How many QCAA Specialist Mathematics Alternative Sequence questions cover Complex numbers?

AusGrader has 7 QCAA Specialist Mathematics Alternative Sequence questions on Complex numbers, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics Alternative Sequence — Complex numbers Questions & Answers

Q1

2022

QCAA

Paper 1

1 mark

Q1

1 mark

Let $z = a + 3i$ and $w = -3 + bi$ , where $a, b \in R$ .

If $z = w$ , then

A

$a = -3, b = -3$

B

$a = -3, b = 3$

C

$a = 3, b = -3$

D

$a = 3, b = 3$

Reveal Answer

A

$a = -3, b = -3$

This is incorrect because equating the imaginary parts of the complex numbers gives $b = 3$ , not $b = -3$ .

B

$a = -3, b = 3$

Correct Answer

This is correct because two complex numbers are equal if and only if their real parts are equal ( $a = -3$ ) and their imaginary parts are equal ( $3 = b$ ).

C

$a = 3, b = -3$

This is incorrect because it reverses the signs. Equating the real parts gives $a = -3$ and equating the imaginary parts gives $b = 3$ .

D

$a = 3, b = 3$

This is incorrect because equating the real parts of the complex numbers gives $a = -3$ , not $a = 3$ .

Q4

2022

QCAA

Paper 1

1 mark

Q4

1 mark

Given $z = a + bi$ , an expression representing $\frac{(z - \bar{z})(z + \bar{z})}{4z\bar{z}}$ is

A

$\frac{ab}{a^2 - b^2}$

B

$\frac{abi}{a^2 - b^2}$

C

$\frac{ab}{a^2 + b^2}$

D

$\frac{abi}{a^2 + b^2}$

Reveal Answer

A

$\frac{ab}{a^2 - b^2}$

This option incorrectly evaluates $z\bar{z}$ as $a^2 - b^2$ and misses the imaginary unit $i$ from $z - \bar{z} = 2bi$ .

B

$\frac{abi}{a^2 - b^2}$

While it correctly includes $i$ in the numerator, it incorrectly evaluates the denominator $z\bar{z}$ as $a^2 - b^2$ instead of $a^2 + b^2$ .

C

$\frac{ab}{a^2 + b^2}$

This option correctly evaluates the denominator but misses the imaginary unit $i$ in the numerator, which arises from $z - \bar{z} = 2bi$ .

D

$\frac{abi}{a^2 + b^2}$

Correct Answer

Substituting $z - \bar{z} = 2bi$ , $z + \bar{z} = 2a$ , and $z\bar{z} = a^2 + b^2$ into the expression yields $\frac{(2bi)(2a)}{4(a^2 + b^2)}$ , which simplifies to $\frac{abi}{a^2 + b^2}$ .

Q6

2020

QCAA

Paper 1

1 mark

Q6

1 mark

Given $z = 2 - 2i$ and $w = -3 + i$ , calculate $z^2 - \bar{w}$

A

$3 - 9i$

B

$3 - 7i$

C

$11 - 9i$

D

$11 - 7i$

Reveal Answer

A

$3 - 9i$

Incorrect. This result comes from subtracting $w$ instead of its complex conjugate $\bar{w}$ , which leads to calculating $-8i - (-3 + i) = 3 - 9i$ .

B

$3 - 7i$

Correct Answer

Correct. First, find $z^2 = (2-2i)^2 = 4 - 8i - 4 = -8i$ . The complex conjugate of $w$ is $\bar{w} = -3 - i$ . Subtracting them yields $-8i - (-3 - i) = 3 - 7i$ .

C

$11 - 9i$

Incorrect. This error occurs from incorrectly calculating $z^2$ as $8 - 8i$ (forgetting that $i^2 = -1$ ) and subtracting $w$ instead of $\bar{w}$ .

D

$11 - 7i$

Incorrect. This mistake happens when $z^2$ is incorrectly calculated as $8 - 8i$ by adding the squared terms ( $4 + 4$ ) instead of subtracting them ( $4 - 4$ ).

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 $u^5$ .

Q8

2020

QCAA

Paper 1

1 mark

Q8

1 mark

Express $(1 - i)^3$ in the form $a + bi$ .

A

$-2 - 2i$

B

$-2 + 2i$

C

$2 - 2i$

D

$2 + 2i$

Reveal Answer

A

$-2 - 2i$

Correct Answer

Correct. First, squaring $(1 - i)$ gives $1 - 2i + i^2 = -2i$ . Multiplying this result by $(1 - i)$ yields $-2i(1 - i) = -2i + 2i^2 = -2 - 2i$ .

B

$-2 + 2i$

Incorrect. This is the result of expanding $(1 + i)^3$ , not $(1 - i)^3$ .

C

$2 - 2i$

Incorrect. This answer likely comes from a sign error when evaluating $2i^2$ , mistakenly treating it as $+2$ instead of $-2$ .

D

$2 + 2i$

Incorrect. This error occurs if you incorrectly evaluate $(1 - i)^2$ as $2i$ instead of $-2i$ before multiplying by the remaining $(1 - i)$ .

Q18

2022

QCAA

Paper 2

5 marks

Q18

5 marks

Consider the polynomials $P(z) = z^3 + (i - a)z^2 - 2biz + 3i$ and $Q(z) = z - 2i$ , where $a, b \in R$ .

Given $\frac{P(z)}{Q(z)}$ has a remainder of $a - bi$ , evaluate the reasonableness that $(z - (a - bi))$ is a factor of $P(z)$ .

Reveal Answer

$P(z) = z^3 + (i - a)z^2 - 2biz + 3i$
Using the remainder theorem
$P(2i) = a - bi$

By substitution
$P(2i) = (2i)^3 + (i - a)(2i)^2 - 2bi(2i) + 3i$
$= -8i - 4(i - a) + 4b + 3i$
$= -8i - 4i + 4a + 4b + 3i$
$= 4a + 4b - 9i$

Equating parts:
$a = 4a + 4b$
$-b = -9$

So $a = -12$ and $b = 9$
$\therefore P(z) = z^3 + (i + 12)z^2 - 18iz + 3i$

$P(a - bi) = P(-12 - 9i) \approx 1566 - 285i$

Since $P(-12 - 9i) \neq 0$ , it is not reasonable that $(z - (a - bi))$ is a factor of $P(z)$ .

Marking Criteria

Descriptor	Marks
correctly determines an expression for $P(2i)$ using the remainder theorem	1
correctly determines an expression for $P(2i)$ using substitution into $P(z)$	1
forms two simultaneous equations by equating parts	1
determines $P(a - bi)$ using the values for $a$ and $b$	1
evaluates the reasonableness of the statement using mathematical reasoning	1

QCAA Specialist Mathematics Alternative Sequence Complex numbers

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QCAA Specialist Mathematics Alternative Sequence Complex numbers

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