Complex Numbers — Worked Solutions (HSC Maths Extension 2)
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Worked examples for HSC Maths Extension 2 complex numbers. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Extension 2 — Complex Numbers. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Complex number questions reward fluent switching between Cartesian and polar form — pick whichever makes the operation easiest before you start.
Example 1 — Division in Cartesian form
Question
Express $\dfrac{3 + i}{1 - 2i}$ in the form $x + iy$ where $x, y$ are real.
Solution
Division means multiply top and bottom by the conjugate of the denominator. The conjugate of $1 - 2i$ is $1 + 2i$.
$\dfrac{3 + i}{1 - 2i} \times \dfrac{1 + 2i}{1 + 2i} = \dfrac{(3 + i)(1 + 2i)}{(1 - 2i)(1 + 2i)}$.
Numerator: $(3 + i)(1 + 2i) = 3 + 6i + i + 2i^2 = 3 + 7i - 2 = 1 + 7i$.
Denominator: $(1 - 2i)(1 + 2i) = 1 - (2i)^2 = 1 + 4 = 5$.
So the answer is $\dfrac{1 + 7i}{5} = \dfrac15 + \dfrac{7}{5}i$. Keep the real and imaginary parts split — that's the form they asked for.
We can't divide by a complex number directly, so the standard move is to make the denominator real. We do that by multiplying top and bottom by the conjugate of the bottom — flip the sign of the imaginary part, so $1 - 2i$ becomes $1 + 2i$.
$\dfrac{3 + i}{1 - 2i} \cdot \dfrac{1 + 2i}{1 + 2i}$.
Multiply out the top carefully, remembering $i^2 = -1$: $(3 + i)(1 + 2i) = 3 + 6i + i + 2i^2 = 3 + 7i - 2 = 1 + 7i$.
The bottom is a difference of squares in disguise: $(1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 + 4 = 5$, a real number — which is exactly why we used the conjugate.
Dividing through, $\dfrac{1 + 7i}{5} = \dfrac15 + \dfrac{7}{5}i$. Splitting it into real and imaginary parts gives the requested $x + iy$ form.
Multiply by the conjugate of the denominator.
- Conjugate of $1 - 2i$ is $1 + 2i$
- Numerator: $(3 + i)(1 + 2i) = 3 + 7i + 2i^2 = 1 + 7i$
- Denominator: $(1 - 2i)(1 + 2i) = 1 + 4 = 5$
- $\dfrac{1 + 7i}{5} = \dfrac15 + \dfrac{7}{5}i$
Where the marks go
- 1 mark: Multiplies by the conjugate $1 + 2i$ and expands numerator and denominator correctly
- 1 mark: Correct answer $\frac15 + \frac{7}{5}i$ in the form $x + iy$
Key idea
To divide complex numbers, multiply top and bottom by the conjugate of the denominator; this makes the denominator the real number $|z|^2$.
Example 2 — Polar form and de Moivre's theorem
Question
Let $z = 1 + i\sqrt{3}$. Write $z$ in modulus–argument form and hence evaluate $z^6$, giving your answer in the form $x + iy$.
Solution
Get the modulus and argument, then let de Moivre do the heavy lifting.
Modulus: $|z| = \sqrt{1^2 + (\sqrt3)^2} = \sqrt{1 + 3} = 2$.
Argument: $z$ is in the first quadrant, $\tan\theta = \dfrac{\sqrt3}{1}$, so $\theta = \dfrac{\pi}{3}$.
Thus $z = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)$.
By de Moivre, $z^6 = 2^6\left(\cos 6\cdot\dfrac{\pi}{3} + i\sin 6\cdot\dfrac{\pi}{3}\right) = 64(\cos 2\pi + i\sin 2\pi) = 64(1 + 0) = 64$.
Raising to a power is trivial in polar form — never expand $(1 + i\sqrt3)^6$ by hand.
Powers are painful in Cartesian form but easy in polar form, so let's convert $z = 1 + i\sqrt3$ first.
The modulus is the distance from the origin: $|z| = \sqrt{1^2 + (\sqrt3)^2} = \sqrt{4} = 2$.
For the argument, the point $(1, \sqrt3)$ is in the first quadrant and $\tan\theta = \dfrac{\sqrt3}{1} = \sqrt3$, which gives $\theta = \dfrac{\pi}{3}$ (that's $60^\circ$). So $z = 2\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right)$.
Now de Moivre's theorem says we raise the modulus to the power and multiply the angle by the power: $z^6 = 2^6\left(\cos\dfrac{6\pi}{3} + i\sin\dfrac{6\pi}{3}\right) = 64(\cos 2\pi + i\sin 2\pi)$.
Since $\cos 2\pi = 1$ and $\sin 2\pi = 0$, this is simply $64$. It's purely real — which makes sense, because the angle came all the way back around to the positive real axis.
Convert to polar form, apply de Moivre.
- $|z| = \sqrt{1 + 3} = 2$
- First quadrant, $\tan\theta = \sqrt3 \Rightarrow \theta = \tfrac{\pi}{3}$
- $z = 2\,\mathrm{cis}\,\tfrac{\pi}{3}$
- $z^6 = 2^6\,\mathrm{cis}\,\tfrac{6\pi}{3} = 64\,\mathrm{cis}\,2\pi$
- $\cos 2\pi = 1,\ \sin 2\pi = 0 \Rightarrow z^6 = 64$
Where the marks go
- 1 mark: Correct modulus $|z| = 2$
- 1 mark: Correct argument $\arg z = \frac{\pi}{3}$ with polar form stated
- 1 mark: Applies de Moivre's theorem with correct modulus and angle
- 1 mark: Correct final value $z^6 = 64$ in $x + iy$ form
Key idea
In modulus–argument form, $[r\,\mathrm{cis}\,\theta]^n = r^n\,\mathrm{cis}\,(n\theta)$ (de Moivre), turning powers of complex numbers into simple arithmetic on $r$ and $\theta$.
Frequently asked questions
Step-by-step solutions to exam-style questions on Complex Numbers in HSC Maths Extension 2, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Extension 2 syllabus for Complex Numbers, using the methods and notation expected in the exam.