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.
§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.