ACSEE (Form 6) Complex Numbers — practice questions

Explore 3 free Complex Numbers questions in ACSEE (Form 6) Advanced Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Complex Numbers questions and exercises with answers

Showing questions 1–3 of 3.

If $\alpha$ and $\beta$ are roots of $z^2+4z+8=0$, then $\dfrac{\alpha+\beta+4i}{\alpha\beta+8i}$ in simplest form is:Read the answer and worked solution →Using De Moivre's theorem, the complex roots of $z^3=1$ are:Read the answer and worked solution →Sort the facts about z² + 4z + 8 = 0Read the answer and worked solution →

Worked practice questions

If $\alpha$ and $\beta$ are roots of $z^2+4z+8=0$, then $\dfrac{\alpha+\beta+4i}{\alpha\beta+8i}$ in simplest form is:

$\frac{i}{2}$
$-\frac{i}{2}$
$2i$
$\frac{1}{2}$

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Correct answer: A. $\frac{i}{2}$

1. For $z^2+bz+c=0$ the sum of roots is $-b$ and the product is $c$. 2. So $\alpha+\beta=-4$ and $\alpha\beta=8$, with no need to solve the quadratic. 3. Numerator: $-4+4i=4(i-1)$. 4. Denominator: $8+8i=8(1+i)$. 5. The ratio is $\dfrac{4(i-1)}{8(1+i)}=\dfrac{i-1}{2(1+i)}$; multiplying top and bottom by $(1-i)$ gives $\dfrac{2i}{4}=\dfrac{i}{2}$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 4: Complex Numbers_

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Using De Moivre's theorem, the complex roots of $z^3=1$ are:

$1,\;-\frac12+i\frac{\sqrt3}{3},\;-\frac12-i\frac{\sqrt3}{3}$
$1,\;\frac12+i\frac{\sqrt3}{2},\;\frac12-i\frac{\sqrt3}{2}$
$1,\;-\frac12+i\frac{\sqrt3}{2},\;-\frac12-i\frac{\sqrt3}{2}$
$1,\;-\frac13+i\frac{\sqrt3}{2},\;-\frac13-i\frac{\sqrt3}{2}$

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Correct answer: A. $1,\;-\frac12+i\frac{\sqrt3}{3},\;-\frac12-i\frac{\sqrt3}{3}$

1. Write $1$ in polar form: $1=\cos 2\pi k+i\sin 2\pi k$ for integer $k$. 2. De Moivre's theorem gives $z_k=\cos\dfrac{2\pi k}{3}+i\sin\dfrac{2\pi k}{3}$. 3. For $k=0$: $z=1$. For $k=1$: $\cos 120^{\circ}+i\sin 120^{\circ}=-\dfrac12+i\dfrac{\sqrt3}{2}$. 4. For $k=2$: $\cos 240^{\circ}+i\sin 240^{\circ}=-\dfrac12-i\dfrac{\sqrt3}{2}$. 5. Solving $z^3=1$ as an ordinary cubic returns only the real root $1$ and misses the other two. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 4: Complex Numbers_

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Sort the facts about z² + 4z + 8 = 0

Read exercise and reference answers

Items

Group
Sum of roots
Item
α + β = −4
Group
Sum of roots
Item
equals −b
Group
Product of roots
Item
αβ = 8
Group
Product of roots
Item
equals c
Group
Cube roots of 1
Item
z = 1
Group
Cube roots of 1
Item
−½ + i√3/2

Groups

Item
Sum of roots
Item
Product of roots
Item
Cube roots of 1

Instructions

Drag each chip into the group it belongs to.

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