IB Maths AI HL Complex Numbers Paper 1 & 2 Cartesian form ~7 min read

Introduction to Complex Numbers

An equation like x² = −1 has no real solution — you cannot square root a negative. Mathematicians fixed this by defining a new number, i, whose square is −1. Every number built from a real part and a multiple of i is a complex number.

📘 What you need to know

i is the imaginary unit: i = √−1, so i² = −1.
A complex number has a real part and an imaginary part: z = a + bi, with a, b real.
Re(z) = a and Im(z) = b — the imaginary part does not include the i.
To root a negative, rewrite it as a multiple of √−1, e.g. √−4 = √4 × √−1 = 2i.
The set of all complex numbers is ℂ; a real number is just a complex number with b = 0.
Two complex numbers are equal only if their real parts and imaginary parts both match.

Imaginary numbers and the unit i

The equation x² = −1 has no real solution, because squaring any real number gives a result that is zero or positive. To get past this, mathematicians defined a brand-new number whose square is −1 and called it the imaginary unit, i.

The imaginary unit i = √−1 and i² = −1

Once i exists, the square root of any negative number can be found. Split off the −1 and root what is left: using √(ab) = √a × √b, write √−k = √k × √−1 = (√k)i. For example √−25 = √25 × √−1 = 5i.

Keep the −1 separate. Always pull out √−1 on its own — never split √−4 as √−2 × √−2. Rewrite as √4 × √−1 = 2i, then simplify the remaining surd fully.

Complex numbers in Cartesian form

A complex number has two parts: a real part and an imaginary part. Written in Cartesian form it looks like z = a + bi, where a and b are real numbers. Complex numbers are usually labelled z.

The real part is Re(z) = a and the imaginary part is Im(z) = b. Note carefully: the imaginary part is the coefficient only — for z = 3 − 5i, Im(z) = −5, not −5i.

The complex number z = 4 + 3i on the complex plane. Its real part (4) is read along the real axis; its imaginary part (3) along the imaginary axis.

A complex number can have just one part. If b = 0 it is purely real (e.g. 7); if a = 0 it is purely imaginary (e.g. −8i). The real numbers therefore sit inside the set of complex numbers ℂ.

Equating complex numbers

Two complex numbers are equal if and only if both their real parts are equal and both their imaginary parts are equal. So a + bi = c + di splits into two real equations: a = c and b = d. This lets one complex equation be solved for two unknowns.

Cartesian form & equality z = a + bi ⇒ Re(z) = a, Im(z) = b a + bi = c + di ⇒ a = c and b = d match real with real, imaginary with imaginary — never mix the two

🧭 Recipe — solving an equation with no real roots

Isolate the squared term — get x² or (x + k)² alone on one side.
Square root both sides — and always write the ±.
Split the negative — write √−k as √k × √−1.
Replace √−1 with i and simplify the surd fully.
Write the answer in Cartesian form a + bi, rearranging if needed.

Worked examples

WE 1

Square root of a negative number

Express √−50 in the form bi, where b is an exact value.

split off the −1 √−50 = √50 × √−1 simplify the surd: 50 = 25 × 2 √50 = 5√2 √−50 = 5√2 i 5√2 ≈ 7.07 — always simplify the surd, don’t leave √50.

WE 2

Real and imaginary parts

For z = −6 + 11i, write down Re(z) and Im(z). State whether w = −8i and v = 5 are purely real or purely imaginary.

read off the two parts of z Re(z) = −6, Im(z) = 11 classify w and v Re(z) = −6 · Im(z) = 11 w = −8i has Re = 0 ⇒ purely imaginary. v = 5 has Im = 0 ⇒ purely real.

WE 3

Solving x² = a negative

Solve the equation x² = −45, giving your answers in exact Cartesian form.

square root both sides — keep the ± x = ±√−45 = ±√45 × √−1 simplify: 45 = 9 × 5 √45 = 3√5 x = ±3√5 i two roots, equal and opposite — the ± is essential.

WE 4

A completed-square equation

Solve (x − 5)² = −36, giving your answers in Cartesian form.

square root both sides x − 5 = ±√−36 = ±√36 × √−1 x − 5 = ±6i rearrange into a + bi x = 5 ± 6i the two roots 5 + 6i and 5 − 6i differ only in the sign of the imaginary part.

WE 5

Equating two complex numbers

Find the real numbers p and q such that (p + 5) + (2q − 1)i = −3 + 7i.

match real parts p + 5 = −3 ⇒ p = −8 match imaginary parts 2q − 1 = 7 ⇒ 2q = 8 ⇒ q = 4 p = −8, q = 4 one complex equation gives two real equations — real with real, imaginary with imaginary.

WE 6

Full question: simplify and solve

(a) Simplify √−72. (b) Hence or otherwise, solve 2(x + 1)² + 50 = 0, giving your answers in Cartesian form.

(a) split off the −1 and simplify √−72 = √72 × √−1, 72 = 36 × 2 √−72 = 6√2 i (b) isolate the squared term 2(x + 1)² = −50 ⇒ (x + 1)² = −25 x + 1 = ±√−25 = ±5i (a) 6√2 i · (b) x = −1 ± 5i divide by the 2 first — isolate the bracket before square rooting.

💡 Top tips

i² = −1 is the single most useful fact — it turns every imaginary square back into a real number.
Always write ± when square-rooting — complex roots come in equal-and-opposite (or conjugate) pairs.
The imaginary part is just the coefficient: Im(3 − 5i) = −5, not −5i.
Simplify surds fully: √−72 = 6√2 i, never leave it as √72 i.
Put your GDC in complex (a + bi) mode — it may call Cartesian form “rectangular form”.

⚠ Common mistakes

Writing √−9 = −3 — a negative has no real root; √−9 = 3i.
Including the i in the imaginary part — Im(z) is a real number, the coefficient only.
Dropping the ± when taking a square root — this loses one of the two roots.
Splitting √−4 as √−2 × √−2 — keep the −1 separate: √4 × √−1 = 2i.
Leaving the surd unsimplified — give 5√2 i, not √−50.

Next up: Operations with Complex Numbers — adding, subtracting and multiplying in Cartesian form, plus the surprising pattern in the powers of i, which repeats every four steps.

Need help with AI HL Complex Numbers?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →