QCAA Specialist Mathematics Further complex numbers

15 sample questions with marking guides and sample answers

Q6
2025
VCAA
Paper 2
1 mark
Q6
1 mark

Let z∈Cz \in C.

Given that ∣z∣=1|z| = 1 and z≠1z \neq 1, Re(11−z)\text{Re}\left(\frac{1}{1 - z}\right) is

A

−12-\frac{1}{2}

B

00

C

12\frac{1}{2}

D

32\frac{\sqrt{3}}{2}

Reveal Answer
A

−12-\frac{1}{2}

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

B

00

Incorrect. While the expression 1+z1−z\frac{1+z}{1-z} has a real part of 00 for ∣z∣=1|z|=1, the expression 11−z\frac{1}{1-z} does not.

C

12\frac{1}{2}

Correct Answer

Correct. Letting z=cos⁡θ+isin⁡θz = \cos \theta + i \sin \theta, we get 11−z=(1−cos⁡θ)+isin⁡θ(1−cos⁡θ)2+sin⁡2θ\frac{1}{1-z} = \frac{(1-\cos \theta) + i \sin \theta}{(1-\cos \theta)^2 + \sin^2 \theta}. The real part simplifies to 1−cos⁡θ2−2cos⁡θ=12\frac{1-\cos \theta}{2-2\cos \theta} = \frac{1}{2}.

D

32\frac{\sqrt{3}}{2}

Incorrect. This value might be confused with the imaginary part for specific values of zz (like z=eiπ/3z = e^{i\pi/3}), but the real part is constantly 12\frac{1}{2} for all valid zz.

Q3
2025
QCAA
Paper 2
1 mark
Q3
1 mark

If all roots of the equation z3=1z^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∘120^\circ from the origin, meaning the triangle's internal angles are all 60∘60^\circ, not 90∘90^\circ.

B

an equilateral triangle.

Correct Answer

The three cube roots of unity (11, ω\omega, ω2\omega^2) are equally spaced by 120∘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.

Q6
2020
VCAA
Paper 2
1 mark
Q6
1 mark

For the complex polynomial P(z)=z3+az2+bz+cP(z)=z^3+az^2+bz+c with real coefficients aa, bb and cc, P(−2)=0P(-2)=0 and P(3i)=0P(3i)=0.

The values of aa, bb and cc are respectively

A

−2, 9, −18-2,\ 9,\ -18

B

3, 4, 123,\ 4,\ 12

C

2, 9, 182,\ 9,\ 18

D

−3, −4, 12-3,\ -4,\ 12

E

2, −9, −182,\ -9,\ -18

Reveal Answer
A

−2, 9, −18-2,\ 9,\ -18

This corresponds to the polynomial (z−2)(z2+9)(z-2)(z^2+9), which has a root of z=2z=2 instead of z=−2z=-2.

B

3, 4, 123,\ 4,\ 12

This corresponds to the polynomial (z+3)(z2+4)(z+3)(z^2+4), which has roots of −3-3 and ±2i\pm 2i rather than −2-2 and ±3i\pm 3i.

C

2, 9, 182,\ 9,\ 18

Correct Answer

Since the coefficients are real, the complex roots must occur in conjugate pairs, meaning −3i-3i is also a root. Expanding (z+2)(z−3i)(z+3i)=(z+2)(z2+9)(z+2)(z-3i)(z+3i) = (z+2)(z^2+9) gives z3+2z2+9z+18z^3+2z^2+9z+18, so a=2a=2, b=9b=9, and c=18c=18.

D

−3, −4, 12-3,\ -4,\ 12

This corresponds to the polynomial (z−3)(z2−4)(z-3)(z^2-4), which has real roots 33, 22, and −2-2.

E

2, −9, −182,\ -9,\ -18

This corresponds to the polynomial (z+2)(z2−9)(z+2)(z^2-9), which has real roots −2-2, 33, and −3-3 instead of the required complex roots.

Q8
2024
QCAA
Paper 1
1 mark
Q8
1 mark

Given z=2cis(π3)z = 2\text{cis}\left(\frac{\pi}{3}\right), determine z3z^3.

A

−8-8

B

−6-6

C

66

D

88

Reveal Answer
A

−8-8

Correct Answer

Using De Moivre's Theorem, z3=23cis(3⋅π3)=8cis(π)z^3 = 2^3\text{cis}\left(3 \cdot \frac{\pi}{3}\right) = 8\text{cis}(\pi). Since cis(π)=cos⁡(π)+isin⁡(π)=−1\text{cis}(\pi) = \cos(\pi) + i\sin(\pi) = -1, the result is 8(−1)=−88(-1) = -8.

B

−6-6

This option incorrectly multiplies the modulus by the exponent (2×3=62 \times 3 = 6) instead of raising the modulus to the power (23=82^3 = 8).

C

66

This answer results from two errors: multiplying the modulus by the exponent (2×32 \times 3) and incorrectly evaluating cis(π)\text{cis}(\pi) as 11.

D

88

This option correctly calculates the new modulus (23=82^3 = 8) but incorrectly evaluates cis(π)\text{cis}(\pi) as 11 instead of −1-1.

Q18
2021
QCAA
Paper 2
6 marks
Q18
6 marks

Consider the polynomial P(z)=z3+az2+bz+cP(z) = z^3 + az^2 + bz + c, where a,b,c∈Ra, b, c \in R and z∈Cz \in C.

Two of the roots of P(z)P(z) are also roots of z4+z3+z2+z+1z^4 + z^3 + z^2 + z + 1. The remaining root of P(z)P(z) is z=2z = 2.

Given z5−1=(z−1)(z4+z3+z2+z+1)z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1), determine a possible expression for P(z)P(z).

Leave your answer in expanded form.

Reveal Answer

The roots of z5=1z^5 = 1 are z=cis(2kπ5)z = \text{cis}\left(\frac{2k\pi}{5}\right) where k∈Zk \in Z
Given z5−1=(z−1)(z4+z3+z2+z+1)z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1), the four roots of z4+z3+z2+z+1z^4 + z^3 + z^2 + z + 1 must be the four complex roots of the five 5th roots of unity, z5=1z^5 = 1.

By the conjugate root theorem, the two remaining roots of P(z)P(z) must be a conjugate pair of roots of z5=1z^5 = 1.
One possible pair of roots is cis(2π5)\text{cis}\left(\frac{2\pi}{5}\right) and cis(−2π5)\text{cis}\left(-\frac{2\pi}{5}\right)

Determining a quadratic factor of P(z)P(z)
(z−cis(2π5))(z−cis(−2π5))\left(z - \text{cis}\left(\frac{2\pi}{5}\right)\right)\left(z - \text{cis}\left(-\frac{2\pi}{5}\right)\right)
=z2−(cis(2π5)+cis(−2π5))z+cis(2π5)cis(−2π5)= 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)
=z2−(cos⁡(2π5)+isin⁡(2π5)+cos⁡(−2π5)+isin⁡(−2π5))z+cis(0)= 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)
=z2−2cos⁡(2π5)z+1= z^2 - 2\cos\left(\frac{2\pi}{5}\right)z + 1

P(z)=(z−2)(z2−2cos⁡(2π5)z+1)P(z) = (z-2)\left(z^2 - 2\cos\left(\frac{2\pi}{5}\right)z + 1\right)
=z3−2(cos⁡(2π5)+1)z2+(4cos⁡(2π5)+1)z−2= 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
DescriptorMarks

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
Q3
2025
QCAA
Paper 1
1 mark
Q3
1 mark

The polynomial P(z)=z3+pz2+qz+1P(z) = z^3 + pz^2 + qz + 1 has complex conjugate roots when

A

p=1p = 1 and q=1q = 1

B

p=1p = 1 and q=iq = i

C

p=ip = i and q=1q = 1

D

p=ip = i and q=iq = i

Reveal Answer
A

p=1p = 1 and q=1q = 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=1p=1 and q=1q=1 satisfies this, yielding the roots −1,i,-1, i, and −i-i.

B

p=1p = 1 and q=iq = i

The coefficient q=iq=i is non-real. If the polynomial had complex conjugate roots, the third root would have to be real to make the constant term 11, which would force all coefficients to be real.

C

p=ip = i and q=1q = 1

The coefficient p=ip=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=ip = i and q=iq = i

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

Q6
2021
VCAA
Paper 2
1 mark
Q6
1 mark

If z∈Cz \in C, z≠0z \neq 0 and z2∈Rz^2 \in R, then the possible values of arg(z)\text{arg}(z) are

A

kπ2,k∈Z\frac{k\pi}{2}, k \in Z

B

kπ,k∈Zk\pi, k \in Z

C

(2k+1)π2,k∈Z\frac{(2k+1)\pi}{2}, k \in Z

D

(4k+1)π2,k∈Z\frac{(4k+1)\pi}{2}, k \in Z

E

(4k−1)π2,k∈Z\frac{(4k-1)\pi}{2}, k \in Z

Reveal Answer
A

kπ2,k∈Z\frac{k\pi}{2}, k \in Z

Correct Answer

Let z=reiθz = r e^{i\theta}. Then z2=r2ei2θz^2 = r^2 e^{i2\theta}. For z2z^2 to be real, its imaginary part must be zero, meaning sin⁡(2θ)=0\sin(2\theta) = 0. This implies 2θ=kπ2\theta = k\pi, so θ=kπ2\theta = \frac{k\pi}{2} for any integer kk.

B

kπ,k∈Zk\pi, k \in Z

This only accounts for real values of zz (where z2>0z^2 > 0), but zz can also be purely imaginary (where z2<0z^2 < 0) and still have a real square.

C

(2k+1)π2,k∈Z\frac{(2k+1)\pi}{2}, k \in Z

This only accounts for purely imaginary values of zz (where z2<0z^2 < 0), missing the real values of zz (where z2>0z^2 > 0) which also have real squares.

D

(4k+1)π2,k∈Z\frac{(4k+1)\pi}{2}, k \in Z

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

E

(4k−1)π2,k∈Z\frac{(4k-1)\pi}{2}, k \in Z

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

Q6
2023
SCSA
Paper 1
5 marks
Q6
5 marks

Solve the complex equation z4=2−23iz^4 = 2 - 2\sqrt{3}i giving solutions in the form rcisθr \text{cis} \theta where −π<θ≤π-\pi < \theta \le \pi.

Reveal Answer

∣z4∣=22+(23)2=4+12=4|z^4| = \sqrt{2^2+(2\sqrt{3})^2} = \sqrt{4+12} = 4
Arg(z4)=tan⁡−1(−232)=tan⁡−1(−3)=−π3Arg(z^4) = \tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}

∴Solve z4=4cis(−π3)\therefore \text{Solve } z^4 = 4\text{cis}\left(-\frac{\pi}{3}\right)

∴z=414cis(−π12+k(π2))k=0,1,2,3...(1)\therefore z = 4^{\frac{1}{4}}\text{cis}\left(\frac{-\pi}{12}+k\left(\frac{\pi}{2}\right)\right) \quad k = 0, 1, 2, 3 \quad ... (1)

Roots are:
z0=2cis(−π12)z_0 = \sqrt{2}\text{cis}\left(-\frac{\pi}{12}\right)
z1=2cis(5π12)z_1 = \sqrt{2}\text{cis}\left(\frac{5\pi}{12}\right)
z2=2cis(11π12)z_2 = \sqrt{2}\text{cis}\left(\frac{11\pi}{12}\right)
z3=2cis(−7π12)Note: z3=2cis(17π12) does not satisfy −π<θ≤π.z_3 = \sqrt{2}\text{cis}\left(-\frac{7\pi}{12}\right) \quad \text{Note: } z_3 = \sqrt{2}\text{cis}\left(\frac{17\pi}{12}\right) \text{ does not satisfy } -\pi < \theta \le \pi.

Marking Criteria
DescriptorMarks

states the value for ∣z4∣|z^4| correctly

1

states the value for Arg(z4)Arg(z^4) correctly

1

states the principal solution z0=2cis(−π12)z_0 = \sqrt{2}\text{cis}\left(-\frac{\pi}{12}\right)

1

indicates a separation of π2\frac{\pi}{2} between solution arguments

1

states all solutions correctly using the condition −π<θ≤π-\pi < \theta \le \pi

1
Q7
2024
SCSA
Paper 1
5 marks
Q7

Consider the quartic polynomial R(z)=z4−6z3+17z2−22z+14R(z) = z^4 - 6z^3 + 17z^2 - 22z + 14 and P(z)=z2−2z+2P(z) = z^2 - 2z + 2 where R(z)=P(z)(z2+az+b)R(z) = P(z)(z^2 + az + b).

Q7a
2 marks

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

Reveal Answer

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

Marking Criteria
DescriptorMarks

substitutes z=1+iz = 1+i correctly into P(z)P(z)

1

expands correctly to show that P(1+i)=0P(1+i) = 0

1
Q7b
3 marks

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

Reveal Answer

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

R(z)=z4−6z3+17z2−22z+14=(z2−2z+2)(z2+az+b)R(z) = z^4 - 6z^3 + 17z^2 - 22z + 14 = (z^2 - 2z + 2)(z^2 + az + b)
∴2b=14  ⟹  b=7\therefore 2b = 14 \quad \implies b = 7
−6z3=z2(az)−2z(z2)  ⟹  a=−4-6z^3 = z^2(az) - 2z(z^2) \quad \implies a = -4

Solve R(z)=(z2−2z+2)(z2−4z+7)=0:R(z) = (z^2 - 2z + 2)(z^2 - 4z + 7) = 0:

(z2−2z+2)=0  ⟹  ∴z=1±i(z^2 - 2z + 2) = 0 \implies \therefore z = 1 \pm i
and

(z2−4z+7)=0(z−2)2+3=0∴z=2±3i\begin{align*} (z^2 - 4z + 7) &= 0\\ (z - 2)^2 + 3 &= 0\\ \therefore z &= 2 \pm \sqrt{3}i \end{align*}
Marking Criteria
DescriptorMarks

states that z=1+iz = 1+i and z=1−iz = 1-i are solutions

1

determines that a=−4a = -4 and b=7b = 7 or (z2−4z+7)(z^2 - 4z + 7) is a factor of R(z)R(z)

1

solves the equation z2+ax+b=0z^2 + ax + b = 0 correctly

1
Q12
2022
QCAA
Paper 1
6 marks
Q12
6 marks

Given z1=a+biz_1 = a + bi, z2=c+diz_2 = c + di ∀a,b,c,d∈R\forall a, b, c, d \in R, and z2≠0z_2 \neq 0, prove the identity
∣z1z2∣=∣z1∣∣z2∣\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}

Reveal Answer

Prove ∣z1z2∣=∣z1∣∣z2∣\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}, given z1=a+bi,z2=c+di,z2≠0z_1 = a+bi, z_2 = c+di, z_2 \neq 0

LHS =∣a+bic+di∣= \left| \frac{a+bi}{c+di} \right|
=∣(a+bi)(c−di)(c+di)(c−di)∣= \left| \frac{(a+bi)(c-di)}{(c+di)(c-di)} \right|
=∣ac−adi+bci−bdi2c2+d2∣= \left| \frac{ac - adi + bci - bdi^2}{c^2 + d^2} \right|
=∣(ac+bd)+(bc−ad)ic2+d2∣= \left| \frac{(ac+bd) + (bc-ad)i}{c^2+d^2} \right|
=(ac+bd)2+(bc−ad)2(c2+d2)2= \sqrt{\frac{(ac+bd)^2 + (bc-ad)^2}{(c^2+d^2)^2}}
=(ac)2+2abcd+(bd)2+(bc)2−2abcd+(ad)2(c2+d2)2= \sqrt{\frac{(ac)^2 + 2abcd + (bd)^2 + (bc)^2 - 2abcd + (ad)^2}{(c^2+d^2)^2}}
=(ac)2+(bd)2+(bc)2+(ad)2(c2+d2)2= \sqrt{\frac{(ac)^2 + (bd)^2 + (bc)^2 + (ad)^2}{(c^2+d^2)^2}}
=a2(c2+d2)+b2(c2+d2)(c2+d2)2= \sqrt{\frac{a^2(c^2+d^2) + b^2(c^2+d^2)}{(c^2+d^2)^2}}
=(a2+b2)(c2+d2)(c2+d2)2= \sqrt{\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}}
=a2+b2c2+d2= \sqrt{\frac{a^2+b^2}{c^2+d^2}}
=∣z1∣∣z2∣= \frac{|z_1|}{|z_2|}
= RHS QED

Marking Criteria
DescriptorMarks

Correctly multiplies the numerator and denominator by the complex conjugate of z₂

1

Realises denominator (in simplest form) and expands numerator

1

Determines modulus of the expression

1

Simplifies numerator

1

Factorises numerator

1

Completes the proof using mathematical reasoning

1
Q18
2022
SCSA
Paper 2
7 marks
Q18a
3 marks

Show that for all positive integers nn and complex numbers zz where 0≤θ≤π20 \le \theta \le \frac{\pi}{2},

(zn−cis (θ))(zn+cis (−θ))=z2n−(2isin⁡θ)zn−1(z^n - \text{cis } (\theta))(z^n + \text{cis } (-\theta)) = z^{2n} - (2i \sin \theta) z^n - 1.

Reveal Answer

(zn−cis(θ))(zn+cis(−θ))(z^n - \text{cis}(\theta))(z^n + \text{cis}(-\theta))
=z2n−cis(θ)zn+cis(−θ)zn−cis(θ)cis(−θ)= z^{2n} - \text{cis}(\theta)z^n + \text{cis}(-\theta)z^n - \text{cis}(\theta)\text{cis}(-\theta)
=z2n−(cis(θ)−cis(−θ))zn−cis(0)= z^{2n} - (\text{cis}(\theta) - \text{cis}(-\theta))z^n - \text{cis}(0)
=z2n−(2isin⁡θ)zn−1= z^{2n} - (2i\sin\theta)z^n - 1

Marking Criteria
DescriptorMarks

expands the binomial products correctly to obtain 4 terms

1

uses the property cis(θ)cis(−θ)=cis(0)=1\text{cis}(\theta)\text{cis}(-\theta) = \text{cis}(0) = 1

1

uses the property cis(θ)−cis(−θ)=2isin⁡θ\text{cis}(\theta) - \text{cis}(-\theta) = 2i\sin\theta

1
Q18b
4 marks

Hence, using the result from part (a), obtain all the solutions to the equation z6−(i)z3−1=0z^6 - (i)z^3 - 1 = 0 in exact polar form.

Reveal Answer

To solve z6−(i)z3−1=0i.e. (z3)2−(2(12)i)z3−1=0\begin{align*} \text{To solve }z^6 - (i)z^3 - 1 &= 0\\ \text{i.e. }(z^3)^2 - \left(2\left(\frac{1}{2}\right)i\right)z^3 - 1 &= 0 \end{align*}

∴n=3 and sin⁡θ=12  ⟹   i.e. θ=π6 using 0≤θ≤π2 from part (a).\therefore n = 3 \text{ and } \sin\theta = \frac{1}{2} \implies \text{ i.e. } \theta = \frac{\pi}{6} \text{ using } 0 \le \theta \le \frac{\pi}{2} \text{ from part (a).}

Hence solve (z3−cis(π6))(z3+cis(−π6))=0∴z3=cis(π6)\begin{align*} \text{Hence solve }\left(z^3 - \text{cis}\left(\frac{\pi}{6}\right)\right)\left(z^3 + \text{cis}\left(-\frac{\pi}{6}\right)\right) &= 0\\ \therefore z^3 &= \text{cis}\left(\frac{\pi}{6}\right) \end{align*}

∴z=cis(π18),cis(13π18),cis(25π18) or cis(−11π18)\therefore z = \text{cis}\left(\frac{\pi}{18}\right), \text{cis}\left(\frac{13\pi}{18}\right), \text{cis}\left(\frac{25\pi}{18}\right) \text{ or } \text{cis}\left(-\frac{11\pi}{18}\right)

OR z3=−cis(−π6)=cis(π)cis(−π6)=cis(5π6)z^3 = -\text{cis}\left(-\frac{\pi}{6}\right) = \text{cis}(\pi)\text{cis}\left(-\frac{\pi}{6}\right) = \text{cis}\left(\frac{5\pi}{6}\right)

∴z=cis(5π18),cis(17π18),cis(29π18) or cis(−7π18)\therefore z = \text{cis}\left(\frac{5\pi}{18}\right), \text{cis}\left(\frac{17\pi}{18}\right), \text{cis}\left(\frac{29\pi}{18}\right) \text{ or } \text{cis}\left(-\frac{7\pi}{18}\right)

Marking Criteria
DescriptorMarks

determines the value of nn and θ\theta correctly

1

deduces that z3=cis(π6)z^3 = \text{cis}\left(\frac{\pi}{6}\right) and z3=cis(5π6)z^3 = \text{cis}\left(\frac{5\pi}{6}\right)

1

gives ALL solutions for z3=cis(π6)z^3 = \text{cis}\left(\frac{\pi}{6}\right) correctly

1

gives ALL solutions for z3=cis(5π6)z^3 = \text{cis}\left(\frac{5\pi}{6}\right) correctly

1
Q18
2020
QCAA
Paper 1
6 marks
Q18
6 marks

Consider the function P(z)=2z4+az3+6z2+az+bP(z) = 2z^4 + az^3 + 6z^2 + az + b where a,b∈Z+a, b \in Z^+

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

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

Reveal Answer

P(z)=2z4+az3+6z2+az+bP(z) = 2z^4 + az^3 + 6z^2 + az + b where a,b∈Z+a, b \in Z^+
Given z=−iz = -i is a root of P(z)P(z), then P(−i)=0P(-i) = 0
∴2(−i)4+a(−i)3+6(−i)2+a(−i)+b=0\therefore 2(-i)^4 + a(-i)^3 + 6(-i)^2 + a(-i) + b = 0
2+ai−6−ai+b=02 + ai - 6 - ai + b = 0
−4+b=0-4 + b = 0
b=4b = 4
∴P(z)=2z4+az3+6z2+az+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=iz = i, another factor of P(z)P(z) is (z−i)(z - i).
(z−i)(z+i)=(z2+1)(z - i)(z + i) = (z^2 + 1) is a factor of P(z)P(z)

By inspection,
P(z)=(z2+1)(2z2+az+4)P(z) = (z^2 + 1)(2z^2 + az + 4)

Given all roots of P(z)P(z) have an imaginary component, 2z2+az+42z^2 + az + 4 must have only complex roots.
For complex roots, b2−4ac<0b^2 - 4ac < 0
a2−4×2×4<0a^2 - 4 \times 2 \times 4 < 0
a<32a < \sqrt{32}
So a=1,2,3,4a = 1, 2, 3, 4 or 55 and b=4b = 4

Marking Criteria
DescriptorMarks

correctly applies the factor theorem to determine bb

1

correctly uses the conjugate root of the given root to identify another factor of P(z)P(z)

1

correctly identifies that (z2+1)(z^2 + 1) is a factor of P(z)P(z)

1

determines the remaining quadratic factor in terms of aa

1

applies the complex root requirement to the remaining quadratic factor

1

determines the possible values for aa given a,b∈Z+a, b \in Z^+

1
Q8
2020
VCAA
Paper 2
1 mark
Q8
1 mark

Given that (x+iy)14=a+ib(x+iy)^{14}=a+ib, where xx, yy, aa, b∈Rb\in\mathbb{R}, (y−ix)14(y-ix)^{14} for all values of xx and yy is equal to

A

−a−ib-a-ib

B

b−iab-ia

C

−b+ia-b+ia

D

−a+ib-a+ib

E

b+iab+ia

Reveal Answer
A

−a−ib-a-ib

Correct Answer

Factoring out −i-i gives y−ix=−i(x+iy)y-ix = -i(x+iy). Raising this to the 14th power yields (−i)14(x+iy)14=−1(a+ib)=−a−ib(-i)^{14}(x+iy)^{14} = -1(a+ib) = -a-ib.

B

b−iab-ia

This would be the result if (−i)14(-i)^{14} evaluated to −i-i, which only happens for powers like n=15n=15, not n=14n=14.

C

−b+ia-b+ia

This would be the result if (−i)14(-i)^{14} evaluated to ii, which occurs for powers like n=13n=13, but (−i)14=−1(-i)^{14} = -1.

D

−a+ib-a+ib

This incorrectly applies the −1-1 multiplier only to the real part aa, rather than distributing it to both terms to get −a−ib-a-ib.

E

b+iab+ia

This assumes the transformation results in multiplying by ii, but (−i)14=(i2)7=(−1)7=−1(-i)^{14} = (i^2)^7 = (-1)^7 = -1, not ii.

Q3
2020
VCAA
Paper 1
3 marks
Q3
3 marks

Find the cube roots of 12−12i\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i. Express your answers in polar form using principal values of the argument.

Reveal Answer

cis(−3π4),cis(−π12),cis(7π12)\text{cis}\left(-\frac{3\pi}{4}\right), \text{cis}\left(-\frac{\pi}{12}\right), \text{cis}\left(\frac{7\pi}{12}\right)

Marking Criteria
DescriptorMarks

Correctly expresses the complex number in polar form: cis(−π4)\text{cis}\left(-\frac{\pi}{4}\right)

1

Applies a valid method to find the cube roots, such as De Moivre's theorem or a geometric approach

1

Correctly states all three cube roots in polar form using principal values: cis(−3π4),cis(−π12),cis(7π12)\text{cis}\left(-\frac{3\pi}{4}\right), \text{cis}\left(-\frac{\pi}{12}\right), \text{cis}\left(\frac{7\pi}{12}\right)

1
Q1
2024
QCAA
Paper 2
1 mark
Q1
1 mark

Given that z=−2+3iz = -2 + 3i is a root of z3+az+b=0z^3 + az + b = 0, where a,b∈Ra, b \in R, another root is

A

−2−3i-2 - 3i

B

−2+3i-2 + 3i

C

2−3i2 - 3i

D

2+3i2 + 3i

Reveal Answer
A

−2−3i-2 - 3i

Correct Answer

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

B

−2+3i-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

2−3i2 - 3i

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

D

2+3i2 + 3i

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

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