IB Maths AA HL · Unit 1: Number and Algebra
IB Maths AA HL Complex Numbers Questions
Exam-style IB Maths AA HL complex numbers questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 41 questions
- Paper 1: 36
- Paper 2: 5
- 7 easy
- 13 medium
- 15 hard
- 6 starter
- 3 worked examples
Practise Complex Numbers questions →
AA HL formula booklet
What's examined in AA HL complex numbers
The question bank covers these complex numbers question types (number of questions in brackets):
- De Moivre's Theorem & Roots (20)
- Complex Arithmetic & Forms (12)
- Loci & Polynomials (9)
Key formulas
- Complex numbers — polar (Cartesian → polar)
- \(z = a + bi = r\,\mathrm{cis}\,\theta = re^{i\theta}\)
- Modulus & argument
- \(r = |z| = \sqrt{a^2 + b^2},\ \theta = \arg z\)
In the same notation as the IB formula booklet. All AA HL formulas →
Complex Numbers worked examples
Worked example 1: Finding the modulus and argument · easy
Express the complex number $z = -1 + \sqrt{3}i$ in the polar form $r\text{cis}\theta$, where $-\pi < \theta \le \pi$.
1. Calculate the modulus $r = |z| = \sqrt{(-1)^2 + (\sqrt{3})^2}$.
2. Evaluate the sum to get $r = \sqrt{1 + 3} = 2$.
3. Determine the reference angle $\alpha = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.
4. Identify the quadrant on the Argand diagram; since the real part is negative and imaginary part is positive, it lies in Quadrant 2.
5. Compute the principal argument $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
6. State the final polar form: $\mathbf{z = 2\text{cis}\left(\frac{2\pi}{3}\right)}$.
Examiner tip: A common mistake is simply plugging $\arctan(-\sqrt{3}/1)$ into the calculator and blindly accepting the negative fourth-quadrant angle without checking the coordinates on an Argand diagram sketch.
Worked example 2: Applying De Moivre's theorem to powers · medium
Use De Moivre's Theorem to find the exact value of $(1 - i)^8$.
1. Convert the base $1 - i$ into polar form. The modulus is $r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
2. Find the argument. The point $(1, -1)$ is in the fourth quadrant, so $\theta = -\frac{\pi}{4}$.
3. Apply De Moivre's Theorem: $(r\text{cis}\theta)^n = r^n \text{cis}(n\theta)$.
4. Substitute the values into the theorem: $(\sqrt{2})^8 \text{cis}\left(8 \times -\frac{\pi}{4}\right)$.
5. Simplify the modulus and argument: $16\text{cis}(-2\pi)$.
6. Evaluate the exact Cartesian value by expanding: $16(\cos(-2\pi) + i\sin(-2\pi)) = 16(1 + 0i) = \mathbf{16}$.
Examiner tip: Students often try to expand high powers of complex binomials using Pascal's triangle, which is incredibly time-consuming and error-prone compared to immediately converting to polar form and utilizing De Moivre's theorem.
Worked example 3: Finding complex roots of equations · hard
Solve the equation $z^3 = 8i$, giving your answers in the Euler form $re^{i\theta}$ where $-\pi < \theta \le \pi$.
1. Express $8i$ in Euler form. Its modulus is $8$ and it lies on the positive imaginary axis, so $8i = 8e^{i\frac{\pi}{2}}$.
2. Set up the general equation for roots by adding $2k\pi i$ to the exponent to account for full rotations: $z^3 = 8e^{i\left(\frac{\pi}{2} + 2k\pi\right)}$.
3. Take the cube root of both sides to find the general root: $z = 8^{1/3} e^{i\left(\frac{\pi/2 + 2k\pi}{3}\right)} = 2e^{i\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)}$.
4. Substitute $k = 0$ to find the first root: $z_1 = \mathbf{2e^{i\frac{\pi}{6}}}$.
5. Substitute $k = 1$ to find the second root: $z_2 = 2e^{i\left(\frac{\pi}{6} + \frac{4\pi}{6}\right)} = \mathbf{2e^{i\frac{5\pi}{6}}}$.
6. Substitute $k = -1$ to ensure the principal argument domain $[-\pi, \pi]$ for the third root: $z_3 = 2e^{i\left(\frac{\pi}{6} - \frac{4\pi}{6}\right)} = \mathbf{2e^{-i\frac{\pi}{2}}}$.
Examiner tip: Forgetting to add $2k\pi$ to the argument before dividing by $n$ is the most frequent error, resulting in students only finding one single complex root instead of finding all $n$ distinct roots symmetrically spaced around the origin.
Try these IB Maths AA HL complex numbers questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 2 marks · Paper 1
Consider the complex number \(z = 3 - 4i\). Find its complex conjugate \(z^*\), and calculate its exact modulus \(|z|\).
Question 2 · medium · 3 marks · Paper 1
Given \(z_1 = 4\operatorname{cis}\left(\frac{\pi}{3}\right)\) and \(z_2 = 2\operatorname{cis}\left(\frac{\pi}{6}\right)\), find the product \(z_1 z_2\), leaving your answer in the simplest Cartesian form \(a + bi\).
Question 3 · hard · 6 marks · Paper 2
Find the three distinct complex cube roots of \(w = 27\operatorname{cis}\left(\frac{3\pi}{4}\right)\). Give your answers in the Euler form \(re^{i\theta}\), where \(-\pi < \theta \le \pi\).
All 41 complex numbers questions with mark schemes →
FAQ
How many IB Maths AA HL complex numbers questions are there?
There are 41 exam-style complex numbers questions in the AA HL question bank (Paper 1: 36 · Paper 2: 5), graded 7 easy, 13 medium, 15 hard, 6 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is complex numbers on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 36 · Paper 2: 5. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA HL Unit 1 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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