Complex Numbers is one of the most distinctive topics in IB Mathematics Analysis & Approaches HL — it has no SL counterpart, it stretches across AHL 1.12, 1.13 and 1.14, and it appears every year on Paper 1 and frequently on Paper 2. The challenge is not the algebra itself but the interplay between three different representations (Cartesian, polar, Euler) plus their geometric meaning on the Argand diagram. Students who try to solve everything in Cartesian form burn time; students who switch fluently into Euler form for multiplication, powers, and roots usually finish with marks to spare.
This cheatsheet condenses every formula, identity, trick, and trap from AHL 1.12–1.14 — including De Moivre's theorem, $n$th roots of a complex number, the conjugate root theorem, and the C+iS series — into one revisable page. Scroll to the bottom for the printable PDF, the full notes booklet, and the worked tutorial set.
§1 — Cartesian Form & Basic Operations AHL 1.12
Definition & Arithmetic
General form:$z = a + bi,\; a,b\in\mathbb{R},\; i^2 = -1$
Real part:$a = \operatorname{Re}(z)$
Imaginary part:$b = \operatorname{Im}(z)$
Addition:$(a+bi)+(c+di) = (a+c)+(b+d)i$
Multiplication:$(a+bi)(c+di) = (ac-bd)+(ad+bc)i$
Division:$\dfrac{a+bi}{c+di} = \dfrac{(a+bi)(c-di)}{c^2+d^2}$
Conjugate Properties
Conjugate:$z^* = a - bi$
Modulus squared:$zz^* = |z|^2 = a^2 + b^2$
Sum:$z + z^* = 2a \in \mathbb{R}$
Difference:$z - z^* = 2bi \in i\mathbb{R}$
Product conj.:$(zw)^* = z^* w^*$
Quotient conj.:$\left(\dfrac{z}{w}\right)^* = \dfrac{z^*}{w^*}$
Power conj.:$(z^n)^* = (z^*)^n$
TrickTo divide $a+bi$ by $c+di$: multiply numerator and denominator by $c-di$ (the conjugate of the denominator). The denominator becomes the real number $c^2+d^2$ — clean and fast.
Trap$i^2 = -1$, not $+1$. In multiplication, the $bi \cdot di$ term gives $-bd$, not $+bd$ — forgetting this sign is the single most common A1 lost in this section.
§2 — Modulus, Argument & the Argand Diagram AHL 1.12–1.13
Modulus & Argument
Modulus:$|z| = \sqrt{a^2 + b^2} \geq 0$
Argument:$\theta = \arg z \in (-\pi, \pi]$
Finding $\theta$:Use CAST: $\theta = \arctan(b/a)$, then adjust quadrant.
$|z_1 z_2|$:$|z_1| \cdot |z_2|$
$\arg(z_1 z_2)$:$\arg z_1 + \arg z_2 \pmod{2\pi}$
$|z_1/z_2|$:$|z_1|/|z_2|$
$\arg(z_1/z_2)$:$\arg z_1 - \arg z_2 \pmod{2\pi}$
Key Modulus & Argument Results
Conjugate mod:$|z^*| = |z|$
Conjugate arg:$\arg(z^*) = -\arg z$
Negation mod:$|-z| = |z|$
Negation arg:$\arg(-z) = \arg z \pm \pi$
$|z^n|$:$|z|^n$
$\arg(z^n)$:$n \arg z \pmod{2\pi}$
Triangle inequality:$|z_1 + z_2| \leq |z_1| + |z_2|$
TrapArgument must lie in $(-\pi, \pi]$. Always check the quadrant using the signs of $a$ and $b$ — $\arctan(b/a)$ alone gives the wrong answer for Quadrants 2 and 3.
NoteOn the Argand diagram: $|z_1 - z_2|$ is the geometric distance between the points representing $z_1$ and $z_2$. The locus $|z - z_0| = r$ is a circle, centre $z_0$, radius $r$.
§3 — Polar & Euler Forms AHL 1.13
Three Forms
Cartesian:$z = a + bi$
Polar:$z = r(\cos\theta + i\sin\theta) = r\,\mathrm{cis}\,\theta$
Euler:$z = r e^{i\theta}$
Cartesian → polar:$a = r\cos\theta,\; b = r\sin\theta$
Polar → Cartesian:$r = |z|,\; \theta = \arg z$
Products & Quotients in Polar/Euler
Product:$r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2\, e^{i(\theta_1 + \theta_2)}$
Quotient:$\dfrac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \dfrac{r_1}{r_2}\, e^{i(\theta_1 - \theta_2)}$
Conjugate:$(r e^{i\theta})^* = r e^{-i\theta}$
Negation:$-r e^{i\theta} = r e^{i(\theta \pm \pi)}$
TrickTo multiply or divide complex numbers, convert to Euler form first — multiply/divide moduli and add/subtract arguments. Far faster than expanding Cartesian brackets, especially when raising to a power.
Trap$\mathrm{cis}\,\theta$ is shorthand for $\cos\theta + i\sin\theta$ — not $\cos\theta \cdot i\sin\theta$. Never split it as a product.
NoteEuler's identity: $e^{i\pi} + 1 = 0$ (special case $r=1$, $\theta=\pi$). Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ is in the IB formula booklet.
§4 — Conjugate Root Theorem & Polynomials AHL 1.14
Conjugate Root Theorem. If $p(z) = 0$ has real coefficients and $\alpha \in \mathbb{C}$ is a root, then $\alpha^*$ is also a root.
| Complex roots come in pairs | $(z-\alpha)(z-\alpha^*) = z^2 - 2\operatorname{Re}(\alpha)\,z + |\alpha|^2$ — a real quadratic. |
| Odd-degree real polynomial | Must have at least one real root. |
| Given one complex root $\alpha$ | Write down $\alpha^*$ immediately; find any real root via Vieta's. |
TrickSum of roots $= -b/a$ and product of roots $= c/a$ (Vieta's). Use these to find the missing real root and any unknown coefficients without fully expanding the factored form.
TrapThe conjugate root theorem applies only when all coefficients are real. If any coefficient is complex, $\alpha^*$ is not automatically a root.
§5 — De Moivre's Theorem AHL 1.14
Statement & Applications
De Moivre:$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
Euler form:$(e^{i\theta})^n = e^{in\theta}, \; n\in\mathbb{Q}$
Key identity:$z^n + z^{-n} = 2\cos n\theta$
$z^n - z^{-n} = 2i\sin n\theta$
where:$z = \cos\theta + i\sin\theta = e^{i\theta}$
Trig Expansions via Binomial
Expand $(\cos\theta + i\sin\theta)^n$ by the binomial theorem; equate the real part to $\cos n\theta$ and the imaginary part to $\sin n\theta$:
$\cos 3\theta$:$4\cos^3\theta - 3\cos\theta$
$\sin 3\theta$:$3\sin\theta - 4\sin^3\theta$
$\cos^3\theta$:$\tfrac{1}{4}(\cos 3\theta + 3\cos\theta)$
$\sin^3\theta$:$\tfrac{1}{4}(3\sin\theta - \sin 3\theta)$
TrickTo express $\cos^n\theta$ or $\sin^n\theta$ as a sum of multiple-angle terms: use $z^n + z^{-n} = 2\cos n\theta$, expand $(z + z^{-1})^n$ binomially, then substitute back. This is the standard route on Paper 1.
TrapDe Moivre requires the base to be in the form $\cos\theta + i\sin\theta$ (modulus 1). If $|z| = r \neq 1$, write $z = r e^{i\theta}$ first, then $(re^{i\theta})^n = r^n e^{in\theta}$.
NoteProof by induction for $n \in \mathbb{Z}^+$ is examinable. Know the three steps: base case $n = 1$, assume $n = k$, prove $n = k+1$ using trig addition formulas.
§6 — $n$th Roots of a Complex Number AHL 1.14
The $n$ roots of $w = R e^{i\phi}$
Formula:$z_k = R^{1/n} e^{i(\phi + 2\pi k)/n},\; k = 0, 1, \ldots, n-1$
Each modulus:$R^{1/n}$ (all roots have equal modulus)
Arguments:$\dfrac{\phi}{n},\; \dfrac{\phi + 2\pi}{n},\; \dfrac{\phi + 4\pi}{n}, \ldots$
Angular spacing:$\dfrac{2\pi}{n}$ between consecutive roots.
Geometric picture & sum/product
The $n$ roots form a regular $n$-gon on a circle of radius $R^{1/n}$, centred at the origin.
Sum of roots:$0$ (no $z^{n-1}$ term in $z^n - w = 0$)
Product of roots:$(-1)^{n+1} w$
Roots of unity:$z^n = 1$: $z_k = e^{2\pi i k / n}$, $\sum z_k = 0$.
TrickAlways write $w$ in Euler form $R e^{i\phi}$ first, with $\phi \in (-\pi, \pi]$. The principal root ($k = 0$) then has the smallest positive argument, which makes the regular $n$-gon picture obvious.
TrapThere are exactly $n$ distinct roots — never more, never fewer. Students often stop at 2 or 3 roots for a quartic or quintic. Always count up to $k = n - 1$.
§7 — C+iS Series & Advanced Applications AHL 1.14
The C+iS method
If $C = \sum a_n \cos n\theta$ and $S = \sum a_n \sin n\theta$, form $C + iS = \sum a_n e^{in\theta}$. Recognise the result as a geometric series with ratio $r e^{i\theta}$ (converges for $|r| < 1$); sum it, then separate real and imaginary parts.
$$C + iS = \frac{1}{1 - r e^{i\theta}} = \frac{1 - r e^{-i\theta}}{1 - 2r\cos\theta + r^2}$$
Integrating powers of trig
Use $z^n \pm z^{-n}$ expansions to reduce $\cos^n\theta$ or $\sin^n\theta$ to a sum of multiple angles, then integrate term by term:
$\cos^4\theta$:$\tfrac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)$
$\sin^4\theta$:$\tfrac{1}{8}(\cos 4\theta - 4\cos 2\theta + 3)$
$\cos^6\theta$:$\tfrac{1}{32}(\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)$
Roots of unity trick
For $z^n = 1$: roots are $\omega^k$ where $\omega = e^{2\pi i / n}$.
Key identity:$1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$
Pairs:$\omega^k + \omega^{n-k} = 2\cos\dfrac{2\pi k}{n}$
Use these identities to find exact values of $\cos(2\pi/n)$, $\cos(4\pi/n)$, etc.
TrickFor a C+iS series: write the first three terms in full, identify the ratio $r e^{i\theta}$, apply $\dfrac{a}{1-x}$, then multiply numerator and denominator by the complex conjugate of the denominator to separate $C$ and $S$.
TrapWhen separating $C + iS$ after summing the geometric series, multiply by the complex conjugate of the denominator — not by the denominator itself.
§8 — Exam Attack Plan All sections
When you see this in the question — reach for that:
| Question trigger | Reach for |
|---|---|
| "Write in polar / Euler form" | $r = \sqrt{a^2 + b^2}$, then CAST for $\theta$ |
| "Express as $a + bi$" | Convert to polar, apply De Moivre or expand, convert back |
| "Find modulus and argument of $z^n$" | $|z^n| = |z|^n$, $\arg(z^n) = n\arg z$ (adjust to $(-\pi, \pi]$) |
| "Show $z$ is purely real / imaginary" | Show $\operatorname{Im}(z) = 0$ or $\operatorname{Re}(z) = 0$ |
| "Prove trig identity using De Moivre" | Binomial expand $(\cos\theta + i\sin\theta)^n$, equate parts |
| "Express $\cos^n\theta$ in multiple angles" | Use $z + z^{-1} = 2\cos\theta$, expand $(z + z^{-1})^n$ binomially |
| "Find $n$th roots of $w$" | Write $w = R e^{i\phi}$; apply $z_k = R^{1/n} e^{i(\phi + 2\pi k)/n}$ |
| "Roots on Argand diagram" | Regular $n$-gon on circle of radius $R^{1/n}$; mark all $n$ roots |
| "Sum/product of roots of $z^n = c$" | Sum $= 0$ (no $z^{n-1}$ term); product from Vieta's |
| "One root of real-coeff polynomial is $\alpha$" | State $\alpha^*$ is also a root; find real root via Vieta's |
| "Prove $X$ is purely imaginary / real" | Compute $X + X^*$ or $X - X^*$ |
| "C+iS series" | Form $C + iS$, recognise GP, sum, multiply by conjugate denominator |
| "Smallest $n$ with $z^n \in \mathbb{R}$" | Write $z = r e^{i\theta}$; need $n\theta \in \pi\mathbb{Z}$ |
| "Show two complex numbers are equal" | Equate real and imaginary parts separately |
| "Proof by induction for De Moivre" | Base $n = 1$: trivial. Step: multiply by $e^{i\theta}$, use trig addition. |
Top 5 marks lost — based on past papers
- Wrong quadrant for argument. $\arctan(b/a)$ gives the correct angle only in Q1/Q4. For Q2/Q3, add or subtract $\pi$. Always sketch the point first.
- Not adjusting argument to $(-\pi, \pi]$ after multiplication. After adding or subtracting arguments, always check the result is in range. Subtract $2\pi$ if $> \pi$; add $2\pi$ if $\leq -\pi$.
- Missing roots when solving $z^n = w$. There are exactly $n$ roots. Use $k = 0, 1, \ldots, n-1$. Stopping early loses 2–3 marks every time.
- Forgetting the conjugate root in polynomials. If $a + bi$ is a root and coefficients are real, immediately write $a - bi$. Then use Vieta's — don't try to find the third root by polynomial division unless necessary.
- Induction proof: not applying the trig addition formula correctly. In the inductive step, $(\cos\theta + i\sin\theta)^{k+1} = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)$ — then expand using $\cos(A+B)$ and $\sin(A+B)$. Many students forget the addition step.
Worked Example — IB-Style $n$th Roots Problem
Question (HL Paper 1 style — 7 marks)
Find all complex solutions of $z^4 = -8 + 8\sqrt{3}\,i$, giving each in the form $r e^{i\theta}$ with $\theta \in (-\pi, \pi]$. Sketch the four solutions on an Argand diagram and state the geometric figure they form.
Solution
- Write $w = -8 + 8\sqrt{3}\,i$ in Euler form. $|w| = \sqrt{64 + 192} = 16$ and $w$ lies in Q2 (negative real, positive imaginary), so $\arg w = \pi - \arctan(8\sqrt{3}/8) = \pi - \pi/3 = 2\pi/3$. Hence $w = 16\, e^{i\,2\pi/3}$. (M1)(A1)
- Apply the $n$th-roots formula with $n = 4$, $R = 16$, $\phi = 2\pi/3$: $$z_k = 16^{1/4}\, e^{i(2\pi/3 + 2\pi k)/4} = 2\, e^{i(\pi/6 + \pi k/2)}, \quad k = 0, 1, 2, 3.$$ (M1)
- Substitute each $k$ and adjust to $(-\pi, \pi]$: $z_0 = 2 e^{i\pi/6}$, $z_1 = 2 e^{i\,2\pi/3}$, $z_2 = 2 e^{i\,7\pi/6} = 2 e^{-i\,5\pi/6}$, $z_3 = 2 e^{i\,5\pi/3} = 2 e^{-i\pi/3}$. (A1)(A1)
- All four roots have modulus $2$ and arguments separated by $\pi/2$. Plot them on a circle of radius $2$ centred at the origin. (A1)
- The four points form a square inscribed in the circle of radius $2$. (R1)
Examiner's note: The most common error here is taking $\arg w = -\pi/3$ from $\arctan(-\sqrt{3})$ alone, ignoring that $w$ is in Q2. A quick Argand sketch eliminates the trap. The second most common error is producing only three roots — for $n = 4$ you must use $k = 0, 1, 2, 3$.
Common Student Questions
Why does $\arctan(b/a)$ sometimes give the wrong argument for a complex number?
$\arctan$ only outputs values in $(-\pi/2, \pi/2)$, so it returns the correct angle only when $z$ lies in Quadrant 1 or Quadrant 4. If $z$ is in Q2 or Q3, you must add or subtract $\pi$ to land in the correct quadrant. Always sketch $z$ on the Argand diagram first, identify the quadrant from the signs of $a$ and $b$, then adjust $\arctan(b/a)$ so the final argument sits in $(-\pi, \pi]$.
Do I need to memorise De Moivre's theorem and its proof by induction?
Yes. De Moivre's theorem $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ is in the IB formula booklet, but the proof by induction for $n \in \mathbb{Z}^+$ is examinable. Know the three steps: base case $n = 1$ (trivially true), inductive hypothesis $n = k$, then multiply both sides by $(\cos\theta + i\sin\theta)$ and apply the trig addition formulas to recover $\cos((k+1)\theta) + i\sin((k+1)\theta)$.
How do I find all $n$ roots of a complex equation $z^n = w$?
Write $w$ in Euler form $R e^{i\phi}$ with $\phi \in (-\pi, \pi]$. Then $z_k = R^{1/n} e^{i(\phi + 2\pi k)/n}$ for $k = 0, 1, \ldots, n-1$. There are exactly $n$ distinct roots — never more, never fewer. Each has the same modulus $R^{1/n}$ and the arguments are spaced $2\pi/n$ apart, so the roots form a regular $n$-gon on a circle of radius $R^{1/n}$. Stopping at 2 or 3 roots for a quartic or quintic is the most common error.
When does the conjugate root theorem actually apply?
Only when the polynomial has real coefficients. If $p(z)$ has real coefficients and $\alpha$ is a complex root, then $\alpha^*$ is automatically a root too. If any coefficient is itself complex, this theorem does not apply and $\alpha^*$ may not be a root. Once you know two roots $\alpha$ and $\alpha^*$, multiply $(z - \alpha)(z - \alpha^*)$ to get a real quadratic $z^2 - 2\operatorname{Re}(\alpha)z + |\alpha|^2$, then divide or use Vieta's to find the remaining real root.
What is the C+iS method and when do I use it?
If you need to sum two related series $C = \sum a_n \cos n\theta$ and $S = \sum a_n \sin n\theta$, form $C + iS = \sum a_n e^{in\theta}$. This is a geometric series with ratio $r e^{i\theta}$; apply $\dfrac{a}{1-r}$, then multiply numerator and denominator by the complex conjugate of the denominator to separate real and imaginary parts. The real part is $C$, the imaginary part is $S$. The trick is recognising the geometric pattern — write out the first three terms if you're not sure.
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