How many QCAA Specialist Mathematics questions cover Further complex numbers?

AusGrader has 81 QCAA Specialist Mathematics questions on Further complex numbers, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Specialist Mathematics — Further complex numbers Questions & Answers | AusGrader

Q1

2022

QCAA

Paper 2

1 mark

Q1

1 mark

A solution of the equation $z^2 = ai$ , where $a \ne R$ , is $z = -2 - 2i$ .
The other solution is

A

$-8i$

B

$-2+2i$

C

$2+2i$

D

$8i$

Q8

2021

QCAA

Paper 2

1 mark

Q8

1 mark

The imaginary part of $\left(\text{cis}\left(\frac{\pi}{8}\right)\right)^{-2}$ is

A

$-6.83$

B

$-0.71$

C

$0.71$

D

$1.17$

Q12

2023

QCAA

Paper 2

7 marks

Q12

Consider the complex number $z = -3 + 2i$ .

Q12a

2 marks

Determine $z^3$ using the binomial theorem. Leave your answer in the form $a + bi$ , where $a, b \in R$ .

Descriptor	Marks
correctly uses the binomial theorem	1
expresses $z^3$ in the form of $a+bi$	1

Q12b

1 mark

Convert $z$ into the form $r \text{ cis}(\theta)$ , where $-\pi < \theta \leq \pi$ .

$z = 3.61\text{cis}(2.55)$

Marking Criteria

Descriptor	Marks
correctly converts $z$ into $r\text{cis}(\theta)$ form where $-\pi < \theta \le \pi$	1

Q12c

2 marks

Use the result from Question 12b) to determine $z^3$ using De Moivre's theorem. Leave your answer in the form $r \text{ cis}(\theta)$ , where $-\pi < \theta \leq \pi$ .

Using De Moivre's theorem
$z^3 = (3.61\text{cis}(2.55))^3$
$= 3.61^3\text{cis}(3 \times 2.55)$
$= 46.87\text{cis}(7.66)$
$= 46.87\text{cis}(1.38)$

Marking Criteria

Descriptor	Marks
uses De Moivre's theorem to calculate $z^3$	1
determines $z^3$ in the form of $r\text{cis}(\theta)$ where $-\pi < \theta \le \pi$	1

Q12d

2 marks

Evaluate the reasonableness of your results from Questions 12a) and 12c), noting that the two methods to determine $z^3$ should produce the same result.

$46.87\text{cis}(1.38) = 46.87(\cos(1.38) + i\sin(1.38))$
$= 8.89 + 46.02i$
$\approx 9 + 46i$

The two methods produce approximately the same $z^3$ value. The small variation is a result of rounding used in earlier calculations.

Marking Criteria

Descriptor	Marks
shows mathematical reasoning to convert from $r\text{cis}(\theta)$ form to $a+bi$ form	1
states a decision regarding the reasonableness	1

Q13

2021

QCAA

Paper 1

6 marks

Q13

6 marks

Use $z = a + bi$ and $w = c + di$ , where $a, b, c, d \in R$ , to prove

$|z - w|^2 = |z|^2 + |w|^2 - 2Re(z\bar{w})$

RTP
$|z - w|^2 = |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})$

LHS $= |z - w|^2$
$= |(a + bi) - (c + di)|^2$
$= |(a - c) + (b - d)i|^2$
$= (a - c)^2 + (b - d)^2$
$= a^2 - 2ac + c^2 + b^2 - 2bd + d^2$

RHS $= |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})$
$= |a + bi|^2 + |c + di|^2 \dots$
$\dots - 2\text{Re}((a + bi)(c - di))$
$= a^2 + b^2 + c^2 + d^2 \dots$
$\dots - 2\text{Re}((ac + bd) + (bc - ad)i)$
$= a^2 + b^2 + c^2 + d^2 - 2(ac + bd)$
$= a^2 + b^2 + c^2 + d^2 - 2ac - 2bd$
$= a^2 - 2ac + c^2 + b^2 - 2bd + d^2$
$= \text{LHS}$

Marking Criteria

Descriptor	Marks
correctly expresses $z - w$ in terms of $a, b, c$ and $d$ in Cartesian form	1
expresses $\|z - w\|^2$ in terms of $a, b, c$ and $d$ in expanded form	1
correctly expresses $\|z\|^2 + \|w\|^2$ in terms of $a, b, c$ and $d$ in expanded form	1
correctly expresses $(z \overline{w})$ in terms of $a, b, c$ and $d$	1
completes proof	1
shows logical organisation, communicating key steps	1

Q13

2023

QCAA

Paper 1

5 marks

Q13

5 marks

Given $z \in C$ , where $z \neq 0$ , prove $\frac{|z|}{z\bar{z}} = |z^{-1}|$ .

Method 1

Prove $\left|\frac{z}{z\bar{z}}\right| = |z^{-1}|$
Let $z = a+bi$ , where $a,b \in \mathbb{R}$

$LHS = \left|\frac{z}{z\bar{z}}\right|$
$= \left|\frac{a+bi}{(a+bi)(a-bi)}\right|$
$= \frac{\sqrt{a^2+b^2}}{a^2+b^2}$
$= \frac{1}{\sqrt{a^2+b^2}}$
$= \frac{1}{|a+bi|}$
$= \left|\frac{1}{a+bi}\right|$
$= |z^{-1}|$
$= RHS$

Marking Criteria

Descriptor	Marks
correctly represents $\bar{z}$ in terms of $a$ and $b$	1
correctly represents $\|z\|$ as a radical expression in terms of $a$ and $b$	1
correctly simplifies $z\bar{z}$ in terms of $a$ and $b$	1
determines an expression without an index that represents $\|z\|^{-1}$	1
correctly shows mathematical reasoning to complete proof	1

QCAA Specialist Mathematics Further complex numbers

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