Mixed Complex Number Problems I
By now you have every tool: polar form, de Moivre, $n$th roots, roots of unity, modulus/argument arithmetic and Argand-plane geometry. HSC questions rarely ask you to use just one — they ask you to choose the right combination. This lesson trains the question-reading skill: which technique does this prompt want, and in what order?
List the four "headline" tools from Modules 12 and 13: polar form $r(\cos\theta + i\sin\theta)$, de Moivre's theorem, the formula for $n$th roots, and the $n$th roots of unity. Before checking — beside each, write the one signal in a question that tells you that tool is the right one to use.
HSC mixed questions reward two habits: decode the prompt (what is being asked — a value, a proof, a locus, a sum of roots?) and choose the right form (Cartesian, polar, exponential, geometric on the Argand plane). The form you start in often determines whether the problem is one line or fifteen.
The decode-form-execute flow: (1) name what the question wants (a number, a proof, a set of $z$), (2) pick the form best suited (polar for powers/roots; Cartesian for $\mathrm{Re}/\mathrm{Im}$; geometric for loci and arguments), (3) execute and convert back if needed.
Powers/roots → polar · Re/Im components → Cartesian · $|z - a| = r$ → geometry
Key facts
- Polar form: $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, with $r = |z|$, $\theta = \arg z$
- De Moivre: $z^n = r^n(\cos n\theta + i\sin n\theta)$
- $n$th roots of $w = R e^{i\phi}$: $z_k = R^{1/n}\exp\!\Big(i\,\tfrac{\phi + 2\pi k}{n}\Big)$, $k = 0, 1, \ldots, n-1$
- Sum of $n$th roots of unity equals $0$ (for $n \ge 2$)
Concepts
- Why polar form trivialises powers and roots, and why Cartesian is preferred for real/imaginary extraction
- Why $\arg(zw) = \arg z + \arg w$ and $|zw| = |z||w|$ make polar form multiplicative
- Why geometric interpretation on the Argand plane often beats algebraic manipulation
Skills
- Decide whether to start a problem in Cartesian, polar or geometric form
- Combine de Moivre with binomial expansion to derive trig identities
- Use roots of unity to evaluate sums of trig functions of equally-spaced angles
Most mixed problems contain explicit signals about which tool to use. Learn to read them:
- "Find $z^n$" with integer $n \ge 3$ → convert to polar, apply de Moivre, convert back if Cartesian required.
- "Solve $z^n = w$" → convert $w$ to polar; apply the $n$th root formula; list all $n$ solutions.
- "Show that $\sum \cos\frac{2\pi k}{n} = \ldots$" → use the fact that $\sum_{k=0}^{n-1} e^{2\pi i k/n} = 0$; take real (or imaginary) parts.
- "Find the locus of $z$ such that $|z - a| = |z - b|$" → perpendicular bisector of $a$ and $b$ on the Argand plane (no algebra needed).
- "Show that $\cos n\theta = \ldots$" → binomial expansion of $(\cos\theta + i\sin\theta)^n$, equate with de Moivre, take real part (see L09).
Hook resolved: for $z = \sqrt{3} + i$, $|z| = 2$ and $\arg z = \pi/6$. By de Moivre, $z^6 = 2^6(\cos\pi + i\sin\pi) = -64$. For $w = 1+i$, $|w| = \sqrt{2}$ and $\arg w = \pi/4$, so $w^{12} = (\sqrt{2})^{12}(\cos 3\pi + i\sin 3\pi) = -64$. Therefore $(z/w)^{12} = z^{12}/w^{12} = (-64)^2 / (-64) = -64$. Polar form makes this a five-line answer; Cartesian would be a nightmare.
Powers $z^n$, roots $z^{1/n}$, division → polar form · Sums of trig at equally-spaced angles → roots of unity, sum = 0 · $|z-a| = |z-b|$ → perpendicular bisector (geometry, not algebra) · $\cos n\theta$ identities → binomial of $(\cos\theta + i\sin\theta)^n$
Pause — copy the four signal-to-tool mappings (powers → polar; equally-spaced angle sums → roots of unity; $|z-a|=|z-b|$ → bisector; $\cos n\theta$ identity → binomial of cis) into your book.
Quick check: You are asked to evaluate $(1 - i\sqrt{3})^{10}$. Which is the most efficient first step?
We just saw the signal-to-tool map: powers/roots/division → polar; sums at equally-spaced angles → roots of unity (sum $= 0$); $|z-a|=|z-b|$ → perpendicular bisector. That raises a question: how do roots of unity generate trig identities involving sums of cosines? This card answers it → take $\omega = e^{2\pi i/n}$, use $1+\omega+\cdots+\omega^{n-1} = 0$, and equate real and imaginary parts.
Let $\omega = e^{2\pi i/n}$ be a primitive $n$th root of unity. The complete set $\{1, \omega, \omega^2, \ldots, \omega^{n-1}\}$ has three pillar properties:
- Sum is zero ($n \ge 2$): $1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$ (geometric series with ratio $\omega$ summed via $\frac{\omega^n - 1}{\omega - 1}$).
- They factor $z^n - 1$: $z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k)$.
- Real parts are $\cos\frac{2\pi k}{n}$, imaginary parts are $\sin\frac{2\pi k}{n}$.
Combining (1) and (3): taking real parts of $1 + \omega + \cdots + \omega^{n-1} = 0$ gives $\sum_{k=0}^{n-1} \cos\frac{2\pi k}{n} = 0$; imaginary parts give the matching $\sin$ sum.
$\omega = e^{2\pi i/n}$ is a primitive $n$th root of unity · $1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0$ — the foundation identity · $z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1})$ — factorisation over $\mathbb{C}$ · Real part of the sum → $\sum \cos = 0$; imaginary part → $\sum \sin = 0$
Pause — copy $\omega = e^{2\pi i/n}$, the sum identity $1+\omega+\cdots+\omega^{n-1} = 0$, the factorisation $z^n-1 = \prod(z-\omega^k)$, and the consequence $\sum\cos = 0$, $\sum\sin = 0$ into your book.
Did you get this? True or false: if $\omega = e^{2\pi i/7}$, then $\omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = -1$.
Worked examples · 3 in a row, reveal as you go
Find the exact value of $(1 + i\sqrt{3})^{8}$ in Cartesian form, using polar form and de Moivre's theorem.
Let $\omega = e^{2\pi i/5}$. By considering $1 + \omega + \omega^2 + \omega^3 + \omega^4$, show that $\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2}$.
On the Argand plane, find the locus of points $z$ satisfying $|z - 2| = |z + 2i|$. State its equation in $x$–$y$ form, and identify the geometric figure.
Fill the gap: If $z = 2\big(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\big)$, then $z^6 = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the locus $|z - 3| = |z - 3i|$ is a circle of radius $\sqrt{9 + 9} = 3\sqrt{2}$.
Activities · practice with the ideas
Evaluate $(\sqrt{3} - i)^{10}$ in exact Cartesian form using polar form and de Moivre's theorem.
Solve $z^4 = -16$ for all complex $z$. Give answers in Cartesian form.
Let $\omega = e^{2\pi i/6}$. Show that $\cos\tfrac{2\pi}{6} + \cos\tfrac{4\pi}{6} + \cos\tfrac{6\pi}{6} + \cos\tfrac{8\pi}{6} + \cos\tfrac{10\pi}{6} = -1$.
Sketch the locus of $z$ in the Argand plane satisfying $|z - 1 - 2i| = 3$. Identify the figure.
Show that if $z$ is on the unit circle ($|z| = 1$), then $z + \tfrac{1}{z} = 2\cos(\arg z)$.
Odd one out: Three of these techniques are well-suited to "find $(1 - i)^{12}$". Which one is NOT?
Earlier you compared the polar and Cartesian approaches to evaluating $(z/w)^{12}$ with $z = \sqrt{3} + i$ and $w = 1 + i$.
Polar form turned a brutal Cartesian computation into a short calculation: $|z/w| = 2/\sqrt{2} = \sqrt{2}$, $\arg(z/w) = \pi/6 - \pi/4 = -\pi/12$, and $(z/w)^{12} = (\sqrt{2})^{12}(\cos(-\pi) + i\sin(-\pi)) = 2^6 \cdot (-1) = -64$. The lesson: whenever a problem mentions powers, roots, products or quotients with non-trivial integer powers, the answer to "which form?" is almost always polar.
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. Use polar form and de Moivre's theorem to find the exact Cartesian value of $(-1 + i)^{10}$. (3 marks)
Q2. Solve $z^3 = 8i$, giving all solutions in Cartesian form. (4 marks)
Q3. Let $\omega = e^{2\pi i/7}$. (a) Explain why $1 + \omega + \omega^2 + \cdots + \omega^6 = 0$. (b) Hence show that $\cos\tfrac{2\pi}{7} + \cos\tfrac{4\pi}{7} + \cos\tfrac{6\pi}{7} = -\tfrac{1}{2}$. (4 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $|\sqrt{3} - i| = 2$, $\arg = -\pi/6$. $(\sqrt{3} - i)^{10} = 2^{10}(\cos(-10\pi/6) + i\sin(-10\pi/6)) = 1024(\cos(-5\pi/3) + i\sin(-5\pi/3)) = 1024(\tfrac{1}{2} + i\tfrac{\sqrt{3}}{2}) = 512 + 512i\sqrt{3}$.
2. $-16 = 16 e^{i\pi}$. Fourth roots: $z_k = 2\exp\!\big(i\tfrac{\pi + 2\pi k}{4}\big)$, $k = 0, 1, 2, 3$. Arguments $\pi/4, 3\pi/4, 5\pi/4, 7\pi/4$. Cartesian: $z_0 = \sqrt{2} + i\sqrt{2}$, $z_1 = -\sqrt{2} + i\sqrt{2}$, $z_2 = -\sqrt{2} - i\sqrt{2}$, $z_3 = \sqrt{2} - i\sqrt{2}$.
3. $1 + \omega + \cdots + \omega^5 = 0$ (sum of 6th roots of unity). Take real parts: $1 + \sum_{k=1}^{5} \cos\tfrac{2\pi k}{6} = 0$, so the sum without the leading $1$ is $-1$.
4. $|z - (1 + 2i)| = 3$ means distance from $z$ to $(1, 2)$ is $3$. Circle, centre $(1, 2)$, radius $3$.
5. $z = e^{i\theta}$ (unit circle), so $1/z = \bar z = e^{-i\theta}$. Hence $z + 1/z = e^{i\theta} + e^{-i\theta} = 2\cos\theta = 2\cos(\arg z)$.
Q1 (3 marks): $|-1 + i| = \sqrt{2}$, $\arg = 3\pi/4$ (second quadrant) [1]. $(-1+i)^{10} = (\sqrt{2})^{10}\big(\cos\tfrac{30\pi}{4} + i\sin\tfrac{30\pi}{4}\big) = 32\big(\cos\tfrac{15\pi}{2} + i\sin\tfrac{15\pi}{2}\big)$ [1]. Reduce $15\pi/2 = 7\pi + \pi/2 \equiv -\pi/2$ mod $2\pi$: $32(\cos(-\pi/2) + i\sin(-\pi/2)) = 32(0 - i) = -32i$ [1].
Q2 (4 marks): $8i = 8 e^{i\pi/2}$ [1]. Cube roots: $z_k = 2\exp\!\big(i\tfrac{\pi/2 + 2\pi k}{3}\big)$ for $k = 0, 1, 2$ [1]. Arguments: $\pi/6, 5\pi/6, 3\pi/2$ [1]. Cartesian: $z_0 = 2(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}) = \sqrt{3} + i$; $z_1 = -\sqrt{3} + i$; $z_2 = -2i$ [1].
Q3 (4 marks): (a) The 7th roots of unity are exactly $1, \omega, \omega^2, \ldots, \omega^6$. They are the seven roots of $z^7 - 1 = 0$ [1]; the coefficient of $z^6$ is $0$, so by Vieta's their sum is $0$. Equivalently, $\sum_{k=0}^{6} \omega^k = \tfrac{\omega^7 - 1}{\omega - 1} = 0$ since $\omega^7 = 1$ but $\omega \neq 1$ [1]. (b) Take real parts: $1 + \sum_{k=1}^{6} \cos\tfrac{2\pi k}{7} = 0$. Pair $\omega^k$ with $\omega^{7-k} = \overline{\omega^k}$, giving equal cosines: $1 + 2\big(\cos\tfrac{2\pi}{7} + \cos\tfrac{4\pi}{7} + \cos\tfrac{6\pi}{7}\big) = 0$ [1], so $\cos\tfrac{2\pi}{7} + \cos\tfrac{4\pi}{7} + \cos\tfrac{6\pi}{7} = -\tfrac{1}{2}$ [1].
Five timed questions mixing polar form, de Moivre, roots of unity and Argand geometry. 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 questions. Lighter alternative to the boss.
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