$ and write the result in the form $a + bi$ .

Exercise 2Basic

For which real value of $m$ is $z = (m^{2} - 3 m - 4) + (m^{2} - 1) i$ purely imaginary?

Exercise 3Advanced

The complex number $z$ satisfies $z + 2 \overset{z}{ˉ} = 6 - i$ . Find $z$ (in the form $a + bi$ ).

Exercise 4Challenge

Compute $i + i^{2} + i^{3} + \dots + i^{2026}$ .

Try this

Why

One product, one quotient: (1+2i)(2−i) and (4+3i)÷(1+2i)

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Is “real part 0” enough for a purely imaginary number?

Using √a·√b = √(ab) for negative numbers