Complex Numbers (AI HL)
Complex numbers extend the real number line into a plane by introducing \(i\), where \(i^2=-1\), so that every quadratic equation has a solution. This topic covers writing a complex number in Cartesian form \(a+bi\), finding its modulus and argument, converting to polar or exponential (Euler) form, and using these forms to add, multiply and interpret complex numbers geometrically on the Argand diagram.
What the syllabus says
This topic maps onto two points in the official IB Applications & Interpretation syllabus, both HL-only.
| Code | Syllabus content |
|---|---|
| AHL1.12 | Complex numbers: the number \(i\) such that \(i^2=-1\). Cartesian form \(z=a+bi\); the terms real part, imaginary part, conjugate, modulus and argument. The complex plane and Argand diagrams. Complex numbers as solutions to quadratic equations \(ax^2+bx+c=0\) with real coefficients where \(b^2-4ac<0\). Calculating sums, differences, products and quotients by hand and with technology, and powers in Cartesian form with technology. |
| AHL1.13 | Modulus-argument (polar) form \(z=r(\cos\theta+i\sin\theta)=r\,\mathrm{cis}\,\theta\) and exponential (Euler) form \(z=re^{i\theta}\). Conversion between Cartesian, polar and exponential forms, by hand and with technology. Calculating products, quotients and integer powers in polar or exponential form. Adding sinusoidal functions with the same frequency but different phase shifts. Geometric interpretation: multiplication as a rotation and stretch on the Argand diagram. |
Both points are AHL content, examinable only at AI HL, not at AI 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 a complex number in Cartesian form?
A complex number in Cartesian form is written \(z=a+bi\), where \(a\) is the real part, \(b\) is the imaginary part, and \(i\) satisfies \(i^2=-1\). It can be plotted as a single point (or vector) on the Argand diagram, with \(a\) along the horizontal axis and \(b\) along the vertical.
e.g. \(z=3+4i\) has real part \(3\) and imaginary part \(4\).
What is the modulus of a complex number?
The modulus \(|z|\) is the distance from the origin to \(z\) on the Argand diagram - essentially the length of the vector, found with Pythagoras' theorem: \(|z|=\sqrt{a^2+b^2}\).
e.g. \(z=3+4i\): \(|z|=\sqrt{9+16}=\sqrt{25}=5\).
What is the argument of a complex number?
The argument \(\arg(z)\) is the angle, usually in radians, that the vector \(z\) makes with the positive real axis, measured anticlockwise. It's found from \(\tan\theta=b/a\), adjusted for the correct quadrant.
e.g. \(z=\sqrt3+i\): \(\arg z=\arctan(1/\sqrt3)=\pi/6\).
What is polar form?
Polar form, \(z=r(\cos\theta+i\sin\theta)=r\,\mathrm{cis}\,\theta\), describes a complex number by its modulus \(r\) and argument \(\theta\) instead of its real and imaginary parts - useful because multiplying and dividing become simple with this form.
e.g. \(z=3+4i\) has \(r=5\), so \(z=5\,\mathrm{cis}(0.927)\) (angle in radians, to 3 s.f.).
What is exponential (Euler) form?
Exponential form, \(z=re^{i\theta}\), is polar form written using Euler's relation \(e^{i\theta}=\cos\theta+i\sin\theta\). It's the same number as polar form, just more compact, and it makes powers of complex numbers especially quick to compute.
e.g. \(z=2e^{i\pi/2}=2(\cos90^\circ+i\sin90^\circ)=2i\).
Key formulas
Complex numbers have two ways to write the same value, and each is suited to different operations. The tables below summarise all the key relationships at a glance.
Formula reference
The three forms of a complex number are all on the official formula booklet. The modulus and multiplication/division rules in polar form follow from index laws and Pythagoras, so aren't listed as separate formulas.
| Formula | Used for | Booklet? |
|---|---|---|
| \(z=a+bi\) | Cartesian form | ✓ Yes |
| \(z=r(\cos\theta+i\sin\theta)=r\,\mathrm{cis}\,\theta\) | Polar form | ✓ Yes |
| \(z=re^{i\theta}\) | Exponential (Euler) form | ✓ Yes |
| \(|z|=\sqrt{a^2+b^2}\) | Modulus, from Cartesian form | Not in the formula booklet - prior knowledge (Pythagoras) |
| \(z_1z_2=r_1r_2\,\mathrm{cis}(\theta_1+\theta_2)\) | Product in polar form | Not in the formula booklet - derived from index laws |
Cartesian vs polar/exponential
The two representations aren't competing methods - each is the natural tool for a different operation, which is exactly why the syllabus expects you to convert fluently between them.
| Feature | Cartesian \(a+bi\) | Polar / exponential \(r\,\mathrm{cis}\,\theta\) |
|---|---|---|
| Best for | Addition and subtraction | Multiplication, division, powers |
| Addition | Add real and imaginary parts separately | Awkward - convert to Cartesian first |
| Multiplication | Expand brackets, use \(i^2=-1\) | Multiply moduli, add arguments |
| What it shows directly | Position as a point \((a,b)\) | Distance \(r\) and direction \(\theta\) from the origin |
Cartesian form
Cartesian form is where most complex-number questions start, and it's the natural form for addition, subtraction and solving quadratics.
Arithmetic
\[(a+bi)\pm(c+di) = (a\pm c)+(b\pm d)i\]
Add or subtract real and imaginary parts separately - treat \(i\) like an unknown, but remember \(i^2=-1\) when multiplying.
Not in the formula booklet - prior knowledgeComplex roots of a quadratic
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
When \(b^2-4ac<0\), the square root is of a negative number, giving a conjugate pair of complex roots.
✓ In the formula bookletConjugate
\[\overline{a+bi} = a-bi\]
Reflects \(z\) in the real axis - same modulus, opposite-sign argument. Used to simplify division by a complex number.
Not in the formula booklet - prior knowledgePolar & exponential form
Once a complex number is in polar or exponential form, multiplying, dividing and raising to a power become arithmetic on \(r\) and \(\theta\) instead of algebra.
Products & quotients
\[z_1z_2=r_1r_2\,\mathrm{cis}(\theta_1+\theta_2),\quad \frac{z_1}{z_2}=\frac{r_1}{r_2}\,\mathrm{cis}(\theta_1-\theta_2)\]
Multiply (or divide) the moduli and add (or subtract) the arguments.
Not in the formula booklet - derived from index lawsInteger powers
\[z^n = r^n\,\mathrm{cis}(n\theta)\]
Raise the modulus to the power \(n\) and multiply the argument by \(n\) - far quicker than repeated Cartesian multiplication.
Not in the formula booklet - derived from index lawsGeometric interpretation
Adding complex numbers is vector addition on the Argand diagram. Multiplying by \(z=r\,\mathrm{cis}\,\theta\) rotates a point by \(\theta\) and stretches it by a factor of \(r\).
Not a formula - geometric resultWorked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Given \(z_1 = 2+3i\) and \(z_2 = 4-i\), find \(z_1+z_2\) and \(|z_1+z_2|\).
Worked solution
\(z_1+z_2 = 6+2i.\) M1 A1
\(z_1+z_2 = 6+2i.\)
\(|6+2i| = \sqrt{36+4} = \sqrt{40}\) M1
\(\approx 6.32.\) A1
Write \(z = 1 + i\) in the form \(re^{i\theta}\), \(\theta\) in radians:
(a)(i) State the value of \(r\).
(a)(ii) State the value of \(\theta\).
(b) Verify by converting your polar form back to Cartesian.
Worked solution
(a)(i) \(r = \sqrt{1+1} = \sqrt2.\) M1
(a)(ii) \(\theta = \tan^{-1}(1) = \tfrac{\pi}{4}.\) A1 So \(z = \sqrt2\,e^{i\pi/4}.\) A1
(b) \(\sqrt2\cos\tfrac{\pi}{4}=1,\ \sqrt2\sin\tfrac{\pi}{4}=1.\) R1
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 \(i^2=-1\) when expanding brackets. Multiplying \((a+bi)(c+di)\) produces an \(i^2\) term that must be replaced with \(-1\), not left as \(i^2\) or dropped entirely.
- Using \(\arctan(b/a)\) without checking the quadrant. A calculator's \(\tan^{-1}\) only returns angles between \(-\pi/2\) and \(\pi/2\), so for \(z\) in the second or third quadrant you must add or subtract \(\pi\) to get the correct argument.
- Adding complex numbers in polar form directly. Only multiplication, division and powers are simple in polar form - to add or subtract, convert back to Cartesian first.
- Forgetting complex roots come in conjugate pairs. If a real quadratic has one complex root \(a+bi\), the other root is automatically \(a-bi\) - there's no need to solve for it separately.
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.
Useful for verifying that a quadratic really does have complex roots - graphing \(f(x)=ax^2+bx+c\) shows no real \(x\)-intercepts when \(b^2-4ac<0\), confirming the roots must be complex.
- Graph \(f(x)\) first so you can see how many real 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: If the numeric solver reports no real solution near your guess and the graph shows no \(x\)-intercepts, that's your confirmation the roots are a complex conjugate pair - solve them with the quadratic formula instead.
Most GDCs also have a dedicated complex-number mode (often called a+bi mode) for adding, multiplying and dividing complex numbers directly, plus functions such as abs( and angle( to read off the modulus and argument without doing the Pythagoras/arctan working by hand. The exact menu path differs by model, so see the full GDC guide for the steps on yours.
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 Cartesian and polar form of a complex number?
Cartesian form, \(z=a+bi\), is best for adding and subtracting complex numbers. Polar (or exponential) form, \(z=r\,\mathrm{cis}\,\theta\) or \(z=re^{i\theta}\), is best for multiplying, dividing and raising to a power, since you just combine the moduli and arguments instead of expanding brackets.
Why do complex roots of a quadratic always come in conjugate pairs?
For a quadratic with real coefficients, the quadratic formula gives roots of the form \(p\pm q\sqrt{\text{negative number}}\). Because the coefficients are real, the plus and minus branches are complex conjugates of each other - if \(a+bi\) is a root, so is \(a-bi\).
How do I convert between polar and exponential form?
They're the same thing written two ways. Polar form \(r\,\mathrm{cis}\,\theta = r(\cos\theta+i\sin\theta)\) becomes exponential form \(re^{i\theta}\) by Euler's relation \(e^{i\theta}=\cos\theta+i\sin\theta\) - the modulus \(r\) and argument \(\theta\) don't change, only the notation does.
Can I use my GDC for complex numbers?
Yes - most GDCs have a dedicated complex-number mode (often called a+bi mode) for adding, multiplying, dividing and converting complex numbers directly, plus functions to find the modulus and argument automatically. 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 AI HL syllabus unit, in case you want to keep going.