How many QCAA Specialist Mathematics Alternative Sequence questions cover Complex arithmetic and algebra?

AusGrader has 12 QCAA Specialist Mathematics Alternative Sequence questions on Complex arithmetic and algebra, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics Alternative Sequence — Complex arithmetic and algebra Questions & Answers

Q1

2022

QCAA

Paper 2

1 mark

Q1

1 mark

Given $w = \text{cis}\left(\frac{\pi}{2}\right)$ and $z = \text{cis}\left(\frac{\pi}{4}\right)$ , determine $\frac{w}{z}$ .

A

$\text{cis}\left(\frac{\pi}{4}\right)$

B

$\text{cis}\left(\frac{\pi}{2}\right)$

C

$\text{cis}\left(\frac{1}{2}\right)$

D

$\text{cis}(2)$

Reveal Answer

A

$\text{cis}\left(\frac{\pi}{4}\right)$

Correct Answer

Correct. When dividing complex numbers in polar form, you subtract their arguments: $\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$ .

B

$\text{cis}\left(\frac{\pi}{2}\right)$

Incorrect. This is the value of $w$ , not the quotient $\frac{w}{z}$ . Remember to subtract the argument of $z$ from the argument of $w$ .

C

$\text{cis}\left(\frac{1}{2}\right)$

Incorrect. This results from incorrectly dividing the arguments as $\frac{\pi/4}{\pi/2}$ instead of subtracting them.

D

$\text{cis}(2)$

Incorrect. This results from incorrectly dividing the arguments as $\frac{\pi/2}{\pi/4} = 2$ instead of subtracting them.

Q13

2020

QCAA

Paper 1

5 marks

Q13a

3 marks

Given $z^2 + 4z + 13 = (z - h)^2 + k$ , where $h, k \in R$ , determine the values of $h$ and $k$ .

Reveal Answer

$z^2+4z+13$
$=z^2+4z+4-4+13$
$=(z+2)^2+9$
Equating terms with $(z-h)^2+k$
$h=-2$ and $k=9$

Marking Criteria

Descriptor	Marks
correctly completes the square	1
determines value of $h$	1
determines value of $k$	1

Q13b

2 marks

Using the result from 13a), or otherwise, solve $z^2 + 4z + 13 = 0$ where $z \in C$ . Express your answer in simplest form.

Reveal Answer

Method 1
$(z+2)^2+9=0$
$z+2=\pm\sqrt{-9}$
$z=-2\pm 3i$

Marking Criteria

Descriptor	Marks
expresses result from a) as a linear equation in terms of $z$	1
determines both solutions in simplest form	1

Q8

2020

QCAA

Paper 2

1 mark

Q8

1 mark

Let $u = 1 + i$ and $v = -12 + 5i$

$Re(u^5 - |v|)$ is

A

-17

B

-4

C

8

D

9

Reveal Answer

A

-17

Correct Answer

First, find $u^5 = (1+i)^5 = -4 - 4i$ . Then, find the magnitude $|v| = \sqrt{(-12)^2 + 5^2} = 13$ . The expression becomes $-4 - 4i - 13 = -17 - 4i$ , which has a real part of $-17$ .

B

-4

This is only the real part of $u^5$ . You must also subtract the magnitude of $v$ to find the correct real part of the entire expression.

C

8

This answer results from calculation errors in either expanding the complex binomial $u^5$ or finding the magnitude $|v|$ .

D

9

This incorrect answer likely comes from subtracting the real part of $u^5$ ( $-4$ ) from $|v|$ ( $13$ ), rather than subtracting $|v|$ from the real part of $u^5$ .

Q16

2022

QCAA

Paper 1

5 marks

Q16

A quadratic equation with real coefficients has a solution of $z = \sqrt{3} - i$ .

Q16a

3 marks

Given the coefficient of the $z^2$ term is $\sqrt{3}$ , use the conjugate root to determine the equation. Express your answer in expanded form.

Reveal Answer

Method 1
Since the coefficients of the equation are real, the other solution is $z = \sqrt{3} + i$ .

The quadratic equation is
$\sqrt{3}\left(z - (\sqrt{3} - i)\right)\left(z - (\sqrt{3} + i)\right) = 0$
$\sqrt{3}\left(z^2 - (\sqrt{3} - i)z - (\sqrt{3} + i)z + (\sqrt{3} - i)(\sqrt{3} + i)\right) = 0$
$\sqrt{3}(z^2 - 2\sqrt{3}z + 4) = 0$
$\sqrt{3}z^2 - 6z + 4\sqrt{3} = 0$

Marking Criteria

Descriptor	Marks
correctly identifies the other solution of the quadratic equation	1
expresses two solutions within a quadratic equation	1
expresses a quadratic equation in expanded form with the coefficient of the $z^2$ term equal to $\sqrt{3}$	1

Q16b

2 marks

Use the quadratic formula to verify your result from Question 16a).

Reveal Answer

Using the quadratic formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 4(\sqrt{3})(4\sqrt{3})}}{2\sqrt{3}}$
$= \frac{6 \pm \sqrt{-12}}{2\sqrt{3}} = \frac{3 \pm \sqrt{3}i}{\sqrt{3}}$
$= \frac{(3 \pm \sqrt{3}i)\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{3(\sqrt{3} \pm i)}{3}$
$= \sqrt{3} \pm i$

The result is verified.

Marking Criteria

Descriptor	Marks
substitutes results from 16a) into the quadratic formula	1
shows mathematical reasoning to verify both solutions	1

QCAA Specialist Mathematics Alternative Sequence Complex arithmetic and algebra

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