How many SCSA Mathematics Specialist questions cover Complex numbers?

AusGrader has 100 SCSA Mathematics Specialist questions on Complex numbers, all with instant AI grading and detailed marking feedback.

SCSA (WACE) Mathematics Specialist — 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

$1.17$

Q13

2020

SCSA

Paper 2

4 marks

Q13

4 marks

Solve the equation $z^4 = 8\sqrt{3} + 8i$ giving exact solutions in the form $r\text{cis}\theta$ where $-\pi < \theta \le \pi$ .

$z^4 = 8\sqrt{3} + 8i$

$r^2 = \left(8\sqrt{3}\right)^2 + 8^2 = 8^2(3) + 8^2 = 4(8^2) \implies \therefore r = 2(8) = 16$

$\tan \theta = \frac{8}{8\sqrt{3}} \implies \therefore \theta = \frac{\pi}{6}$

$\text{Hence solve } z^4 = 16cis\left(\frac{\pi}{6}\right)$

$\text{Solutions are given by } z = 16^{\frac{1}{4}}cis\left(\frac{\frac{\pi}{6} + 2\pi k}{4}\right) \text{ where } k = 0, 1, 2, 3 \text{ .}$

$\text{Solutions are: }$

$z_0 = 2cis\left(\frac{\pi}{24}\right)$

$z_1 = 2cis\left(\frac{13\pi}{24}\right)$

$z_2 = 2cis\left(\frac{13\pi}{24} + \frac{12\pi}{24}\right) = 2cis\left(-\frac{23\pi}{24}\right)$

$z_3 = 2cis\left(-\frac{11\pi}{24}\right)$

Marking Criteria

Descriptor	Marks
determines the modulus correctly for $8\sqrt{3} + 8i$	1
determines the argument correctly for $8\sqrt{3} + 8i$	1
states one solution as $z = 2cis\left(\frac{\pi}{24}\right)$	1
states the correct arguments for the other 3 solutions (using $-\pi < \theta \le \pi$ )	1

Q11

2020

SCSA

Paper 2

5 marks

Q11

Let $z$ , $w$ and $u$ be complex numbers where:

$w = (1 + i)\overline{z} \quad \text{Arg}(w) = \frac{\pi}{3} \quad |w| =2 \quad u = \frac{z}{2 - 2i}$

Q11a

3 marks

Determine $\text{Arg}(u)$ exactly.

\begin{align*} Arg(w) &= Arg(1+i) + Arg(\overline{z})\\ \frac{\pi}{3} &= \frac{\pi}{4} - Arg(z)\\ \therefore Arg(z) &= -\frac{\pi}{12} \end{align*}

\begin{align*} Arg(u) &= Arg(z) - Arg(2-2i)\\ &= -\frac{\pi}{12} - \left(-\frac{\pi}{4}\right)\\ &= \frac{\pi}{6} \end{align*}

Marking Criteria

Descriptor	Marks
expresses relationships between arguments correctly	1
determines $Arg(z)$ correctly	1
determines $Arg(u)$ correctly	1

Q11b

2 marks

Determine $|u|$ exactly.

\begin{align*} |w| &= |1+i| \times |\overline{z}|\\ 2 &= \sqrt{2} \times |z| \\ \therefore |z| &= \sqrt{2} \end{align*}

$|u| = \frac{|z|}{|2-2i|} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2}$

Marking Criteria

Descriptor	Marks
expresses relationship between the modulus of numbers correctly	1
determines $\|u\|$ correctly	1

Q8

2020

SCSA

Paper 1

3 marks

Q8

3 marks

Consider the complex sum: $\sum_{n=1}^{2020} n i^n = 1i^1 + 2i^2 + 3i^3 + \dots + 2020i^{2020}$

Express the value of this sum in the form $r \text{cis} \, \theta$ where $-\pi < \theta \le \pi$ .

\begin{align*} \sum_{n=1}^4 n i^n &= 1(i)^1 + 2(i)^2 + 3(i)^3 + 4(i)^4 = i - 2 - 3i + 4 = 2 - 2i\\ \sum_{n=5}^8 n i^n &= 5(i)^5 + 6(i)^6 + 7(i)^7 + 8(i)^8 = 5i - 6 - 7i + 8 = 2 - 2i \\ \sum_{n=9}^{12} n i^n &= 9(i)^9 + 10(i)^{10} + 11(i)^{11} + 12(i)^{12} = 9i - 10 - 11i + 12 = 2 - 2i \\ \end{align*}

\begin{align*} \text{Hence }\sum_{n=1}^{2020} n i^n &= (2-2i) + (2-2i) + (2-2i) + \dots \quad \text{505 terms}\\ &= 505(2-2i) \\ &= 1010 - 1010i \\ &= 1010\sqrt{2} \text{ cis}\left(-\frac{\pi}{4}\right) \quad \text{or} \quad \frac{2020}{\sqrt{2}} \text{cis}\left(-\frac{\pi}{4}\right) \end{align*}

Marking Criteria

Descriptor	Marks
evaluates the sum of the first 4 terms correctly	1
generalises that the sum of the first 4 terms repeats 505 times	1
simplifies correctly in the form $r \text{ cis } \theta$	1

SCSA Mathematics Specialist Complex numbers

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