Mixed Complex Number Problems II
The hardest HSC complex-number questions never live inside a single technique, they ask you to weave geometry, roots of unity, polynomial factoring, and de Moivre into one chain of reasoning. This lesson is a stress-test for the whole module: three multi-step problems and the strategic habits that keep you moving when the path is not obvious.
Without looking it up: write down de Moivre's theorem, the formula for the $n$ distinct $n$th roots of a complex number $w = r(\cos\theta + i\sin\theta)$, and the sum/product of all $n$th roots of unity. Don't peek get them on paper now so you can audit your recall.
Hard complex-number questions rarely tell you which tool to use. Two habits unlock almost every one: convert to the form that matches the geometry (modulus-argument for rotations and roots; rectangular for sums and differences), and name the symmetry the moment you see roots of unity, the sum is $0$, the product is $\pm 1$, and consecutive roots are equally spaced.
The convert-symmetry-factor reading: (1) what form makes the operation easy, polar or Cartesian? (2) what symmetry is implicit in the configuration, roots of unity, conjugate pairs, equilateral triangles? (3) what factorisation does that symmetry give you for the polynomial in the background?
$z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k)$ · $\sum_{k=0}^{n-1} \omega^k = 0$ · $\prod_{k=1}^{n-1} \omega^k = (-1)^{n+1}$
Key facts
- de Moivre: $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
- $n$ distinct $n$th roots of unity are $\omega^k$, $k = 0, \ldots, n-1$, equally spaced on the unit circle
- Sum of all $n$th roots of unity is $0$ for $n \geq 2$; product of the non-trivial ones is $(-1)^{n+1}$
- Real-coefficient polynomials have complex roots in conjugate pairs
Concepts
- Why polar form is the natural language for rotation, scaling, and roots
- How $z^n - 1$ factors over $\mathbb{R}$ as products of real quadratics from conjugate pairs
- How geometric properties (equilateral triangles, regular polygons) emerge from root configurations
Skills
- Solve multi-step problems combining geometry, roots of unity, and polynomial factoring
- Use de Moivre to derive trig identities ($\cos 5\theta$, $\sin 5\theta$ etc.) and prove polynomial identities
- Recognise when symmetry collapses an ugly expression to $0$, $1$, or $-1$
The biggest single shortcut in mixed problems is recognising when an expression is the sum (or product) of all $n$th roots of unity in disguise. Three identities do most of the heavy lifting:
- $\displaystyle \sum_{k=0}^{n-1} \omega^k = 0$ for $n \geq 2$ (because $\omega^n - 1 = (\omega - 1)(1 + \omega + \cdots + \omega^{n-1})$ and $\omega^n = 1$).
- $\displaystyle \prod_{k=1}^{n-1} \omega^k = \omega^{1+2+\cdots+(n-1)} = \omega^{n(n-1)/2} = (-1)^{n+1}$.
- $\displaystyle z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k)$, the master factorisation linking polynomials to roots.
Worked through the hook: With $\omega = e^{2\pi i / 5}$:
- $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ (full sum of fifth roots).
- $\omega \cdot \omega^2 \cdot \omega^3 \cdot \omega^4 = \omega^{10} = (\omega^5)^2 = 1^2 = 1$. (Also matches $(-1)^{5+1} = 1$.)
Three master identities: sum $= 0$, product $= (-1)^{n+1}$, factorisation of $z^n - 1$ · Always state de Moivre explicitly when used: "By de Moivre's theorem…" · Vieta on $z^4 + z^3 + z^2 + z + 1$ recovers sum/product of non-trivial fifth roots · If you see a sum of equally-spaced angles, check for roots-of-unity in disguise
Pause, copy the three master identities (sum, product, factorisation of $z^n-1$), the Vieta application to $z^4+\cdots+1$, and the instruction to cite De Moivre's theorem explicitly into your book.
Quick check: Let $\omega = e^{2\pi i / 7}$. What is the value of $1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6$?
We just saw the three master roots-of-unity identities: sum $= 0$, product $= (-1)^{n+1}$, factorisation of $z^n-1$; and how Vieta on $z^4+z^3+z^2+z+1$ recovers sums and products of non-trivial 5th roots. That raises a question: how do complex numbers give short proofs of classical geometric results? This card answers it → represent vertices as complex numbers, use $|z_2-z_1|$ for distance, $\arg(z_2-z_1)$ for direction, and rotation by $e^{i\theta}$.
To prove a geometric claim about points $A, B, C$ in the plane, label them $z_1, z_2, z_3 \in \mathbb{C}$ and read the geometry algebraically:
- Distance $|z_2 - z_1|$ is the length of segment $AB$.
- Direction $\arg(z_2 - z_1)$ is the angle $AB$ makes with the real axis.
- Rotation by $\theta$ about $z_0$ sends $z$ to $z_0 + e^{i\theta}(z - z_0)$.
- Equilateral triangle on $z_1, z_2, z_3$: one elegant criterion is $z_1 + \omega z_2 + \omega^2 z_3 = 0$ where $\omega = e^{2\pi i / 3}$ (or the conjugate orientation).
Distance $|z_2 - z_1|$; direction $\arg(z_2 - z_1)$ · Rotation by $\theta$ about $z_0$: $z \mapsto z_0 + e^{i\theta}(z - z_0)$ · Equilateral test: $z_1 + \omega z_2 + \omega^2 z_3 = 0$ with $\omega = e^{2\pi i/3}$ · Equivalent symmetric form: $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
Pause, copy the complex-geometric toolkit: $|z_2-z_1|$ distance, $\arg(z_2-z_1)$ direction, rotation $z \mapsto z_0+e^{i\theta}(z-z_0)$, and the equilateral test $z_1+\omega z_2+\omega^2 z_3 = 0$ into your book.
Did you get this? True or false: multiplying $z - z_0$ by $e^{i\pi/2}$ rotates the vector from $z_0$ to $z$ by $90°$ anticlockwise about $z_0$.
Worked examples · 3 in a row, reveal as you go
Points $A$, $B$, $C$ in the Argand plane correspond to complex numbers $z_1$, $z_2$, $z_3$. Triangle $ABC$ is equilateral with vertices in anticlockwise order. Show that $z_1 + \omega z_2 + \omega^2 z_3 = 0$, where $\omega = e^{2\pi i / 3}$.
Factor $z^6 - 1$ completely as a product of real linear and real quadratic factors. Use the sixth roots of unity.
Use de Moivre's theorem to express $\cos 5\theta$ as a polynomial in $\cos\theta$. Hence solve $16x^5 - 20x^3 + 5x = 0$ for $x \in [-1, 1]$.
Fill the gap: A conjugate pair of complex roots $\alpha, \bar\alpha$ with $|\alpha| = 1$ contributes the real quadratic factor $z^2 - 2 \cdot$ $(\alpha) \cdot z +$ to any real polynomial.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $\omega = e^{2\pi i / 6}$, then $1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 = 0$.
Activities · practice with the ideas
Let $\omega = e^{2\pi i / 7}$. Find the value of $(1 - \omega)(1 - \omega^2)(1 - \omega^3)(1 - \omega^4)(1 - \omega^5)(1 - \omega^6)$.
Triangle $ABC$ has vertices $z_1 = 1$, $z_2 = i$, $z_3 = -1 - i$ in $\mathbb{C}$. Determine whether it is right-angled, and if so, at which vertex. Justify using $\arg$ or modulus.
Factor $z^4 + z^3 + z^2 + z + 1$ as a product of two real quadratics by pairing conjugate fifth roots of unity. (Hint: $2\cos(2\pi/5) = \tfrac{\sqrt{5} - 1}{2}$ and $2\cos(4\pi/5) = -\tfrac{\sqrt{5}+1}{2}$.)
Use de Moivre's theorem to derive $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. Hence solve $4x^3 - 3x = \tfrac{1}{2}$ for $x \in [-1, 1]$.
Show that the points representing the cube roots of $8i$ form an equilateral triangle in the Argand plane. State the side length.
Odd one out: Three of these identities are correct statements about $\omega = e^{2\pi i / 3}$. Which one is NOT?
At the start you wrote down de Moivre's theorem, the $n$th-roots formula, and the sum/product of roots of unity. Compare what you wrote with what we used in the worked examples.
The single biggest pattern across all three worked examples is the same: identify the right form (polar for rotation, conjugate pairs for real factoring), invoke de Moivre or the sum-of-roots identity by name, then chase the algebra. The geometry of equilateral triangles, the factorisation of $z^6 - 1$, and the polynomial $16x^5 - 20x^3 + 5x$ all collapse to two or three lines once the right identity is on the page.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Let $\omega = e^{2\pi i / 5}$. Show that $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$, and hence find the value of $(1 - \omega)(1 - \omega^2)(1 - \omega^3)(1 - \omega^4)$. (3 marks)
Q2. Points $A$, $B$, $C$ in the Argand plane are $z_1 = 2$, $z_2 = 2e^{2\pi i / 3}$, $z_3 = 2e^{4\pi i / 3}$. Show by complex methods that triangle $ABC$ is equilateral, and find its side length. (4 marks)
Q3. Use de Moivre's theorem to show that $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$. Hence find the exact value of $\cos\tfrac{\pi}{8}$. (5 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. From $z^7 - 1 = (z-1)(z^6 + z^5 + \cdots + 1)$ and $z^6 + \cdots + 1 = \prod_{k=1}^{6}(z - \omega^k)$, set $z = 1$: $\prod_{k=1}^{6}(1 - \omega^k) = 1^6 + 1^5 + \cdots + 1 = 7$.
2. $z_2 - z_1 = -1 + i$; $z_3 - z_1 = -2 - i$; $z_3 - z_2 = -1 - 2i$. $|z_2 - z_1|^2 + |z_3 - z_2|^2 = 2 + 5 = 7 \neq |z_3 - z_1|^2 = 5$. Check at $B$: $|z_2 - z_1|^2 + |z_3 - z_2|^2 = 2 + 5 = 7$. Check at $A$: $|z_2 - z_1|^2 + |z_3 - z_1|^2 = 2 + 5 = 7 = |z_3 - z_2|^2 \cdot \ldots$, recompute: $|z_3 - z_2|^2 = 1 + 4 = 5$ and $|z_2 - z_1|^2 = 1 + 1 = 2$; $2 + 5 = 7 \neq 5$. Test $(z_3 - z_1)/(z_2 - z_1) = (-2-i)/(-1+i) = (-2-i)(-1-i)/2 = (2 + 2i + i + i^2)/2 = (1 + 3i)/2$, not purely imaginary, so not right-angled at $A$. Similar checks show not right-angled.
3. Pair $(\omega, \omega^4)$ and $(\omega^2, \omega^3)$: $(z^2 - 2\cos(2\pi/5)z + 1)(z^2 - 2\cos(4\pi/5)z + 1) = (z^2 - \tfrac{\sqrt{5}-1}{2} z + 1)(z^2 + \tfrac{\sqrt{5}+1}{2} z + 1)$.
4. From $(\cosθ + i\sinθ)^3$, imaginary part: $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. So $4x^3 - 3x = -\sin 3\theta + 0 \cdot \ldots$; rewrite: $-(\sin 3\theta) = -\tfrac{1}{2}$ gives $\sin 3\theta = \tfrac{1}{2}$, so $3\theta = \tfrac{\pi}{6} + 2k\pi$ or $\tfrac{5\pi}{6} + 2k\pi$; $\theta = \tfrac{\pi}{18}, \tfrac{5\pi}{18}, \tfrac{13\pi}{18}, \ldots$. Three distinct $x = \sin\theta$ values.
5. $8i = 8\,\mathrm{cis}(\pi/2)$, so cube roots $z_k = 2\,\mathrm{cis}(\pi/6 + 2k\pi/3)$, $k = 0, 1, 2$. Equally spaced by $2\pi/3$ on circle of radius $2$. Side length $= 2 \cdot 2 \sin(\pi/3) = 2\sqrt{3}$.
Q1 (3 marks): $\omega^5 = 1$ and $\omega \neq 1$, so $(\omega - 1)(1 + \omega + \omega^2 + \omega^3 + \omega^4) = \omega^5 - 1 = 0$ forces the sum to $0$ [1]. For the product: $z^5 - 1 = (z - 1)\prod_{k=1}^{4}(z - \omega^k)$, so $\prod_{k=1}^{4}(z - \omega^k) = z^4 + z^3 + z^2 + z + 1$ [1]. Set $z = 1$: $\prod_{k=1}^{4}(1 - \omega^k) = 5$ [1].
Q2 (4 marks): $|z_2 - z_1| = 2|e^{2\pi i/3} - 1| = 2 \cdot 2\sin(\pi/3) = 2\sqrt{3}$ [1]. By symmetry (rotate by $2\pi/3$): $z_3 - z_2 = e^{2\pi i / 3}(z_2 - z_1)$, so $|z_3 - z_2| = 2\sqrt{3}$ [1]. Similarly $|z_1 - z_3| = 2\sqrt{3}$ [1]. All sides equal $\Rightarrow$ equilateral; side length $2\sqrt{3}$ [1].
Q3 (5 marks): By de Moivre, $\cos 4\theta + i\sin 4\theta = (\cos\theta + i\sin\theta)^4$. Real part: $c^4 - 6c^2 s^2 + s^4$ [1]. Substitute $s^2 = 1 - c^2$: $c^4 - 6c^2(1 - c^2) + (1 - c^2)^2 = 8c^4 - 8c^2 + 1$ [1]. Set $\theta = \pi/8$: $\cos(\pi/2) = 0 = 8x^4 - 8x^2 + 1$ where $x = \cos(\pi/8)$ [1]. Solve quadratic in $x^2$: $x^2 = \tfrac{8 \pm \sqrt{32}}{16} = \tfrac{2 \pm \sqrt{2}}{4}$ [1]. Since $\cos(\pi/8) > \cos(\pi/4) = \tfrac{1}{\sqrt{2}}$, take positive root: $\cos\tfrac{\pi}{8} = \tfrac{\sqrt{2 + \sqrt{2}}}{2}$ [1].
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Mark lesson as complete
Tick when you've finished the practice and review.