Complex numbers are the AHL gateway topic in IB Mathematics Applications & Interpretation HL. AI treats them less as abstract algebra and more as a modelling tool: a single complex number captures both a size and a direction, which is exactly what you need for AC electrical circuits (voltage and current as rotating phasors) and for the long-run behaviour of coupled systems (whose spirals fall straight out of complex eigenvalues). A calculator is always allowed, so the marks live in choosing the right form and reading the answer correctly — not in hand-grinding arithmetic.
This cheatsheet condenses every form, rule, De Moivre power, root and application from AHL 1.12–1.15 onto one page, then flags the traps that quietly cost method marks. The concise PDF is at the bottom — free to download.
§1 — Three forms of a complex number AHL 1.12
A complex number pairs a real part $a$ with an imaginary part $b$, where $i^2=-1$. The same number wears three interchangeable forms:
where $r=|z|$ is the modulus and $\theta=\arg z$ is the argument. Convert Cartesian → polar with $r=\sqrt{a^2+b^2}$, $\theta=\tan^{-1}\frac{b}{a}$ (fix the quadrant); polar → Cartesian with $a=r\cos\theta$, $b=r\sin\theta$.
§2 — Modulus & argument on the Argand diagram AHL 1.12
On the Argand diagram $z=x+yi$ is a point (or vector) from the origin.
The calculator's $\tan^{-1}$ only ever returns a first- or fourth-quadrant angle, so you must add or subtract $180^\circ$ by hand when $z$ sits in the second or third quadrant.
§3 — Adding, subtracting & multiplying AHL 1.12
Add and subtract componentwise: $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i$.
Multiplication is ordinary bracket expansion, then replace $i^2$ with $-1$ — which is why the real part is a subtraction, $ac-bd$.
§4 — The conjugate & division AHL 1.12
To write a quotient in $x+yi$ form, multiply top and bottom by the conjugate of the denominator — this makes the denominator real:
$$\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}$$
§5 — Multiply & divide in polar form AHL 1.12
Polar form turns multiplication into arithmetic on the modulus and the angle:
Geometrically, multiplying by $z$ rotates by $\arg z$ and scales by $|z|$ — the idea behind phasors and rotations in §8.
§6 — De Moivre's theorem: powers AHL 1.13
So $|z^n|=|z|^n$ and $\arg(z^n)=n\theta$. To find a power: get $r$ and $\theta$, raise $r$ to the power, multiply the angle by $n$, then convert back if the answer is wanted in $a+bi$ form.
§7 — Roots of a complex number AHL 1.13
A non-zero complex number has exactly $n$ distinct $n$-th roots, equally spaced around a circle of radius $r^{1/n}$:
$$z^{1/n}=r^{1/n}\operatorname{cis}\!\left(\frac{\theta+360^\circ k}{n}\right),\qquad k=0,1,\dots,n-1$$
§8 — Application: AC circuits & phasors AHL 1.13
In an alternating-current circuit the impedance is a complex number $Z=R+Xi$, where $R$ is resistance and $X$ is reactance:
Two out-of-phase signals $A_1\angle\phi_1$ and $A_2\angle\phi_2$ combine as phasors: convert each to $A\cos\phi+iA\sin\phi$, add in Cartesian form, then take the modulus for the resultant amplitude.
§9 — Application: complex eigenvalues & oscillation AHL 1.15
A $2\times2$ transition or feedback matrix $A$ drives a coupled system. Its eigenvalues solve the characteristic equation:
$$\lambda^2-(\operatorname{tr}A)\lambda+\det A=0,\qquad \lambda=\frac{\operatorname{tr}A\pm\sqrt{(\operatorname{tr}A)^2-4\det A}}{2}$$
When the discriminant is negative the eigenvalues form a complex conjugate pair $\lambda=\alpha\pm\beta i$ and every trajectory spirals: the real part $\alpha$ sets growth ($\alpha>0$) or decay ($\alpha<0$), and the oscillation has period
§10 — Exam attack plan All sections
| Question trigger | Reach for |
|---|---|
| "Find $|z|$ and $\arg z$" | $\sqrt{x^2+y^2}$, then $\tan^{-1}(y/x)$ — sketch to fix the quadrant |
| "Write in the form $a+bi$" (from a fraction) | Multiply by the conjugate; denominator $\to c^2+d^2$ |
| "Find $z^n$" | Polar form + De Moivre: $r^n\operatorname{cis}(n\theta)$ |
| "Solve $z^n=c$" / "$n$-th roots" | $r^{1/n}\operatorname{cis}\!\frac{\theta+360^\circ k}{n}$, $k=0..n-1$ |
| "Impedance / current / phase" | $Z=R+Xi$; $|Z|=\sqrt{R^2+X^2}$; $|I|=|V|/|Z|$ |
| "Combine two AC signals" | Phasors: to Cartesian, add, back to modulus |
| "Period of the oscillation" (matrix system) | Complex eigenvalues $\alpha\pm\beta i$; $T=2\pi/\beta$ |
Worked Example — AI HL-Style AC Circuit
Question (AI HL Paper 2 style — 6 marks)
A series AC circuit has resistance $R=8\,\Omega$ and reactance $X=15\,\Omega$, so its impedance is $Z=8+15i$. A voltage of magnitude $|V|=240$ V is applied.
(a) Find the magnitude of the impedance $|Z|$.
(b) Find the phase angle $\arg Z$, in degrees.
(c) Find the magnitude of the current, $|I|=|V|/|Z|$.
Solution
- (a) $|Z|=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\,\Omega$. (M1)(A1) (8-15-17 is a Pythagorean triple — spot it and skip the decimals.)
- (b) $\arg Z=\tan^{-1}\!\dfrac{15}{8}=61.9^\circ$ (3 s.f.). Positive $\Rightarrow$ inductive circuit. (M1)(A1)
- (c) $|I|=\dfrac{|V|}{|Z|}=\dfrac{240}{17}=14.1$ A (3 s.f.). (M1)(A1)
Examiner's note: in AI, keep the correct units ($\Omega$, degrees, A) and round to 3 s.f. unless told otherwise — a numerically right answer with a missing unit still drops the accuracy mark. Do not add $R$ and $X$ arithmetically ($8+15=23$ is wrong); the impedance magnitude is the modulus.
Common Student Questions
When should I switch to polar / Euler form?
Why is the real part of a product $ac-bd$ and not $ac+bd$?
My calculator's argument is in the wrong quadrant — what do I do?
Do the amplitudes of two AC signals just add up?
For a spiralling system, which part of the eigenvalue gives the period?
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