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].
Five timed multi-step questions weaving geometry, roots of unity and de Moivre. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick mixed-complex-number questions. Lighter alternative to the boss.
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