How many VCAA Specialist Mathematics questions cover Algebra, number and structure?

AusGrader has 197 VCAA Specialist Mathematics questions on Algebra, number and structure, all with instant AI grading and detailed marking feedback.

VCE Specialist Mathematics — Algebra, number and structure Questions & Answers

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 $-\pi < \theta \le \pi$ .

Reveal Answer

\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

Q6

2020

VCAA

Paper 2

1 mark

Q6

1 mark

For the complex polynomial $P(z)=z^3+az^2+bz+c$ with real coefficients $a$ , $b$ and $c$ , $P(-2)=0$ and $P(3i)=0$ .

The values of $a$ , $b$ and $c$ are respectively

A

$-2,\ 9,\ -18$

B

$3,\ 4,\ 12$

C

$2,\ 9,\ 18$

D

$-3,\ -4,\ 12$

E

$2,\ -9,\ -18$

Reveal Answer

A

$-2,\ 9,\ -18$

This corresponds to the polynomial $(z-2)(z^2+9)$ , which has a root of $z=2$ instead of $z=-2$ .

B

$3,\ 4,\ 12$

This corresponds to the polynomial $(z+3)(z^2+4)$ , which has roots of $-3$ and $\pm 2i$ rather than $-2$ and $\pm 3i$ .

C

$2,\ 9,\ 18$

Correct Answer

Since the coefficients are real, the complex roots must occur in conjugate pairs, meaning $-3i$ is also a root. Expanding $(z+2)(z-3i)(z+3i) = (z+2)(z^2+9)$ gives $z^3+2z^2+9z+18$ , so $a=2$ , $b=9$ , and $c=18$ .

D

$-3,\ -4,\ 12$

This corresponds to the polynomial $(z-3)(z^2-4)$ , which has real roots $3$ , $2$ , and $-2$ .

E

$2,\ -9,\ -18$

This corresponds to the polynomial $(z+2)(z^2-9)$ , which has real roots $-2$ , $3$ , and $-3$ instead of the required complex roots.

Q4

2023

VCAA

Paper 2

1 mark

Q4

1 mark

If $z = a + bi$ , where $a, b \in R$ and $a > b > 0$ , then $\text{Arg}(z + i\bar{z})$ is equal to

A

$-\frac{3\pi}{4}$

B

$-\frac{\pi}{4}$

C

$-\frac{3\pi}{4}$ or $\frac{\pi}{4}$

D

$\frac{\pi}{4}$

E

$\frac{5\pi}{4}$

Reveal Answer

A

$-\frac{3\pi}{4}$

This would be the argument if $a+b < 0$ , placing the complex number in the third quadrant, but we are given $a > b > 0$ .

B

$-\frac{\pi}{4}$

This would be the argument if the imaginary part was negative, but $z + i\bar{z} = (a+b) + (a+b)i$ has a positive imaginary part.

C

$-\frac{3\pi}{4}$ or $\frac{\pi}{4}$

Since $a > b > 0$ , the sum $a+b$ is strictly positive, meaning the argument is uniquely determined and not a set of possible values.

D

$\frac{\pi}{4}$

Correct Answer

Substituting $z = a+bi$ gives $z + i\bar{z} = (a+b) + (a+b)i$ . Since $a+b > 0$ , this lies in the first quadrant with equal real and imaginary parts, giving an argument of $\frac{\pi}{4}$ .

E

$\frac{5\pi}{4}$

This angle corresponds to the third quadrant, which would require both the real and imaginary parts to be negative, contradicting $a > b > 0$ .

Q3

2025

QCAA

Paper 1

1 mark

Q3

1 mark

The polynomial $P(z) = z^3 + pz^2 + qz + 1$ has complex conjugate roots when

A

$p = 1$ and $q = 1$

B

$p = 1$ and $q = i$

C

$p = i$ and $q = 1$

D

$p = i$ and $q = i$

Reveal Answer

A

$p = 1$ and $q = 1$

Correct Answer

For a polynomial with a real constant term to have complex conjugate roots, all of its coefficients must be real. Setting $p=1$ and $q=1$ satisfies this, yielding the roots $-1, i,$ and $-i$ .

B

$p = 1$ and $q = i$

The coefficient $q=i$ is non-real. If the polynomial had complex conjugate roots, the third root would have to be real to make the constant term $1$ , which would force all coefficients to be real.

C

$p = i$ and $q = 1$

The coefficient $p=i$ is non-real. The complex conjugate root theorem requires all coefficients to be real for non-real roots to occur in conjugate pairs.

D

$p = i$ and $q = i$

Both $p=i$ and $q=i$ are non-real coefficients. A polynomial with non-real coefficients and a real constant term cannot have complex conjugate roots.

Q6

2025

VCAA

Paper 2

1 mark

Q6

1 mark

Let $z \in C$ .

Given that $|z| = 1$ and $z \neq 1$ , $\text{Re}\left(\frac{1}{1 - z}\right)$ is

A

$-\frac{1}{2}$

B

$0$

C

$\frac{1}{2}$

D

$\frac{\sqrt{3}}{2}$

Reveal Answer

A

$-\frac{1}{2}$

Incorrect. This negative value might result from a sign error when multiplying by the complex conjugate or evaluating the denominator.

B

$0$

Incorrect. While the expression $\frac{1+z}{1-z}$ has a real part of $0$ for $|z|=1$ , the expression $\frac{1}{1-z}$ does not.

C

$\frac{1}{2}$

Correct Answer

Correct. Letting $z = \cos \theta + i \sin \theta$ , we get $\frac{1}{1-z} = \frac{(1-\cos \theta) + i \sin \theta}{(1-\cos \theta)^2 + \sin^2 \theta}$ . The real part simplifies to $\frac{1-\cos \theta}{2-2\cos \theta} = \frac{1}{2}$ .

D

$\frac{\sqrt{3}}{2}$

Incorrect. This value might be confused with the imaginary part for specific values of $z$ (like $z = e^{i\pi/3}$ ), but the real part is constantly $\frac{1}{2}$ for all valid $z$ .

Q5

2023

VCAA

Paper 1

4 marks

Q5a

2 marks

Find the three solutions of the equation $z^3 = -8i$ , where $z \in C$ . Give your answers in Cartesian form.

Reveal Answer

$z^3=-8\operatorname{cis}\left(\frac{\pi}{2}+2k\pi\right),\ k=0,1,2$

$\Rightarrow z=-2\operatorname{cis}\left(\frac{\pi}{6}+\frac{2k\pi}{3}\right)$
$=-\sqrt{3}-i,\ \sqrt{3}-i,\ 2i$

Marking Criteria

Descriptor	Marks
Finds all three correct solutions in Cartesian form	2
Provides some relevant progress, such as expressing $-8i$ in polar form or finding one correct solution	1
None of the above	0

Q5b

1 mark

Let $z_1$ and $z_2$ be two of the solutions of the equation $z^3 = -8i$ .

Show that the length of the straight line segment between the points represented by $z_1$ and $z_2$ is $2\sqrt{3}$ .

Reveal Answer

Consider the distance between any two points. For example, $(-\sqrt{3},-1)$ and $(\sqrt{3},-1)$ .

Distance is $\sqrt{(2\sqrt{3})^2+0^2}=2\sqrt{3}$ .

Marking Criteria

Descriptor	Marks
Shows the correct distance calculation between two roots to obtain $2\sqrt{3}$ .	1

Q5c

1 mark

Hence, or otherwise, determine the area of the triangle whose vertices are the points representing the solutions of the equation $z^3 = -8i$ .

Reveal Answer

$A=\frac{1}{2}\times 2\sqrt{3}\times 2\sqrt{3}\times \sin(60^\circ)=3\sqrt{3}$

Marking Criteria

Descriptor	Marks
Correctly determines the area of the triangle as $3\sqrt{3}$ .	1

Q6

2024

VCAA

Paper 2

1 mark

Q6

1 mark

The equation $\text{Arg}(z + 2 - i) = \frac{\pi}{3}$ defines a ray in the complex plane. Identifying complex numbers $x + yi$ with the points $(x, y)$ , this ray lies on the line with equation $y = mx + c$ .

The values of $m$ and $c$ are respectively

A

$\frac{1}{\sqrt{3}}$ and $1 + 2\sqrt{3}$

B

$\sqrt{3}$ and $1 - 2\sqrt{3}$

C

$\frac{1}{\sqrt{3}}$ and $2 + \sqrt{3}$

D

$\sqrt{3}$ and $1 + 2\sqrt{3}$

E

$\sqrt{3}$ and $2 + \sqrt{3}$

Reveal Answer

A

$\frac{1}{\sqrt{3}}$ and $1 + 2\sqrt{3}$

This option incorrectly uses a slope of $\frac{1}{\sqrt{3}}$ , which corresponds to an argument of $\frac{\pi}{6}$ rather than $\frac{\pi}{3}$ .

B

$\sqrt{3}$ and $1 - 2\sqrt{3}$

This option has the correct slope but the wrong y-intercept, likely resulting from incorrectly interpreting the translation as $z - 2 - i$ instead of $z + 2 - i$ .

C

$\frac{1}{\sqrt{3}}$ and $2 + \sqrt{3}$

This option uses the incorrect slope of $\frac{1}{\sqrt{3}}$ (from $\tan(\frac{\pi}{6})$ ) and an incorrect y-intercept.

D

$\sqrt{3}$ and $1 + 2\sqrt{3}$

Correct Answer

Substituting $z = x + yi$ gives $\text{Arg}((x + 2) + (y - 1)i) = \frac{\pi}{3}$ . This means $\frac{y - 1}{x + 2} = \tan(\frac{\pi}{3}) = \sqrt{3}$ , which simplifies to the line $y = \sqrt{3}x + 1 + 2\sqrt{3}$ .

E

$\sqrt{3}$ and $2 + \sqrt{3}$

While the slope is correct, the y-intercept is wrong. This likely results from an arithmetic error when expanding $\sqrt{3}(x + 2) + 1$ .

Q7

2024

SCSA

Paper 1

5 marks

Q7

Consider the quartic polynomial $R(z) = z^4 - 6z^3 + 17z^2 - 22z + 14$ and $P(z) = z^2 - 2z + 2$ where $R(z) = P(z)(z^2 + az + b)$ .

Q7a

2 marks

Show that $(z - 1 - i)$ is a factor of $P(z)$ .

Reveal Answer

$P(1+i) = (1+i)^2 - 2(1+i) + 2 = 2i - 2 - 2i + 2 = 0$
$\therefore P(1+i) = 0$
Hence $(z-1-i)$ is a factor of $P(z)$ .

Marking Criteria

Descriptor	Marks
substitutes $z = 1+i$ correctly into $P(z)$	1
expands correctly to show that $P(1+i) = 0$	1

Q7b

3 marks

Solve the equation $R(z) = 0$ .

Reveal Answer

$z = 1-i$ is also a root of $P(z)$ and $R(z)$ .

$R(z) = z^4 - 6z^3 + 17z^2 - 22z + 14 = (z^2 - 2z + 2)(z^2 + az + b)$
$\therefore 2b = 14 \quad \implies b = 7$
$-6z^3 = z^2(az) - 2z(z^2) \quad \implies a = -4$

Solve $R(z) = (z^2 - 2z + 2)(z^2 - 4z + 7) = 0:$

$(z^2 - 2z + 2) = 0 \implies \therefore z = 1 \pm i$
and

\begin{align*} (z^2 - 4z + 7) &= 0\\ (z - 2)^2 + 3 &= 0\\ \therefore z &= 2 \pm \sqrt{3}i \end{align*}

Marking Criteria

Descriptor	Marks
states that $z = 1+i$ and $z = 1-i$ are solutions	1
determines that $a = -4$ and $b = 7$ or $(z^2 - 4z + 7)$ is a factor of $R(z)$	1
solves the equation $z^2 + ax + b = 0$ correctly	1

Q8

2024

QCAA

Paper 1

1 mark

Q8

1 mark

Given $z = 2\text{cis}\left(\frac{\pi}{3}\right)$ , determine $z^3$ .

A

$-8$

B

$-6$

C

$6$

D

$8$

Reveal Answer

A

$-8$

Correct Answer

Using De Moivre's Theorem, $z^3 = 2^3\text{cis}\left(3 \cdot \frac{\pi}{3}\right) = 8\text{cis}(\pi)$ . Since $\text{cis}(\pi) = \cos(\pi) + i\sin(\pi) = -1$ , the result is $8(-1) = -8$ .

B

$-6$

This option incorrectly multiplies the modulus by the exponent ( $2 \times 3 = 6$ ) instead of raising the modulus to the power ( $2^3 = 8$ ).

C

$6$

This answer results from two errors: multiplying the modulus by the exponent ( $2 \times 3$ ) and incorrectly evaluating $\text{cis}(\pi)$ as $1$ .

D

$8$

This option correctly calculates the new modulus ( $2^3 = 8$ ) but incorrectly evaluates $\text{cis}(\pi)$ as $1$ instead of $-1$ .

Q3

2025

QCAA

Paper 2

1 mark

Q3

1 mark

If all roots of the equation $z^3 = 1$ are plotted on an Argand plane and then joined by straight lines, the shape formed is

A

a right-angled triangle.

B

an equilateral triangle.

C

an isosceles triangle.

D

a scalene triangle.

Reveal Answer

A

a right-angled triangle.

The roots are separated by an angle of $120^\circ$ from the origin, meaning the triangle's internal angles are all $60^\circ$ , not $90^\circ$ .

B

an equilateral triangle.

Correct Answer

The three cube roots of unity ( $1$ , $\omega$ , $\omega^2$ ) are equally spaced by $120^\circ$ on the unit circle, forming a regular polygon with 3 sides, which is an equilateral triangle.

C

an isosceles triangle.

While an equilateral triangle is a special type of isosceles triangle, "equilateral" is the most precise and accurate description of the shape formed since all three sides are equal.

D

a scalene triangle.

A scalene triangle has sides of all different lengths, but the distances between any two cube roots of unity are identical.

Q1

2024

SCSA

Paper 1

3 marks

Q1

3 marks

The complex number $z = r \operatorname{cis} \theta = 3 + bi$ , where $\tan \theta = \sqrt{2}$ . Determine the exact values for $r$ and $b$ .

Reveal Answer

Given $z = r \text{ cis } \theta$ where $\tan \theta = \sqrt{2} \quad \therefore \frac{b}{3} = \sqrt{2}$ i.e. $b = 3\sqrt{2}$

$r^2 = 3^2 + b^2$

$r^2 = 9 + \left(3\sqrt{2}\right)^2 = 9 + 9(2) \quad \therefore r^2 = 27$

i.e. $r = \sqrt{27} = 3\sqrt{3}$

Marking Criteria

Descriptor	Marks
states $b = 3\sqrt{2}$	1
states $r = \sqrt{27}$ or $3\sqrt{3}$	1
justifies the determination of both $b$ and $r$	1

Q6

2021

VCAA

Paper 2

1 mark

Q6

1 mark

If $z \in C$ , $z \neq 0$ and $z^2 \in R$ , then the possible values of $\text{arg}(z)$ are

A

$\frac{k\pi}{2}, k \in Z$

B

$k\pi, k \in Z$

C

$\frac{(2k+1)\pi}{2}, k \in Z$

D

$\frac{(4k+1)\pi}{2}, k \in Z$

E

$\frac{(4k-1)\pi}{2}, k \in Z$

Reveal Answer

A

$\frac{k\pi}{2}, k \in Z$

Correct Answer

Let $z = r e^{i\theta}$ . Then $z^2 = r^2 e^{i2\theta}$ . For $z^2$ to be real, its imaginary part must be zero, meaning $\sin(2\theta) = 0$ . This implies $2\theta = k\pi$ , so $\theta = \frac{k\pi}{2}$ for any integer $k$ .

B

$k\pi, k \in Z$

This only accounts for real values of $z$ (where $z^2 > 0$ ), but $z$ can also be purely imaginary (where $z^2 < 0$ ) and still have a real square.

C

$\frac{(2k+1)\pi}{2}, k \in Z$

This only accounts for purely imaginary values of $z$ (where $z^2 < 0$ ), missing the real values of $z$ (where $z^2 > 0$ ) which also have real squares.

D

$\frac{(4k+1)\pi}{2}, k \in Z$

This only represents positive purely imaginary numbers, which is an incomplete set of solutions since $z$ can be any real or purely imaginary number.

E

$\frac{(4k-1)\pi}{2}, k \in Z$

This only represents negative purely imaginary numbers, which is an incomplete set of solutions since $z$ can be any real or purely imaginary number.

Q18

2021

QCAA

Paper 2

6 marks

Q18

6 marks

Consider the polynomial $P(z) = z^3 + az^2 + bz + c$ , where $a, b, c \in R$ and $z \in C$ .

Two of the roots of $P(z)$ are also roots of $z^4 + z^3 + z^2 + z + 1$ . The remaining root of $P(z)$ is $z = 2$ .

Given $z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1)$ , determine a possible expression for $P(z)$ .

Leave your answer in expanded form.

Reveal Answer

The roots of $z^5 = 1$ are $z = \text{cis}\left(\frac{2k\pi}{5}\right)$ where $k \in Z$
Given $z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1)$ , the four roots of $z^4 + z^3 + z^2 + z + 1$ must be the four complex roots of the five 5th roots of unity, $z^5 = 1$ .

By the conjugate root theorem, the two remaining roots of $P(z)$ must be a conjugate pair of roots of $z^5 = 1$ .
One possible pair of roots is $\text{cis}\left(\frac{2\pi}{5}\right)$ and $\text{cis}\left(-\frac{2\pi}{5}\right)$

Determining a quadratic factor of $P(z)$
$\left(z - \text{cis}\left(\frac{2\pi}{5}\right)\right)\left(z - \text{cis}\left(-\frac{2\pi}{5}\right)\right)$
$= z^2 - \left(\text{cis}\left(\frac{2\pi}{5}\right) + \text{cis}\left(-\frac{2\pi}{5}\right)\right)z + \text{cis}\left(\frac{2\pi}{5}\right)\text{cis}\left(-\frac{2\pi}{5}\right)$
$= z^2 - \left(\cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right) + \cos\left(-\frac{2\pi}{5}\right) + i\sin\left(-\frac{2\pi}{5}\right)\right)z + \text{cis}(0)$
$= z^2 - 2\cos\left(\frac{2\pi}{5}\right)z + 1$

$P(z) = (z-2)\left(z^2 - 2\cos\left(\frac{2\pi}{5}\right)z + 1\right)$
$= z^3 - 2\left(\cos\left(\frac{2\pi}{5}\right) + 1\right)z^2 + \left(4\cos\left(\frac{2\pi}{5}\right) + 1\right)z - 2$

Marking Criteria

Descriptor	Marks
Correctly determines the roots of z^5 = 1	1
Correctly recognises one possible pair of roots	1
Determines a quadratic factor of P(z) in factorised form	1
Expresses determined quadratic factor of P(z) in expanded form	1
Uses the factor of z = 2 to express P(z) in factorised form	1
Determines P(z) in expanded form	1

Q18

2020

QCAA

Paper 1

6 marks

Q18

6 marks

Consider the function $P(z) = 2z^4 + az^3 + 6z^2 + az + b$ where $a, b \in Z^+$

One of the roots of $P(z)$ is $z = -i$

Determine the possible value/s for $a$ and $b$ such that all remaining roots of $P(z)$ have an imaginary component.

Reveal Answer

$P(z) = 2z^4 + az^3 + 6z^2 + az + b$ where $a, b \in Z^+$
Given $z = -i$ is a root of $P(z)$ , then $P(-i) = 0$
$\therefore 2(-i)^4 + a(-i)^3 + 6(-i)^2 + a(-i) + b = 0$
$2 + ai - 6 - ai + b = 0$
$-4 + b = 0$
$b = 4$
$\therefore P(z) = 2z^4 + az^3 + 6z^2 + az + 4$

Given that the coefficients of the polynomial are real, another root is $z = i$ , another factor of $P(z)$ is $(z - i)$ .
$(z - i)(z + i) = (z^2 + 1)$ is a factor of $P(z)$

By inspection,
$P(z) = (z^2 + 1)(2z^2 + az + 4)$

Given all roots of $P(z)$ have an imaginary component, $2z^2 + az + 4$ must have only complex roots.
For complex roots, $b^2 - 4ac < 0$
$a^2 - 4 \times 2 \times 4 < 0$
$a < \sqrt{32}$
So $a = 1, 2, 3, 4$ or $5$ and $b = 4$

Marking Criteria

Descriptor	Marks
correctly applies the factor theorem to determine $b$	1
correctly uses the conjugate root of the given root to identify another factor of $P(z)$	1
correctly identifies that $(z^2 + 1)$ is a factor of $P(z)$	1
determines the remaining quadratic factor in terms of $a$	1
applies the complex root requirement to the remaining quadratic factor	1
determines the possible values for $a$ given $a, b \in Z^+$	1

Q1

2024

QCAA

Paper 2

1 mark

Q1

1 mark

Given that $z = -2 + 3i$ is a root of $z^3 + az + b = 0$ , where $a, b \in R$ , another root is

A

$-2 - 3i$

B

$-2 + 3i$

C

$2 - 3i$

D

$2 + 3i$

Reveal Answer

A

$-2 - 3i$

Correct Answer

According to the Complex Conjugate Root Theorem, if a polynomial has real coefficients ( $a, b \in \mathbb{R}$ ), complex roots occur in conjugate pairs. Therefore, the conjugate of $-2 + 3i$ , which is $-2 - 3i$ , must also be a root.

B

$-2 + 3i$

This is the root already provided in the question. The question asks for "another" root, specifically the one implied by the properties of polynomials with real coefficients.

C

$2 - 3i$

This is incorrect because the real part of the conjugate should remain $-2$ . The complex conjugate of $x + yi$ is $x - yi$ , so only the sign of the imaginary part changes.

D

$2 + 3i$

This is incorrect because it changes the sign of the real part rather than the imaginary part. The correct conjugate of $-2 + 3i$ is $-2 - 3i$ .

VCAA Specialist Mathematics Algebra, number and structure

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