Home › Free Resources › Maths Worksheets › IB Maths AA HL – Topic 1: Number and Algebra
Free Maths Worksheet

IB Maths AA HL – Topic 1: Number and Algebra

Complex numbers, proof by induction, systems of equations, partial fractions.

WhatsApp Mr. AhmadBook a trial lesson (50% off)Free worksheets
12 questions  ·  64 marks  ·  answers included  ·  IB Mathematics: Analysis and Approaches HL
Download the PDFAsk about a question

This worksheet covers:

  • Sequences and series
  • Binomial expansions
  • Complex numbers (Cartesian, polar and Euler form, De Moivre)
  • Proof by induction
  • Systems of linear equations
  • Partial fractions

Sections A and B are non-calculator (Paper 1 style); Section C allows a GDC. Allow about 90 minutes.

Questions

Section A – Skills

Non-calculator.
1

Let $z = 3 – 4\text{i}$ and $w = 1 – 2\text{i}$.(a) Find $|z|$.(b) Find $z^*w$ in the form $a + b\text{i}$.(c) Express $\dfrac{3 + 4\text{i}}{w}$ in the form $a + b\text{i}$.

Show answer
(a) $|z| = 5$(b) $z^* w = (3+4\text{i})(1-2\text{i}) = 11 – 2\text{i}$(c) $\dfrac{(3+4\text{i})(1+2\text{i})}{5} = -1 + 2\text{i}$
[6]
2

Write $-1 + \sqrt{3}\,\text{i}$ in the form $r\,\text{e}^{\text{i}\theta}$ where $r > 0$ and $-\pi < \theta \le \pi$.

Show answer
$2\,\text{e}^{\frac{2\pi}{3}\text{i}}$
[3]
3

Express $\dfrac{3x + 5}{(x+1)(x+3)}$ in partial fractions.

Show answer
$\dfrac{1}{x+1} + \dfrac{2}{x+3}$
[4]
4

Find the coefficient of $x^3$ in the expansion of $(2 – x)^7$.

Show answer
$\binom{7}{3}\, 2^4 (-1)^3 = -560$
[3]
5

Solve $\log_2 x + \log_2 (x – 2) = 3$.

Show answer
$x(x-2) = 8 \Rightarrow x^2 – 2x – 8 = 0 \Rightarrow x = 4$ (reject $x = -2$)
[4]
6

Solve the system of equations$x + y + z = 6$$2x – y + z = 3$$x + 2y – z = 2$.

Show answer
$x = 1,\ y = 2,\ z = 3$
[4]

Section B – Exam style

Non-calculator. Show full working.
7

Prove by mathematical induction that $\displaystyle\sum_{r=1}^{n} r\,2^{r} = (n-1)2^{n+1} + 2$ for all $n \in \mathbb{Z}^+$.

Show answer
Base case $n=1$: LHS $= 2$, RHS $= 0 + 2 = 2$. Assume true for $n = k$. Then $\sum_{r=1}^{k+1} r2^r = (k-1)2^{k+1} + 2 + (k+1)2^{k+1} = 2k\,2^{k+1} + 2 = k\,2^{k+2} + 2 = ((k+1)-1)2^{(k+1)+1} + 2$. Hence true for $n = k+1$; by induction true for all $n \in \mathbb{Z}^+$.
[7]
8

(a) Use De Moivre’s theorem to show that $(1 + \text{i})^{10} = 32\text{i}$.(b) Find the three cube roots of $8\text{i}$, giving your answers in the form $r\,\text{cis}\,\theta$ with $0 \le \theta < 2\pi$.

Show answer
(a) $1+\text{i} = \sqrt{2}\,\text{cis}\frac{\pi}{4}$, so $(1+\text{i})^{10} = 2^5\,\text{cis}\frac{10\pi}{4} = 32\,\text{cis}\frac{\pi}{2} = 32\text{i}$(b) $2\,\text{cis}\frac{\pi}{6},\ 2\,\text{cis}\frac{5\pi}{6},\ 2\,\text{cis}\frac{3\pi}{2}$ (i.e. $\sqrt{3}+\text{i},\ -\sqrt{3}+\text{i},\ -2\text{i}$)
[7]
9

Find the first four terms of the expansion of $(1 + 2x)^{-\frac{1}{2}}$ in ascending powers of $x$, and state the values of $x$ for which the expansion is valid.

Show answer
$1 – x + \frac{3}{2}x^2 – \frac{5}{2}x^3 + \dots$, valid for $|x| < \frac{1}{2}$
[5]
10

The equation $z^2 – 6z + 13 = 0$ has roots $\alpha$ and $\beta$ where $\text{Im}(\alpha) > 0$.(a) Find $\alpha$ and $\beta$.(b) Find $\alpha^2 + \beta^2$.(c) Find $\arg \alpha$ in the form $\arctan k$.

Show answer
(a) $\alpha = 3 + 2\text{i},\ \beta = 3 – 2\text{i}$(b) $(\alpha+\beta)^2 – 2\alpha\beta = 36 – 26 = 10$(c) $\arg\alpha = \arctan\frac{2}{3}$
[6]

Section C – Challenge

GDC allowed.
11

Consider the system$x + y + z = 3$$x + 2y + 3z = 6$$x + 3y + kz = 9$where $k \in \mathbb{R}$.(a) Show that the system has a unique solution if and only if $k eq 5$, and find that solution.(b) Describe the solution set when $k = 5$, giving a general solution in terms of a parameter $\lambda$.

Show answer
(a) Row-reduce: $y + 2z = 3$, $2y + (k-1)z = 6 \Rightarrow (k-5)z = 0$. Unique solution iff $k e 5$: $z = 0,\ y = 3,\ x = 0$.(b) $k = 5$: infinitely many solutions (a line): $z = \lambda,\ y = 3 – 2\lambda,\ x = \lambda$.
[8]
12

A geometric series has first term $3$ and sum to infinity $12$.(a) Find the common ratio $r$.(b) Find the sum of the first five terms, $S_5$.(c) Find the least value of $n$ for which $S_n > 11.9$.

Show answer
(a) $r = \frac{3}{4}$(b) $S_5 = \dfrac{3(1 – (3/4)^5)}{1/4} = 9.15$ (exact $\frac{2343}{256}$)(c) $12(1 – (0.75)^n) > 11.9 \Rightarrow (0.75)^n < \frac{1}{120} \Rightarrow n > 16.6$, so $n = 17$
[7]

Related worksheets

More worksheets in IB Mathematics: Analysis & Approaches:

IB Maths AA SL – Topic 1: Number and AlgebraIB Maths AA SL – Topic 2: FunctionsIB Maths AA SL – Topic 3: Geometry and TrigonometryIB Maths AA SL – Topic 4: Statistics and ProbabilityIB Maths AA SL – Topic 5: CalculusIB Maths AA HL – Topic 2: FunctionsIB Maths AA HL – Topic 3: Geometry and TrigonometryIB Maths AA HL – Topic 4: Statistics and Probability

All maths worksheets  ·  Full resource library  ·  Maths tuition with Mr. Ahmad

Get new maths worksheets by email

One short email when new worksheets are added. No spam, unsubscribe any time.



    Book a trial lesson (50% off)

    Try a full first lesson with Mr. Ahmad at 50% off before you commit to anything. It is a real lesson, not a sales call: bring a topic the student is stuck on and we will work through it together.

    • Online from anywhere in the world, or in person across Dubai
    • You get honest feedback on where the student stands and what to fix first
    • No contract — if it is not the right fit, that is fine

    Prefer to talk? Message on WhatsApp and you will hear back within 24 hours.



      Stuck on this topic?

      Book a trial lesson with Mr. Ahmad at 50% off and bring this worksheet with you — online worldwide or in person in Dubai.

      WhatsApp nowBook a trial lesson (50% off)
      Scroll to Top