Complex Numbers (AA HL)
Complex numbers extend the real number line to a full plane, letting every polynomial equation have a solution. This topic covers writing a complex number in Cartesian, modulus-argument and Euler form, converting between them, and using De Moivre's theorem to raise complex numbers to powers and find their roots - including the roots of unity that sit at the heart of several classic proofs.
What the syllabus says
This topic maps onto three points in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| AHL1.12 | Complex numbers: the number \(i\), where \(i^2=-1\). Cartesian form \(z=a+bi\); the terms real part, imaginary part, conjugate, modulus and argument. The complex plane (also known as the Argand diagram). |
| AHL1.13 | Modulus-argument (polar) form \(z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta\) and Euler form \(z=re^{i\theta}\). Converting between Cartesian, polar and Euler forms is expected. Sums, products and quotients in each form, and their geometric interpretation. |
| AHL1.14 | Complex conjugate roots of quadratic and polynomial equations with real coefficients occur in conjugate pairs. De Moivre's theorem and its extension to rational exponents. Powers and roots of complex numbers. |
Complex numbers are AHL-only content, not examined at AA SL.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is the modulus of a complex number?
The modulus \(|z|\) is the distance from the origin to the point \(z=a+bi\) on the Argand diagram, found using Pythagoras on the real and imaginary parts. It's always a non-negative real number, and it's the "\(r\)" in modulus-argument and Euler form.
e.g. For \(z=3+4i\): \(|z|=\sqrt{3^2+4^2}=\sqrt{25}=5.\)
What is the argument of a complex number?
The argument \(\arg z\) is the angle the line from the origin to \(z\) makes with the positive real axis, measured anticlockwise, usually given in the range \(-\pi<\theta\le\pi.\) You need to check which quadrant \(z\) lies in before trusting a raw \(\arctan\) value.
e.g. For \(z=-1+\sqrt3\,i\) (second quadrant): \(\arg z = \pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}.\)
What is the complex conjugate?
The conjugate \(\bar z\) of \(z=a+bi\) is \(a-bi\) - the reflection of \(z\) in the real axis. Conjugates are essential because \(z\bar z=|z|^2\) is always real, which is how you divide complex numbers, and because non-real roots of real-coefficient polynomials always come in conjugate pairs.
e.g. For \(z=3+4i\): \(z\bar z = (3+4i)(3-4i) = 9+16 = 25 = |z|^2.\)
What does De Moivre's theorem say?
De Moivre's theorem states that \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\) for any integer \(n\). In modulus-argument form this means raising \(z=r\,\text{cis}\,\theta\) to the power \(n\) just raises \(r\) to the power \(n\) and multiplies \(\theta\) by \(n\) - far quicker than repeated Cartesian multiplication.
e.g. \((2\,\text{cis}\,30^\circ)^3 = 2^3\,\text{cis}(3\times30^\circ) = 8\,\text{cis}\,90^\circ = 8i.\)
What are the nth roots of a complex number?
Every non-zero complex number has exactly \(n\) distinct \(n\)th roots, evenly spaced by angle \(\tfrac{2\pi}{n}\) around a circle of radius \(r^{1/n}\) on the Argand diagram. You find them by adding multiples of \(2\pi\) to the argument before dividing by \(n\), since angles that differ by \(2\pi\) describe the same complex number.
e.g. The cube roots of \(-8=8\,\text{cis}\,\pi\) are \(2\,\text{cis}\dfrac{\pi}{3},\ 2\,\text{cis}\,\pi\ (=-2),\ 2\,\text{cis}\!\left(-\dfrac{\pi}{3}\right).\)
Key formulas
Seven formulas cover almost every question on this topic. The two tables below summarise all of them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The three forms of a complex number, De Moivre's theorem and the nth-roots formula are all on the official formula booklet; the modulus and Cartesian definitions are assumed prior knowledge.
| Formula | Used for | Booklet? |
|---|---|---|
| \(i^2=-1\) | Definition of \(i\) | Not in booklet - definition |
| \(z=a+bi,\ \bar z=a-bi\) | Cartesian form and conjugate | Not in booklet - prior knowledge |
| \(|z|=\sqrt{a^2+b^2}\) | Modulus of \(z=a+bi\) | Not in booklet - derived from Pythagoras |
| \(z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta\) | Modulus-argument (polar) form | ✓ Yes |
| \(z=re^{i\theta}\) | Euler form | ✓ Yes |
| \(z^n=r^n\,\text{cis}(n\theta)\) | De Moivre's theorem | ✓ Yes |
| \(z^{1/n}=r^{1/n}\,\text{cis}\!\left(\dfrac{\theta+2k\pi}{n}\right),\ k=0,1,\ldots,n-1\) | \(n\)th roots of a complex number | ✓ Yes |
Cartesian form vs polar/Euler form
Neither form is "better" - each makes a different operation easy, so choosing the right one saves a lot of algebra.
| Feature | Cartesian form \(a+bi\) | Polar / Euler form \(r\,\text{cis}\,\theta\) |
|---|---|---|
| Best for | Addition and subtraction | Multiplication, division, powers and roots |
| Addition | Add real and imaginary parts directly | Awkward - convert to Cartesian first |
| Multiplication | Expand brackets, use \(i^2=-1\) | Multiply moduli, add arguments |
| Powers | Repeated expansion - slow for large \(n\) | De Moivre: raise \(r\), multiply \(\theta\) by \(n\) |
Operations in Cartesian form
Cartesian form treats \(i\) like an unknown, with the one extra rule \(i^2=-1.\)
Addition & subtraction
\[(a+bi)\pm(c+di) = (a\pm c)+(b\pm d)i\]
Combine real parts together and imaginary parts together.
Multiplication
\[(a+bi)(c+di) = (ac-bd)+(ad+bc)i\]
Expand as normal, then replace \(i^2\) with \(-1\) and collect terms.
Division
\[\frac{z_1}{z_2} = \frac{z_1\bar z_2}{z_2\bar z_2}\]
Multiply top and bottom by the conjugate of the denominator so the denominator becomes real.
Converting to polar and Euler form
Every conversion starts from the same two pieces of information: the modulus and the argument.
Find the modulus
\(|z|=\sqrt{a^2+b^2}\) - always non-negative, the "\(r\)" value.
Find the argument
Sketch the point, then find \(\theta\) with \(\tan\theta=\tfrac ba\), adjusting for the correct quadrant.
Write in polar form
\(z=r(\cos\theta+i\sin\theta)=r\,\text{cis}\,\theta.\)
Write in Euler form
\(z=re^{i\theta}\) - the same \(r\) and \(\theta\), just written with the exponential.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Let \(z_1=3+2i\) and \(z_2=1-4i\).
(a)(i) Find the real part of \(z_1+z_2\).
(a)(ii) Find the imaginary part.
(b)(i) Find the real part of \(z_1 z_2\).
(b)(ii) Find the imaginary part.
(c) Find \(|z_1|\).
Worked solution
(a)(i) \(4-2i.\) Real part \(=4.\) A1
(a)(ii) Imaginary part \(=-2.\) A1
(b)(i) \((3+2i)(1-4i)=3-12i+2i-8i^2=3-10i+8=11-10i.\) Real part \(=11.\) M1
(b)(ii) Imaginary part \(=-10.\) A1
(c) \(|z_1|=\sqrt{9+4}=\sqrt{13}.\) A1
Let \(z=-1+\sqrt3\,i\).
(a)(i) Find \(|z|\).
(a)(ii) Find \(\arg z\).
(b) Write \(z\) in the form \(r\,\text{cis}\,\theta\).
Worked solution
(a)(i) \(|z|=\sqrt{1+3}=2.\) M1
(a)(ii) Second quadrant: \(\arg z=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}.\) A1
(b) \(z=2\,\text{cis}\dfrac{2\pi}{3}.\) A1
\(z=2\,\text{cis}\dfrac{2\pi}{3}.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Forgetting to check the quadrant when finding the argument. A raw \(\arctan(b/a)\) only gives the right angle directly in the first and fourth quadrants - always sketch the point first.
- Sign errors expanding brackets. When multiplying \((a+bi)(c+di)\), it's easy to forget \(i^2=-1\) turns a \(+\) term into a \(-\) term - always replace \(i^2\) explicitly before collecting.
- Giving an argument outside the principal range. Arguments should usually be given with \(-\pi<\theta\le\pi\) - an angle like \(\tfrac{5\pi}{3}\) should be rewritten as \(-\tfrac{\pi}{3}.\)
- Treating the modulus as additive. \(|z_1+z_2|\) is not the same as \(|z_1|+|z_2|\) - modulus only behaves simply under multiplication and division, not addition.
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
For very large or very small numbers - avoids typing long strings of zeros and prevents rounding errors, useful when a high power like \(z^5\) or \(z^{10}\) produces real and imaginary parts in the hundreds or thousands.
- Scientific notation means \(a \times 10^n\), e.g. \(3.2 \times 10^8\) or \(4.5 \times 10^{-3}.\)
- Use 2nd → , (EE) to enter the ×10 part: type 3.2 2nd , 8 to enter \(3.2\times10^8.\) Do NOT type ×10^ separately.TI-84
- Use the EE key (or type ×10^ from the keyboard template). Or just type 3.2×10^8 using the ^ key.Nspire
- Use the ×10ˣ key (EXP key) - type 3.2 then EXP then 8. Do NOT type ×10^ manually.Casio
- To display answers in scientific notation: on TI-84 press MODE and choose SCI; on Casio set the display mode in SET UP.
Tip: A common mistake is typing ×10^ instead of using the EE/EXP key - this gives ×10×... (multiplication, then a power) rather than proper scientific notation.
Faster and safer than algebra for messy equations - useful for checking a trigonometric identity produced by De Moivre's theorem (e.g. confirming \(\cos3\theta=4\cos^3\theta-3\cos\theta\) at several values of \(\theta\)) before or after you prove it.
- Graph \(f(x)\) first so you can see how many solutions exist and roughly where they are.
- Rearrange so everything is on one side: \(f(x)=0\) - or graph both sides as separate functions and find intersections.
- MATH → Solver: enter the expression, type a starting guess close to one root, press ALPHA + ENTER. Move the guess to near a different root and repeat for each solution.TI-84
- Type nSolve(f(x)=0, x, guess) - include a guess or interval e.g. nSolve(f(x)=0, x, 2) or nSolve(f(x)=0, x, {1,5}) to target a specific root.Nspire
- Run-Matrix → SolveN(f(x), x) returns all real roots at once; or use the Equation app for a visual approach.Casio
- Always verify each solution by substituting back into the original equation.
Tip: The solver finds ONE root near your starting guess - change the guess to find others. The graph shows you how many to expect.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Complex numbers questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between modulus-argument form and Euler form?
They describe exactly the same complex number - modulus-argument form writes it as \(r(\cos\theta+i\sin\theta) = r\,\text{cis}\,\theta\), while Euler form writes it as \(re^{i\theta}.\) Euler form is more compact and makes multiplying, dividing and raising to powers even faster, since exponent rules apply directly.
How do I find the argument of a complex number in the correct quadrant?
Sketch the point on an Argand diagram first. \(\arctan\) of (imaginary part / real part) only gives the correct angle directly in the first and fourth quadrants - in the second and third quadrants you need to adjust by adding or subtracting \(\pi\), based on the sketch.
What does De Moivre's theorem let me do?
De Moivre's theorem states \((\cos\theta+i\sin\theta)^n = \cos n\theta+i\sin n\theta\), so raising a complex number to a power just means raising the modulus to that power and multiplying the argument by it - far faster than repeated Cartesian multiplication.
Can I use my GDC for complex numbers questions?
Yes, on Paper 2 - most calculators have an \(a+bi\) mode that adds, multiplies and finds the modulus and argument of complex numbers directly. On Paper 1 you'll need every technique above without a calculator. See the GDC guide for model-specific instructions.
Sub-topics
Complex Numbers broken down into its individual skills, each with its own focused page.
Related topics
More Number & Algebra topics from the same AA HL syllabus unit, in case you want to keep going.