Worked Solutions
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.
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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 $\theta$.