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hscscience Ext 2 · Y12
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Module 12 Synthesis & Exam Technique

The final lesson of Complex Numbers I. You will now stitch together every piece of the module — rectangular form, conjugates, modulus and argument, polar form, the geometry of multiplication and division, and loci — and learn how to deploy them under HSC exam conditions: stating the form clearly, sketching the Argand plane for loci, and choosing the right representation for the right question.

Today's hook — An HSC question asks: "Find all $z$ such that $|z - 2| = |z + 2i|$." Without computing, what shape is this locus? Which two points is $z$ equidistant from? In which form (rectangular or polar) would you tackle this most efficiently? Justify after card 05.
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Recall — your gut answer first
+5 XP warm-up

Without notes, list the three "forms" of a complex number used in this module and one situation in which each is the best representation. Be specific — when does rectangular beat polar, and vice versa?

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02
The two moves for exam complex-number questions
+5 XP to read

HSC complex-number questions are rarely about computation alone — they reward two strategic habits: choose the representation that matches the operation (addition/subtraction $\to$ rectangular; multiplication/division/powers $\to$ polar), and sketch the Argand plane whenever a locus or geometric condition appears (a clear, labelled diagram is often worth a mark by itself).

The state-form / sketch-locus habit: (1) at the top of your working, write "$z = a + bi$" or "$z = r(\cos\theta + i\sin\theta)$" — whichever you will use, (2) for loci, draw the Argand plane with axes labelled, mark the fixed points $z_0$ that appear, and (3) translate the algebraic condition into a geometric description before solving.

$|z - z_0| = r$ → circle  ·  $\arg(z - z_0) = \alpha$ → ray  ·  $|z - z_1| = |z - z_2|$ → perpendicular bisector

State form a+bi or r e^{iθ} Sketch Argand Solve algebra Diagram + form choice = half the marks
$z = a + bi \;=\; r(\cos\theta + i\sin\theta)$
Rectangular for + and −
Addition, subtraction, equating real/imaginary parts, conjugates: rectangular form is fastest. $z_1 + z_2 = (a_1 + a_2) + (b_1 + b_2)i$.
Polar for $\times$, $\div$, $^n$
Multiplication multiplies moduli and adds arguments; division divides moduli and subtracts arguments; powers raise modulus to $n$ and multiply argument by $n$.
Sketch every locus
For any condition involving $|z - z_0|$ or $\arg(z - z_0)$: draw labelled axes, mark fixed points, sketch the resulting curve. Markers often award a mark just for the diagram.
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What you'll master
Know

Key facts

  • $z = a + bi$ (rectangular); $z = r(\cos\theta + i\sin\theta) = r e^{i\theta}$ (polar/exponential)
  • $|z| = \sqrt{a^2 + b^2}$; $\arg z = \arctan(b/a)$ adjusted by quadrant; $-\pi < \arg z \leq \pi$
  • Conjugate $\bar z = a - bi$; $z\bar z = |z|^2$; $\overline{z_1 z_2} = \bar z_1 \bar z_2$
  • Product: $|z_1 z_2| = |z_1||z_2|$, $\arg(z_1 z_2) = \arg z_1 + \arg z_2$
Understand

Concepts

  • Why polar form makes products and quotients trivial (modulus & argument behave independently)
  • Why $|z - z_0| = r$ is a circle and $\arg(z - z_0) = \alpha$ is a ray
  • Why $|z - z_1| = |z - z_2|$ is the perpendicular bisector of $z_1 z_2$
Can do

Skills

  • Switch between rectangular and polar form fluently, in both directions
  • Identify and sketch standard loci on the Argand plane
  • Choose the most efficient form for the operation at hand under exam time pressure
04
Key terms (Module 12 vocabulary)
Rectangular (Cartesian) form$z = a + bi$ where $a = \Re(z)$, $b = \Im(z)$. Best for addition/subtraction and for equating real and imaginary parts.
Polar form$z = r(\cos\theta + i\sin\theta)$ where $r = |z| \geq 0$ is the modulus and $\theta = \arg z$ is the argument. Best for multiplication, division and powers.
Exponential form$z = r e^{i\theta}$, equivalent to polar via Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$. Compact for products, quotients and rotations.
Conjugate $\bar z$$\overline{a + bi} = a - bi$. Reflects $z$ in the real axis. Satisfies $z\bar z = |z|^2$, $\overline{z_1 + z_2} = \bar z_1 + \bar z_2$, $\overline{z_1 z_2} = \bar z_1 \bar z_2$.
Modulus $|z|$$|z| = \sqrt{a^2 + b^2}$. Distance from $z$ to the origin on the Argand plane. $|z_1 z_2| = |z_1||z_2|$; $|z_1/z_2| = |z_1|/|z_2|$.
Argument $\arg z$The angle $z$ makes with the positive real axis, measured anticlockwise. Principal value: $-\pi < \arg z \leq \pi$. $\arg(z_1 z_2) = \arg z_1 + \arg z_2$ (mod $2\pi$).
Locus on the Argand planeThe set of all $z$ satisfying a given condition. Standard shapes: $|z - z_0| = r$ (circle), $\arg(z - z_0) = \alpha$ (ray), $|z - z_1| = |z - z_2|$ (perpendicular bisector), $\Re(z) = a$ (vertical line), $\Im(z) = b$ (horizontal line).
MEX-N1NESA outcome (Complex Numbers I): performs operations using the various representations of complex numbers and uses them to solve problems.
05
The Module 12 framework, on one page
core concept · synthesis

Every Module 12 question can be tackled by choosing one of three lenses:

  1. Algebraic (rectangular). $z = a + bi$. Use when adding, subtracting, equating real/imaginary parts, finding conjugates, or testing rectangular conditions like $\Re(z) = 1$.
  2. Geometric (modulus & argument / polar). $z = r(\cos\theta + i\sin\theta)$. Use when multiplying, dividing, raising to powers, computing rotations, or working with $|z - z_0|$ and $\arg(z - z_0)$.
  3. Geometric (loci on the Argand plane). Sketch first; describe the curve in words; only then write algebra. Useful whenever a condition involves $z$ as a moving point.

Loci library — memorise these:

  • $|z - z_0| = r$ — circle, centre $z_0$, radius $r$.
  • $|z - z_0| < r$ — open disc; $|z - z_0| \leq r$ — closed disc.
  • $\arg(z - z_0) = \alpha$ — ray from $z_0$ (excluding $z_0$) at angle $\alpha$ to the positive real axis.
  • $|z - z_1| = |z - z_2|$ — perpendicular bisector of segment from $z_1$ to $z_2$.
  • $\Re(z) = a$ — vertical line $x = a$; $\Im(z) = b$ — horizontal line $y = b$.
  • $|z| = a|z - z_0|$ (with $a \neq 1$) — circle (Apollonius); $a = 1$ gives perpendicular bisector instead.

Worked through the hook. $|z - 2| = |z + 2i|$ asks "which $z$ are equidistant from $2$ and $-2i$?" Geometrically, this is the perpendicular bisector of the segment from $(2, 0)$ to $(0, -2)$. The midpoint is $(1, -1)$; the segment has slope $1$, so the bisector has slope $-1$ and passes through $(1, -1)$: $y - (-1) = -1(x - 1)$, i.e. $y = -x$, i.e. $x + y = 0$, or in complex form $\Re(z) + \Im(z) = 0$.

The strategic point. Doing this algebraically — let $z = x + iy$, square both sides, expand — also works, but takes three times the writing. Recognising the locus first is the time-saver. In the HSC, time is marks.

Three forms: rectangular $a + bi$, polar $r(\cos\theta + i\sin\theta)$, exponential $r e^{i\theta}$ · +/− → rectangular; $\times$, $\div$, $^n$ → polar · $|z - z_0| = r$ circle; $\arg(z - z_0) = \alpha$ ray; $|z - z_1| = |z - z_2|$ perpendicular bisector · Sketch the Argand plane before doing algebra on a locus

Pause — copy the three-form summary (rectangular, polar, exponential), the operation-to-form mapping ($+/-$ → rect; $\times,\div,z^n$ → polar), and the three main locus types into your book.

Quick check: Which of the following loci on the Argand plane is a perpendicular bisector?

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HSC exam approach — three habits
core concept · exam technique

We just saw the Module 12 framework: rectangular for $+/-$; polar for $\times, \div, z^n$; the loci $|z-z_0|=r$, $\arg(z-z_0)=\alpha$, $|z-z_1|=|z-z_2|$; always sketch first. That raises a question: what three habits differentiate a full-mark response from a near-miss in an HSC exam? This card answers it → state the form, sketch the Argand plane, and choose polar for products/powers and rectangular for sums.

Marking guides for HSC complex-number questions reward three habits beyond the algebra itself:

  • State the form clearly. Write "$z = a + bi$ where $a, b \in \mathbb{R}$" or "$z = r(\cos\theta + i\sin\theta)$" at the top. This signals to the marker which approach you are taking, and prevents you from drifting between forms.
  • Sketch the Argand plane for any locus. Even if the question doesn't say "sketch", drawing a small labelled diagram clarifies your thinking. Axes labelled $\Re(z)$, $\Im(z)$; fixed points $z_0$ marked; the locus curve drawn solid (or dashed if strict inequality).
  • Conversion strategy. If the question gives you rectangular and asks for an argument or power, convert to polar first. If it gives you polar and asks for $\Re(z)$ or $\Im(z)$, expand back to rectangular. The "wrong form for the operation" is the single most common source of wasted marks.
$$z_1 z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)], \quad \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)].$$
The marker's view. A correctly-stated form + an accurate sketch + a one-line geometric description routinely earns 1–2 marks before any algebra is shown. Conversely, jumping straight into algebra in the wrong form can cost the same number of marks even when the final answer is right.

State the form: "$z = a + bi$, $a, b \in \mathbb{R}$" or "$z = r(\cos\theta + i\sin\theta)$" · Sketch Argand plane for any locus or geometric condition · Polar form for products/quotients/powers; rectangular for sums/differences · $z_1 z_2$: multiply moduli, add arguments. $z_1/z_2$: divide moduli, subtract arguments

Pause — copy the three HSC habits (state form, sketch, choose polar/rect by operation) and the product/quotient rules ($z_1 z_2$: multiply moduli, add arguments; $z_1/z_2$: divide moduli, subtract arguments) into your book.

Did you get this? True or false: if $z_1 = 2(\cos 30^{\circ} + i\sin 30^{\circ})$ and $z_2 = 3(\cos 45^{\circ} + i\sin 45^{\circ})$, then $z_1 z_2 = 6(\cos 75^{\circ} + i\sin 75^{\circ})$.

PROBLEM 1 · CONVERT & COMPUTE

Let $z_1 = 1 + i\sqrt{3}$ and $z_2 = \sqrt{3} - i$. Convert each to polar form, then compute $z_1 z_2$ and $z_1/z_2$ in polar form, and convert back to rectangular.

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$|z_1| = \sqrt{1 + 3} = 2$; $\arg z_1 = \arctan(\sqrt{3}/1) = \pi/3$ (first quadrant). So $z_1 = 2(\cos(\pi/3) + i\sin(\pi/3))$. Similarly $|z_2| = \sqrt{3 + 1} = 2$; $\arg z_2 = \arctan(-1/\sqrt{3}) = -\pi/6$ (fourth quadrant). So $z_2 = 2(\cos(-\pi/6) + i\sin(-\pi/6))$.
When finding the argument, always check the quadrant — $\arctan$ alone gives an answer in $(-\pi/2, \pi/2)$, which is correct for quadrants I and IV but wrong for II and III.
PROBLEM 2 · SKETCH A LOCUS

Sketch the locus of $z$ satisfying both $|z - 2i| = 2$ and $0 \leq \arg(z) \leq \pi/2$. Describe the resulting set geometrically.

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$|z - 2i| = 2$ is a circle centred at $z_0 = 2i$ (the point $(0, 2)$) with radius $2$. Sketch on the Argand plane: this circle passes through the origin and $(0, 4)$ and is tangent to the real axis at $0$.
First disentangle the conditions one at a time; sketch each in turn on labelled axes.
PROBLEM 3 · POLAR-FORM POWER

Use polar form to compute $(1 + i)^8$. Express the answer in rectangular form.

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Convert: $|1 + i| = \sqrt{2}$, $\arg(1 + i) = \pi/4$. So $1 + i = \sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))$.
Powers in rectangular form (e.g. $(1+i)^8$ by repeated multiplication) take eight multiplications. Polar reduces it to one.

Fill the gap: If $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$, then $z_1 z_2 = $ $\,[\cos($$) + i\sin(\theta_1 + \theta_2)]$.

Trap 01
Wrong quadrant for the argument
$\arctan(b/a)$ alone gives an answer in $(-\pi/2, \pi/2)$. If $z$ is in quadrant II or III, you must add or subtract $\pi$ to get the correct principal argument. Always sketch the point to check the quadrant before reporting $\arg z$.
Trap 02
Using the wrong form
Computing $(\cos 60^{\circ} + i\sin 60^{\circ})(\cos 30^{\circ} + i\sin 30^{\circ})$ by expanding the brackets when you could just add arguments. Or trying to find $z + \bar z$ in polar form when rectangular gives $2a$ instantly. Always pause: which form fits this operation?
Trap 03
Sketching loci without labelling
A locus sketch with unlabelled axes, no centre marked, and no radius written down often loses the mark even if the curve is correct. Always label $\Re(z)$, $\Im(z)$, the fixed point $z_0$, the radius, and indicate solid (closed) or dashed (strict inequality) curves.

Did you get this? True or false: the principal argument of $z = -1 - i$ is $-3\pi/4$.

Work mode · how are you completing this lesson?
1

Express $z = -2 + 2i\sqrt{3}$ in polar form, then compute $z^4$ in rectangular form.

2

Sketch the locus of $z$ satisfying $|z - 1| = |z + 3i|$ on the Argand plane, and find its Cartesian equation.

3

If $z_1 = 3(\cos 50^{\circ} + i\sin 50^{\circ})$ and $z_2 = 2(\cos 20^{\circ} + i\sin 20^{\circ})$, find $z_1/z_2$ in polar form and convert to rectangular.

4

Find all $z$ such that $z^2 = -16$ — give your answer in rectangular form. (Hint: use polar.)

5

Sketch on the Argand plane the region defined by $|z - 1 - i| \leq 1$ AND $\arg(z - 1 - i) \in [0, \pi/2]$. Describe the shape.

Odd one out: Three of these are best handled in polar form. Which one is fastest in rectangular form?

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Revisit your thinking · close the module

Earlier you predicted the shape of $|z - 2| = |z + 2i|$.

It is the perpendicular bisector of the segment from $(2, 0)$ to $(0, -2)$: midpoint $(1, -1)$, gradient $-1$, Cartesian equation $y = -x$ (i.e. $x + y = 0$). Recognising the locus geometrically — rather than expanding $|z - 2|^2 = |z + 2i|^2$ — saves a page of algebra in the exam. This is Module 12 in one slogan: before you compute, choose the form that matches the question. Polar for products, powers and arguments; rectangular for sums, differences and conjugates; geometric for loci. With those three lenses, every Complex Numbers I question becomes a strategic choice followed by short, clean working.

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer
ApplyBand 32 marks

Q1. Express $z = -\sqrt{3} + i$ in polar form, giving the argument in the range $-\pi < \arg z \leq \pi$. (2 marks)

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ApplyBand 43 marks

Q2. Sketch the locus of $z$ satisfying $|z - 2| = |z - 2i|$ on the Argand plane. Find its Cartesian equation and describe the shape. (3 marks)

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AnalyseBand 53 marks

Q3. Let $z_1 = 1 + i$ and $z_2 = \sqrt{3} - i$. Use polar form to find $z_1^6 / z_2^4$, giving your answer in rectangular form. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $z = -2 + 2i\sqrt{3}$. $|z| = \sqrt{4 + 12} = 4$. $z$ is in quadrant II; reference angle $= \arctan(2\sqrt{3}/2) = \pi/3$, so $\arg z = \pi - \pi/3 = 2\pi/3$. $z = 4(\cos(2\pi/3) + i\sin(2\pi/3))$. $z^4 = 4^4 (\cos(8\pi/3) + i\sin(8\pi/3)) = 256(\cos(2\pi/3) + i\sin(2\pi/3)) = 256(-1/2 + i\sqrt{3}/2) = -128 + 128i\sqrt{3}$.

2. Midpoint of $(1, 0)$ and $(0, -3)$: $(1/2, -3/2)$. Segment slope $= -3$; bisector slope $= 1/3$. Equation: $y + 3/2 = (1/3)(x - 1/2)$ $\Rightarrow$ $3y + 9/2 = x - 1/2$ $\Rightarrow$ $x - 3y = 5$.

3. $z_1/z_2 = (3/2)(\cos 30^{\circ} + i\sin 30^{\circ}) = (3/2)(\sqrt{3}/2 + i/2) = 3\sqrt{3}/4 + (3/4)i$.

4. $-16 = 16(\cos \pi + i\sin \pi)$. $z = 4(\cos(\pi/2 + k\pi) + i\sin(\pi/2 + k\pi))$, $k = 0, 1$. $z_0 = 4(\cos(\pi/2) + i\sin(\pi/2)) = 4i$; $z_1 = 4(\cos(3\pi/2) + i\sin(3\pi/2)) = -4i$. So $z = \pm 4i$.

5. The region is the quarter-disc obtained by intersecting (i) the closed disc of radius $1$ centred at $1 + i$ with (ii) the closed first-quadrant sector based at $1 + i$. Shape: a quarter disc, with its right angle at the vertex $1 + i$, bounded by the segment from $1 + i$ to $2 + i$ along the real-axis direction and from $1 + i$ to $1 + 2i$ along the imaginary-axis direction, with the curved arc joining $2 + i$ and $1 + 2i$.

Q1 (2 marks): $|z| = \sqrt{3 + 1} = 2$ [1]. $z$ lies in quadrant II ($\Re < 0$, $\Im > 0$); reference angle $\arctan(1/\sqrt{3}) = \pi/6$, so $\arg z = \pi - \pi/6 = 5\pi/6$. Therefore $z = 2(\cos(5\pi/6) + i\sin(5\pi/6))$ [1].

Q2 (3 marks): The condition $|z - 2| = |z - 2i|$ means $z$ is equidistant from $(2, 0)$ and $(0, 2)$; locus is the perpendicular bisector of the segment from $(2, 0)$ to $(0, 2)$ [1]. Midpoint $(1, 1)$; segment slope $-1$; bisector slope $1$; Cartesian equation $y = x$ [1]. Sketch: line $y = x$ on labelled axes, with the two fixed points marked [1].

Q3 (3 marks): $z_1 = \sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))$, $z_2 = 2(\cos(-\pi/6) + i\sin(-\pi/6))$ [1]. $z_1^6 = (\sqrt{2})^6 (\cos(3\pi/2) + i\sin(3\pi/2)) = 8(-i) = -8i$. $z_2^4 = 16(\cos(-2\pi/3) + i\sin(-2\pi/3)) = 16(-1/2 - i\sqrt{3}/2) = -8 - 8i\sqrt{3}$ [1]. $\dfrac{z_1^6}{z_2^4} = \dfrac{-8i}{-8 - 8i\sqrt{3}} = \dfrac{-8i}{-8(1 + i\sqrt{3})} = \dfrac{i}{1 + i\sqrt{3}} = \dfrac{i(1 - i\sqrt{3})}{1 + 3} = \dfrac{i + \sqrt{3}}{4} = \dfrac{\sqrt{3}}{4} + \dfrac{i}{4}$ [1].

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Boss battle · The Complex Plane Champion
earn bronze · silver · gold

Five timed questions spanning the full module — rectangular, polar, conjugates, products, quotients and loci. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering quick complex-number questions. Lighter alternative to the boss.

Mark lesson as complete · Module 12 finished

Tick when you've finished the practice and review. This completes Complex Numbers I.

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