Complex numbers: definition, arithmetic and polar form

IB Mathematics AI HL· 25 min read

1. Definition and Cartesian Form★★☆☆☆⏱ 8 min

📘 Definition

Complex Number (Cartesian Form)

z = a + bi

A number of the form $a + bi$ , where $a$ and $b$ are real numbers, and $i^{2} = - 1$ . $a = Re (z)$ (real part) and $b = Im (z)$ (imaginary part).

Example:

For $z = 3 - 2 i$ , $Re (z) = 3$ , $Im (z) = - 2$

All complex numbers can be plotted as points $(a, b)$ on the complex plane, with the horizontal axis as the real axis and the vertical axis as the imaginary axis.

📐 Worked Example

Find $Re (z)$ and $Im (z)$ for $z = i (4 - 2 i)$ , after simplifying to Cartesian form.

1
Expand the product, using $i^{2} = - 1$ :
2
$i(4 - 2i) = 4i - 2i^2 = 4i - 2(-1) = 2 + 4i$
3
Match to the form $a + bi$ to get the components:
4
$Re (z) = 2$ , $Im (z) = 4$

Exam tip:

Always simplify your expression fully before identifying real and imaginary parts, and do not forget the sign of the imaginary part.

2. Arithmetic of Complex Numbers (Cartesian Form)★★★☆☆⏱ 10 min

Arithmetic operations on complex numbers follow the same rules as real numbers, with the additional simplification $i^{2} = - 1$ . Addition, subtraction, and multiplication are straightforward, while division requires the use of the complex conjugate.

📘 Definition

Complex Conjugate

\bar{z}

For $z = a + bi$ , the conjugate is $\overset{z}{ˉ} = a - bi$ . The product $z \overset{z}{ˉ} = a^{2} + b^{2}$ , which is always a positive real number. This property lets us rationalize denominators when dividing complex numbers.

Example:

Conjugate of $1 - i$ is $1 + i$ , and $(1 - i) (1 + i) = 2$

📐 Worked Example

Simplify $\frac{3 + 2 i}{1 - 2 i}$ to Cartesian form $a + bi$ .

1
1. Get the conjugate of the denominator: $\overline{1 - 2 i} = 1 + 2 i$
2
1. Multiply numerator and denominator by the conjugate to keep the expression equivalent:
3
$\frac{(3 + 2i)(1 + 2i)}{(1 - 2i)(1 + 2i)}$
4
1. Expand and simplify denominator:
5
$(1)^2 - (2i)^2 = 1 - 4(-1) = 5$
6
1. Expand and simplify numerator:
7
$3(1) + 3(2i) + 2i(1) + 2i(2i) = 3 + 6i + 2i + 4i^2 = 3 + 8i - 4 = -1 + 8i$
8
1. Divide by the denominator to get final result:
9
$\frac{-1 + 8i}{5} = -\frac{1}{5} + \frac{8}{5}i$

3. Polar Form of Complex Numbers★★★☆☆⏱ 12 min

Using the polar coordinate system on the complex plane, we can write any non-zero complex number in terms of its modulus (distance from origin) and argument (angle from positive real axis). This form simplifies multiplication and division of multiple complex numbers.

📘 Definition

Polar (Modulus-Argument) Form

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

Where $r = ∣ z ∣ \geq 0$ (modulus), and $θ = arg (z)$ (argument), usually given in the principal range $- π < θ \leq π$ . For any Cartesian complex number $a + bi$ , $r = a^{2} + b^{2}$ and $tan θ = \frac{b}{a}$ .

📐 Worked Example

Convert $z = - 1 + i$ from Cartesian to polar form.

1
1. Calculate the modulus $r$ :
2
$r = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}$
3
1. Check the quadrant of $z$ : $(- 1, 1)$ is in the second quadrant.
4
1. Calculate $tan θ = \frac{b}{a} = \frac{1}{- 1} = - 1$ . The reference angle is $\frac{π}{4}$ , adjust to the second quadrant:
5
$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
6
1. Write the final polar form:
7
$z = \sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$

For products and quotients of two complex numbers in polar form, we use the properties: if $z_{1} = r_{1} (cos θ_{1} + i sin θ_{1})$ and $z_{2} = r_{2} (cos θ_{2} + i sin θ_{2})$ , then:

$z_{1} z_{2} = r_{1} r_{2} (cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$
$\frac{z _{1}}{z _{2}} = \frac{r _{1}}{r _{2}} (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$

Exam tip:

Always verify the quadrant of your complex number when calculating the argument. The arctangent function on your calculator only returns values for the first and fourth quadrants, so you will need to add $π$ for points in the second and third quadrants.

4. Common Pitfalls

Wrong move:

Stating $Im (3 - 2 i) = 2$ , ignoring the negative sign

Why:

The imaginary part is defined as the coefficient of $i$ , which includes its sign

Correct move:

$Im (3 - 2 i) = - 2$

Wrong move:

When dividing complex numbers, only multiplying the numerator by the conjugate of the denominator

Why:

This changes the value of the expression, as it is equivalent to multiplying the entire fraction by the conjugate

Correct move:

Always multiply both numerator and denominator by the same conjugate to preserve equality

Wrong move:

Using the direct calculator output of $tan^{- 1} (\frac{1}{- 1}) = - \frac{π}{4}$ as the argument for $z = - 1 + i$

Why:

The point $(- 1, 1)$ lies in the second quadrant, while $tan^{- 1}$ only outputs angles between $- \frac{π}{2}$ and $\frac{π}{2}$

Correct move:

Adjust the output by adding $π$ to get the correct argument $\frac{3 π}{4}$

Wrong move:

Writing polar form with a negative modulus $r = - 2$

Why:

Modulus is a distance from the origin, so it is always non-negative. Negative signs are absorbed into the argument

Correct move:

Use $r \geq 0$ and add $π$ to the argument to account for negative coordinates

5. Quick Reference Cheatsheet

Concept	Cartesian Form $z = a + bi$	Polar Form $z = r (cos θ + i sin θ)$
Modulus	$a^{2} + b^{2}$	$r$
Conjugate	$a - bi$	$r (cos (- θ) + i sin (- θ))$
Addition	Add real/imaginary parts separately	Convert to Cartesian first
Multiplication	Expand binomials, replace $i^{2} = - 1$	$r_{1} r_{2} (cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$
Division	Multiply by conjugate of denominator	$\frac{r _{1}}{r _{2}} (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$
Argument	$tan θ = \frac{b}{a}$ (adjust for quadrant)	$θ$ (given directly)

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Complex number division simplification
2022 · 2
Cartesian to polar form conversion

What's Next

Mastering the definition, arithmetic, and polar form of complex numbers is a critical prerequisite for the next subtopics in this unit: finding roots of complex numbers using De Moivre's theorem, and solving higher-degree polynomials with complex roots. This topic regularly appears in both Paper 1 (non-calculator) and Paper 2 (calculator) IB exams, and the skills you learn here are also applied to advanced trigonometric identity problems later in the course.