← IB Mathematics: Applications & Interpretation · HL · IB Maths AI revision

IB Diploma Mathematics: Applications & Interpretation HL · Number & Algebra

Mini-Lesson

Number & Algebra (AI HL)

7 missions. Guess, learn, then check. Your place is saved.

Mission 1How many cells laid end to end would cross the Milky Way?
Mission 220 seats in row 1, two more in each row: how many seats in 30 rows?
Mission 3Double the rice on each chessboard square. How much on all 64?
Mission 4£10,000 at 4% for 10 years: what does monthly compounding add?
Mission 5A magnitude 8 earthquake versus a magnitude 6: how much bigger?
Mission 6 $x^{2} = - 1$ x² = −1 has no real solution. Is inventing one just cheating?
Mission 7Why does multiplying by $i$ i rotate a point by 90°?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

How many cells laid end to end would cross the Milky Way?

Make a guess: How many cells laid end to end would cross the Milky Way?

Number and algebra · standard form and approximation

Standard form

A number is in standard form when it is written $a \times 1 0^{k}$ a × 10^k with $1 \leq a < 10$ 1 ≤ a < 10 and $k$ k an integer.
Large numbers have $k > 0$ k > 0: $305000 = 3.05 \times 1 0^{5}$ 305 000 = 3.05 × 10⁵.
Small numbers have $k < 0$ k < 0: $0.000 72 = 7.2 \times 1 0^{- 4}$ 0.000 72 = 7.2 × 10⁻⁴.
Calculator notation such as 5.2E30 earns no credit in IB exams. Write $5.2 \times 1 0^{30}$ 5.2 × 10³⁰.

Quick check

?Which is

0.000 304

written in standard form?

Number and algebra · standard form and approximation

Calculating in standard form

Multiply: multiply the $a$ a parts and add the indices. $(3 \times 1 0^{4}) \times (5 \times 1 0^{6}) = 15 \times 1 0^{10}$ (3 × 10⁴) × (5 × 10⁶) = 15 × 10¹⁰.
Divide: divide the $a$ a parts and subtract the indices: $\frac{1 0 ^{m}}{1 0 ^{n}} = 1 0^{m - n}$ (10^m)/10ⁿ = 10^(m−n).
Then adjust so $1 \leq a < 10$ 1 ≤ a < 10: $15 \times 1 0^{10} = 1.5 \times 1 0^{11}$ 15 × 10¹⁰ = 1.5 × 10¹¹.
Add or subtract: first rewrite both numbers with the same power of 10.
Give final answers exact or to 3 significant figures, the IB default.

Worked example Number and algebra · standard form and approximation

Dividing in standard form

Question

A hard drive holds

2.4 \times 1 0^{12}

bytes. One photo uses

3.2 \times 1 0^{6}

bytes. How many photos fit on the drive? Give your answer in standard form.

1
Write the division: $\frac{2.4 \times 1 0 ^{12}}{3.2 \times 1 0 ^{6}}$ (2.4 × 10¹²)/(3.2 × 10⁶)
2
Divide the numbers: $2.4 \div 3.2 = 0.75$ 2.4 ÷ 3.2 = 0.75
3
Subtract the indices: $1 0^{12 - 6} = 1 0^{6}$ 10¹²⁻⁶ = 10⁶, giving $0.75 \times 1 0^{6}$ 0.75 × 10⁶
4
Adjust so $1 \leq a < 10$ 1 ≤ a < 10: $0.75 \times 1 0^{6} = 7.5 \times 1 0^{5}$ 0.75 × 10⁶ = 7.5 × 10⁵

Answer

7.5 \times 1 0^{5}

photos (750,000)

⚠️ Watch out

0.75 \times 1 0^{6}

is the right value but not standard form. Making

a

ten times bigger makes the power one smaller.

Calculate

?Write

(2.5 \times 1 0^{- 4}) \times (8 \times 1 0^{9})

in standard form,

a \times 1 0^{k}

. Enter the value of

k

Calculate

?A strand of hair is

8.4 \times 1 0^{- 5}

m wide. A virus particle is

1.2 \times 1 0^{- 7}

m wide. How many virus particles, side by side, would span the width of the hair? Give an exact answer.

Number and algebra · standard form and approximation

Bounds and percentage error

A rounded value hides an interval. If $x = 4.1$ x = 4.1 to 1 dp, then $4.05 \leq x < 4.15$ 4.05 ≤ x < 4.15: the lower bound and upper bound.
Round to a number of decimal places or significant figures, choosing an accuracy that suits the given data.
Percentage error (IB formula booklet), with $v_{A}$ v_A the approximate value and $v_{E}$ v_E the exact value:

ε = \frac{v _{A} - v _{E}}{v _{E}} \times 100%

Estimate first by rounding to 1 significant figure, then check your answer is reasonable (a length can't be negative).

Worked example Number and algebra · standard form and approximation

Maximum percentage error in an area

Question

The radius of a circle is measured as 2.5 cm, to 1 decimal place. Find the maximum percentage error in the area calculated from this measurement.

1
Area from the measurement: $v_{A} = π \times 2. 5^{2} = 19.634 \dots cm^{2}$ v_A = π × 2.5² = 19.634… cm²
2
Bounds for the radius: $2.45 \leq r < 2.55$ 2.45 ≤ r < 2.55, so the true area lies between $π \times 2.4 5^{2} = 18.857 \dots$ π × 2.45² = 18.857… and $π \times 2.5 5^{2} = 20.428 \dots$ π × 2.55² = 20.428…
3
Percentage error if $v_{E} = 18.857 \dots$ v_E = 18.857…: $\frac{19.634 \dots - 18.857 \dots}{18.857 \dots} \times 100% = 4.12%$ |(19.634… − 18.857…)/(18.857…)| × 100% = 4.12%
4
Percentage error if $v_{E} = 20.428 \dots$ v_E = 20.428…: $\frac{19.634 \dots - 20.428 \dots}{20.428 \dots} \times 100% = 3.88%$ |(19.634… − 20.428…)/(20.428…)| × 100% = 3.88%

Answer

maximum percentage error

4.12%

(3 sf)

⚠️ Watch out

a 2% error in the radius (0.05 out of 2.5) becomes about a 4% error in the area, because the radius is squared.

Calculate

?Lena estimates

49.7 \times 20.3

by rounding each number to 1 significant figure. Find the percentage error in her estimate. Give your answer to 3 significant figures.

Mission check

?A cell is about

1 0^{- 5}

m wide and the Milky Way about

1 0^{21}

m across. How many cells, end to end, would cross it?

Mission 2

20 seats in row 1, two more in each row: how many seats in 30 rows?

Make a guess: 20 seats in row 1, two more in each row: how many seats in 30 rows?

Number and algebra · arithmetic sequences and series

Arithmetic sequences

An arithmetic sequence adds the same amount each time: the common difference $d = u_{n + 1} - u_{n}$ d = uₙ₊₁ − uₙ.
With first term $u_{1}$ u₁, the $n$ nth term is

u_{n} = u_{1} + (n - 1) d

It is $(n - 1) d$ (n−1)d, not $n d$ nd: from $u_{1}$ u₁ to $u_{n}$ uₙ there are $n - 1$ n − 1 steps.
Plotted against $n$ n, the terms lie on a straight line with gradient $d$ d.

Worked example Number and algebra · arithmetic sequences and series

Finding a term and a term number

Question

An arithmetic sequence begins 7, 11, 15, … (a) Find

u_{40}

. (b) Which term is equal to 203?

1
Identify the first term and common difference: $u_{1} = 7$ u₁ = 7, $d = 11 - 7 = 4$ d = 11 − 7 = 4
2
(a) Substitute $n = 40$ n = 40: $u_{40} = 7 + 39 \times 4 = 163$ u₄₀ = 7 + 39 × 4 = 163
3
(b) Set $u_{n} = 203$ uₙ = 203: $7 + (n - 1) \times 4 = 203$ 7 + (n−1) × 4 = 203
4
Solve: $(n - 1) \times 4 = 196$ (n−1) × 4 = 196, so $n - 1 = 49$ n − 1 = 49 and $n = 50$ n = 50

Answer

(a)

u_{40} = 163

(b) 203 is the 50th term

⚠️ Watch out

in (b) the answer must be a positive integer. A decimal value of

n

would mean the number is not a term of the sequence.

Quick check

?An arithmetic sequence has

u_{1} = 5

and

d = - 3

. Find

u_{20}

Number and algebra · arithmetic sequences and series

Arithmetic series and sigma notation

An arithmetic series is the sum of the terms. The sum of the first $n$ n terms is

S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d) = \frac{n}{2} (u_{1} + u_{n})

The second form pairs first and last terms: each pair has the same total.
Sigma notation: $r = 1 \sum n (3 r + 2)$ Σ_(r=1)ⁿ (3r + 2) means "substitute $r = 1, 2, \dots, n$ r = 1, 2, …, n and add".
A GDC can generate terms and sums, but you must still state $u_{1}$ u₁ and $d$ d.

Worked example Number and algebra · arithmetic sequences and series

Evaluating a sum in sigma notation

Question

Find

r = 1 \sum 25 (3 r + 2)

1
Write the first terms: $r = 1$ r = 1 gives 5, $r = 2$ r = 2 gives 8, so $u_{1} = 5$ u₁ = 5 and $d = 3$ d = 3
2
Find the last term, $r = 25$ r = 25: $u_{25} = 3 \times 25 + 2 = 77$ u₂₅ = 3 × 25 + 2 = 77
3
Use $S_{n} = \frac{n}{2} (u_{1} + u_{n})$ Sₙ = n/2(u₁ + uₙ) with $n = 25$ n = 25: $S_{25} = \frac{25}{2} (5 + 77)$ S₂₅ = 25/2(5 + 77)
4
Evaluate: $12.5 \times 82 = 1025$ 12.5 × 82 = 1025

Answer

1025

⚠️ Watch out

d

is the coefficient of

r

(here 3), but

u_{1}

is not the constant 2. Substitute

r = 1

to get

u_{1} = 5

Calculate

?Ana invests 2,000 euros at 3.5% simple interest per year. Simple interest adds 3.5% of the original amount each year. After how many complete years will her investment first be worth more than 3,000 euros?

Number and algebra · arithmetic sequences and series

When data are nearly arithmetic

Real data rarely go up by exactly the same amount. If the differences are roughly equal, an arithmetic model is reasonable.
Approximate the common difference, for example by the mean difference: $d \approx \frac{last value - first value}{number of steps}$ d ≈ (last value − first value)/(number of steps).
Use the model to predict, then comment: predictions far beyond the data are less reliable.

Calculate

?Rio records the distance of his longest run each week: 5.0, 5.9, 7.1, 7.9, 9.0 km for weeks 1 to 5. Use an arithmetic model with first term 5.0 km and the mean difference of the data to predict his longest run in week 12. Give your answer to 3 significant figures.

Calculate

?In an arithmetic sequence,

u_{3} = 17

and

u_{10} = 52

. Find the sum of the first 20 terms.

Mission check

?Row 1 has 20 seats and each row has 2 more than the one before. How many seats are in 30 rows?

Mission 3

Double the rice on each chessboard square. How much on all 64?

Make a guess: Double the rice on each chessboard square. How much on all 64?

Number and algebra · geometric sequences and series

Geometric sequences

A geometric sequence multiplies by the same number each time: the common ratio $r = \frac{u _{n + 1}}{u _{n}}$ r = (uₙ₊₁)/uₙ.
With first term $u_{1}$ u₁, the $n$ nth term is

u_{n} = u_{1} r^{n - 1}

Again $n - 1$ n − 1 steps: $u_{1}$ u₁ is multiplied by $r$ r a total of $n - 1$ n − 1 times.
$r > 1$ r > 1: growth. $0 < r < 1$ 0 < r < 1: decay. $r < 0$ r < 0: the signs alternate.

Worked example Number and algebra · geometric sequences and series

A decreasing geometric sequence

Question

A geometric sequence begins 80, 60, 45, … Find

u_{8}

. Give your answer to 3 significant figures.

1
Find the common ratio: $r = \frac{60}{80} = 0.75$ r = 60/80 = 0.75 (check: $\frac{45}{60} = 0.75$ 45/60 = 0.75)
2
Use $u_{n} = u_{1} r^{n - 1}$ uₙ = u₁ rⁿ⁻¹ with $n = 8$ n = 8: $u_{8} = 80 \times 0.7 5^{7}$ u₈ = 80 × 0.75⁷
3
Evaluate: $80 \times 0.133 \dots = 10.678 \dots$ 80 × 0.133… = 10.678…

Answer

u_{8} = 10.7

(3 sf)

⚠️ Watch out

r

is next term divided by previous term. Using

\frac{80}{60}

gives a ratio above 1 for a sequence that is shrinking.

Quick check

?A geometric sequence begins 2, −6, 18, … Find

u_{6}

Number and algebra · geometric sequences and series

Geometric series

The sum of the first $n$ n terms, for $r \neq = 1$ r ≠ 1:

S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}

Either form works; the first is tidier when $r > 1$ r > 1.
Sigma notation: $k = 1 \sum n u_{1} r^{k - 1}$ Σ_(k=1)ⁿ u₁ r^(k−1) adds the first $n$ n terms.
Applications: salaries rising by a fixed percentage, populations, the spread of a disease. With a GDC, still state $u_{1}$ u₁ and $r$ r.

Calculate

?Mai's starting salary is 32,000 dollars a year. It rises by 3% each year. Find the total she earns in her first 10 years. Give your answer to the nearest dollar.

dollars

Calculate

?A geometric sequence has

u_{1} = 5

and

r = 2

. Find the first term that is greater than 10,000. Enter its term number

n

Number and algebra · geometric sequences and series

Sum to infinity

If $∣ r ∣ < 1$ |r| < 1, the terms shrink towards 0 and $S_{n}$ Sₙ gets closer and closer to a limit. The series converges.
The sum to infinity is

S_{\infty} = \frac{u _{1}}{1 - r}, ∣ r ∣ < 1

If $∣ r ∣ \geq 1$ |r| ≥ 1 the series has no sum to infinity: it diverges.
Example: $1 + \frac{1}{2} + \frac{1}{4} + \dots = \frac{1}{1 - 0.5} = 2$ 1 + 1/2 + 1/4 + ⋯ = 1/(1 − 0.5) = 2.

Worked example Number and algebra · geometric sequences and series

A bouncing ball

Question

A ball is dropped from a height of 2 m. After each bounce it rises to 70% of the height it fell from. Find the total distance it travels before it comes to rest.

1
First drop: 2 m. After that, each bounce goes up and comes down the same height
2
Bounce heights: $1.4, 0.98, \dots$ 1.4, 0.98, …, geometric with $u_{1} = 1.4$ u₁ = 1.4 and $r = 0.7$ r = 0.7, and $∣ r ∣ < 1$ |r| < 1
3
Sum of the bounce heights: $S_{\infty} = \frac{1.4}{1 - 0.7} = 4.666 \dots$ S_(∞) = 1.4/(1 − 0.7) = 4.666… m
4
Total: $2 + 2 \times 4.666 \dots = 11.333 \dots$ 2 + 2 × 4.666… = 11.333… m

Answer

11.3

m (3 sf), exactly

\frac{34}{3}

⚠️ Watch out

the first drop is only down. Every later height is travelled twice, up and down.

Calculate

?A geometric series has first term 12 and sum to infinity 30. Find the common ratio

r

. Give an exact answer as a decimal.

Mission check

?One grain on the first square, then double on each square. How many grains on all 64 squares?

Mission 4

£10,000 at 4% for 10 years: what does monthly compounding add?

Make a guess: £10,000 at 4% for 10 years: what does monthly compounding add?

Number and algebra · financial mathematics

Compound interest

Simple interest adds the same amount every year (an arithmetic sequence). Compound interest earns interest on the interest (a geometric sequence).
With present value $P V$ PV, nominal annual rate $r %$ r%, compounded $k$ k times a year for $n$ n years, the future value is

F V = P V \times (1 + \frac{r}{100 k})^{kn}

Yearly: $k = 1$ k = 1. Half-yearly: $k = 2$ k = 2. Quarterly: $k = 4$ k = 4. Monthly: $k = 12$ k = 12.
A GDC finance solver gives the same answers.

Worked example Number and algebra · financial mathematics

Compounding quarterly

Question

5,000 euros is invested at 3.6% a year, compounded quarterly. Find its value after 6 years. Give your answer to 2 decimal places.

1
Identify the values: $P V = 5000$ PV = 5000, $r = 3.6$ r = 3.6, $k = 4$ k = 4, $n = 6$ n = 6
2
Rate per quarter: $\frac{3.6}{100 \times 4} = 0.009$ 3.6/(100 × 4) = 0.009, so each quarter multiplies by $1.009$ 1.009
3
Number of quarters: $kn = 4 \times 6 = 24$ kn = 4 × 6 = 24
4
Evaluate: $F V = 5000 \times 1.00 9^{24} = 6199.518 \dots$ FV = 5000 × 1.009²⁴ = 6199.518…

Answer

6,199.52 euros

⚠️ Watch out

both the rate and the power change.

5000 \times 1.03 6^{24}

uses the yearly rate every quarter.

Number and algebra · financial mathematics

Depreciation, inflation and time

Depreciation: a value falls by $r %$ r% each year, so $F V = P V \times (1 - \frac{r}{100})^{n}$ FV = PV × (1 − r/100)ⁿ.
Real value: inflation reduces what money can buy. IB markschemes use the real interest rate ≈ interest rate − inflation rate in the compound interest formula. This is an approximation: dividing by the inflation factor gives slightly less.
To find how long, solve for $n$ n with a table of values, the finance solver or logarithms, then round up to a whole period if the question asks when a value is first reached.

Calculate

?A car is bought for 24,000 dollars. Its value depreciates by 15% each year. Find its value after 4 years. Give your answer to 2 decimal places.

dollars

Calculate

?8,000 pounds is invested at 5% a year, compounded annually. After how many complete years will it first be worth more than 12,000 pounds?

Calculate

?Zoe invests 10,000 dollars for 8 years at 4.5% a year, compounded annually. Inflation is 2.5% a year. Using the rate (interest rate − inflation rate), find the real value of her investment after 8 years. Give your answer to the nearest dollar.

dollars

Number and algebra · financial mathematics

Loans and annuities with a finance solver

Amortization: a loan is repaid by equal payments, each covering the interest and some of the debt.
An annuity is a series of equal payments, such as monthly savings.
In a GDC finance solver (TVM), enter N (number of payments), I% (annual rate), PV, PMT, FV, P/Y and C/Y (payments and compoundings a year); solve for the unknown.
Money paid out is negative, money received positive. In IB exams, payments are at the end of each period.

Worked example Number and algebra · financial mathematics

Monthly loan repayments

Question

Jun borrows 15,000 dollars for 5 years at 6% a year, compounded monthly. He repays the loan with equal monthly payments. Use your GDC to find the monthly payment and the total interest paid.

1
Enter: N $= 60$ = 60, I% $= 6$ = 6, PV $= 15000$ = 15 000, FV $= 0$ = 0, P/Y $=$ = C/Y $= 12$ = 12
2
Solve for PMT: PMT $= - 289.99$ = −289.99 (negative: Jun pays it out)
3
Total repaid: $60 \times 289.992 \dots = 17 399.52$ 60 × 289.992… = 17 399.52
4
Interest: $17 399.52 - 15000 = 2399.52$ 17 399.52 − 15 000 = 2399.52

Answer

289.99 dollars a month; 2,399.52 dollars of interest

⚠️ Watch out

N counts payments, not years. Entering N

= 5

with P/Y

= 12

treats the loan as 5 months long.

Calculate

?Ella pays 200 pounds into a savings account at the end of every month for 10 years. The account pays 3% a year, compounded monthly. Use your GDC to find the value of the account after 10 years. Give your answer to the nearest pound.

pounds

Mission check

?£10,000 earns 4% a year for 10 years. How much more does monthly compounding give than yearly?

Mission 5

A magnitude 8 earthquake versus a magnitude 6: how much bigger?

Make a guess: A magnitude 8 earthquake versus a magnitude 6: how much bigger?

Number and algebra · exponents and logarithms

Laws of exponents

$a^{m} \times a^{n} = a^{m + n}$ a^m × aⁿ = a^(m+n), $a^{m} \div a^{n} = a^{m - n}$ a^m ÷ aⁿ = a^(m−n), $(a^{m})^{n} = a^{mn}$ (a^m)ⁿ = a^mn
$a^{0} = 1$ a⁰ = 1 and $a^{- n} = \frac{1}{a ^{n}}$ a⁻ⁿ = 1/aⁿ. So $5^{3} \times 5^{- 6} = 5^{- 3}$ 5³ × 5⁻⁶ = 5⁻³.
A power applies to every factor in its bracket: $(2 x)^{4} = 16 x^{4}$ (2x)⁴ = 16x⁴. But $2 x^{- 3} = \frac{2}{x ^{3}}$ 2x⁻³ = 2/x³: only $x$ x is raised to $- 3$ −3.
Rational exponents: $a^{\frac{1}{m}} = m a$ a^(1/m) = m√a and $a^{\frac{m}{n}} = (n a)^{m}$ a^(m/n) = (n√a)^m, so $1 6^{\frac{3}{4}} = 2^{3} = 8$ 16^(3/4) = 2³ = 8.

Quick check

?Simplify

\frac{( 3 x ^{2} ) ^{3}}{9 x ^{- 1}}

Number and algebra · exponents and logarithms

Logarithms

A logarithm is an index: $a^{x} = b ⟺ lo g_{a} b = x$ a^x = b ⇔ log_a b = x, for $a > 0$ a > 0, $a \neq = 1$ a ≠ 1 and $b > 0$ b > 0.
$lo g_{10}$ log₁₀ is written $lo g$ log; $lo g_{e}$ log_e is written $ln$ ln. Your GDC evaluates both.
$lo g_{a} a = 1$ log_a a = 1 and $lo g_{a} 1 = 0$ log_a 1 = 0. There is no logarithm of 0 or of a negative number.
On a base-10 log scale, each step of 1 is a factor of 10: $lo g 30 = 1.477 \dots$ log 30 = 1.477… and $lo g 300 = 2.477 \dots$ log 300 = 2.477…
The earthquake magnitude scale is such a logarithmic scale.

Worked example Number and algebra · exponents and logarithms

Solving with $lo g$ log and $ln$ ln

Question

Solve (a)

1 0^{x} = 250

(b)

e^{x} = 7

. Give your answers to 3 significant figures.

1
(a) Rewrite as a logarithm: $x = lo g_{10} 250 = lo g 250$ x = log₁₀ 250 = log 250
2
Evaluate with your GDC: $x = 2.397 \dots$ x = 2.397…
3
(b) Rewrite with base e: $x = lo g_{e} 7 = ln 7$ x = log_e 7 = ln 7
4
Evaluate: $x = 1.945 \dots$ x = 1.945…

Answer

(a)

x = 2.40

(b)

x = 1.95

⚠️ Watch out

1 0^{2.40}

is about 250, so the answer is sensible. Check by substituting back.

Calculate

?The pH of a solution is

pH = - lo g_{10} c

, where

c

is the hydrogen ion concentration in mol dm⁻³. Find the pH when

c = 3.2 \times 1 0^{- 4}

. Give your answer to 3 significant figures.

Calculate

?A bacteria colony is modelled by

N = 500 \times 2^{t}

, where

t

is the time in hours. Find the time when

N = 20000

. Use logarithms or your GDC. Give your answer to 3 significant figures.

hours

Number and algebra · exponents and logarithms

Laws of logarithms

For $a, x, y > 0$ a, x, y > 0:

lo g_{a} x y = lo g_{a} x + lo g_{a} y, lo g_{a} \frac{x}{y} = lo g_{a} x - lo g_{a} y, lo g_{a} x^{m} = m lo g_{a} x

Example: $lo g 24 = lo g 8 + lo g 3$ log 24 = log 8 + log 3, and $lo g 32 = 5 lo g 2$ log 32 = 5 log 2.
The power law brings an unknown index down: take $ln$ ln (or $lo g$ log) of both sides.
In AI exams the base will be 10 or e.

Worked example Number and algebra · exponents and logarithms

An exponential equation

Question

Solve

3^{2 x - 1} = 20

. Give your answer to 3 significant figures.

1
Take $ln$ ln of both sides: $ln (3^{2 x - 1}) = ln 20$ ln (3^(2x−1)) = ln 20
2
Use the power law: $(2 x - 1) ln 3 = ln 20$ (2x − 1)ln 3 = ln 20
3
Divide by $ln 3$ ln 3: $2 x - 1 = \frac{ln 20}{ln 3} = 2.726 \dots$ 2x − 1 = (ln 20)/(ln 3) = 2.726…
4
Solve: $x = \frac{2.726 \dots + 1}{2} = 1.863 \dots$ x = (2.726… + 1)/2 = 1.863…

Answer

x = 1.86

⚠️ Watch out

\frac{l n 20}{l n 3}

is not

ln \frac{20}{3}

. Dividing two logarithms is not the quotient law.

Calculate

?Given that

2 ln 6 - ln 4 = ln k

, find the value of

k

Mission check

?The magnitude scale is logarithmic (base 10). How many times larger is the measured amplitude of a magnitude 8 earthquake than a magnitude 6?

Mission 6

$x^{2} = - 1$ x² = −1 has no real solution. Is inventing one just cheating?

Make a guess: $x^{2} = - 1$ x² = −1 has no real solution. Is inventing one just cheating?

Number and algebra · complex numbers

The number $i$ i and Cartesian form

Define $i$ i by $i^{2} = - 1$ i² = −1. Then square roots of negatives exist: $- 16 = 16 i = 4 i$ √(−16) = √16 i = 4i.
A complex number in Cartesian form is $z = a + bi$ z = a + bi with $a, b \in R$ a, b ∈ ℝ.
$a$ a is the real part, $Re (z) = a$ Re(z) = a. $b$ b is the imaginary part, $Im (z) = b$ Im(z) = b: a real number, with no $i$ i.
Real numbers have $b = 0$ b = 0; purely imaginary numbers have $a = 0$ a = 0.
Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.

Quick check

?What is

Im (3 - 5 i)

Number and algebra · complex numbers

Arithmetic and the conjugate

Add or subtract: combine real parts and imaginary parts separately.
Multiply: expand the brackets as usual, then replace $i^{2}$ i² with $- 1$ −1.
The complex conjugate of $z = a + bi$ z = a + bi is $z^{*} = a - bi$ z^(*) = a − bi. Then $z z^{*} = a^{2} + b^{2}$ zz^(*) = a² + b², which is real.
Divide: multiply the top and bottom by the conjugate of the denominator, so the denominator becomes real.
Your GDC handles complex arithmetic and powers in Cartesian form, e.g. $(1 + i)^{8} = 16$ (1 + i)⁸ = 16.

Worked example Number and algebra · complex numbers

A product and a quotient

Question

Let

z = 2 - i

and

w = 7 + 4 i

. Find

\frac{w}{z}

in the form

a + bi

1
Multiply top and bottom by $z^{*} = 2 + i$ z^(*) = 2 + i: $\frac{( 7 + 4 i ) ( 2 + i )}{( 2 - i ) ( 2 + i )}$ ((7 + 4i)(2 + i))/((2 − i)(2 + i))
2
Expand the top: $14 + 7 i + 8 i + 4 i^{2} = 14 + 15 i - 4 = 10 + 15 i$ 14 + 7i + 8i + 4i² = 14 + 15i − 4 = 10 + 15i
3
Expand the bottom: $4 - i^{2} = 4 + 1 = 5$ 4 − i² = 4 + 1 = 5
4
Divide each part by 5: $\frac{10 + 15 i}{5} = 2 + 3 i$ (10 + 15i)/5 = 2 + 3i

Answer

\frac{w}{z} = 2 + 3 i

. Check:

(2 + 3 i) (2 - i) = 4 - 2 i + 6 i - 3 i^{2} = 7 + 4 i

⚠️ Watch out

the bottom is

4 - i^{2} = 5

, not

4 - 1 = 3

. Replace

i^{2}

with

- 1

every time it appears.

Calculate

?Find the real part of

(4 + 3 i) (2 - 5 i)

Number and algebra · complex numbers

The Argand diagram and modulus

On an Argand diagram (the complex plane), $z = a + bi$ z = a + bi is the point $(a, b)$ (a, b): real axis across, imaginary axis up.
The modulus $∣ z ∣ = a^{2} + b^{2}$ |z| = √(a² + b²) is the distance from the origin to the point. For example, $∣3 + 4 i ∣ = 5$ |3 + 4i| = 5.
The argument $ar g z$ arg z is the angle from the positive real axis to the point (used in the next mission).
$z^{*}$ z^(*) is the reflection of $z$ z in the real axis.
Adding complex numbers adds them like vectors, so $∣ z_{1} - z_{2} ∣$ |z₁ − z₂| is the distance between the points $z_{1}$ z₁ and $z_{2}$ z₂.

Calculate

?The points

P

and

Q

on an Argand diagram represent

z_{1} = 2 + 3 i

and

z_{2} = 5 - i

. Find the distance

PQ

Worked example Number and algebra · complex numbers

A quadratic with a negative discriminant

Question

Solve

z^{2} + 4 z + 13 = 0

1
Find the discriminant: $b^{2} - 4 a c = 16 - 52 = - 36$ b² − 4ac = 16 − 52 = −36, which is negative
2
Use the quadratic formula: $z = \frac{- 4 \pm - 36}{2}$ z = (−4 ± √(−36))/2
3
Write the root of the negative: $- 36 = 6 i$ √(−36) = 6i
4
Simplify: $z = \frac{- 4 \pm 6 i}{2} = - 2 \pm 3 i$ z = (−4 ± 6i)/2 = −2 ± 3i

Answer

z = - 2 + 3 i

z = - 2 - 3 i

, a pair of complex conjugates.

⚠️ Watch out

the graph of

y = x^{2} + 4 x + 13

does not meet the

x

-axis, but the equation still has two complex roots.

Calculate

?The equation

z^{2} + b z + c = 0

, where

b, c \in R

, has a root

1 + 2 i

. Find the value of

c

Mission check

x^{2} = - 1

has no real solution. Is inventing one just cheating?

Mission 7

Why does multiplying by $i$ i rotate a point by 90°?

Make a guess: Why does multiplying by $i$ i rotate a point by 90°?

Number and algebra · complex numbers

Modulus–argument form

Any $z \neq = 0$ z ≠ 0 is fixed by its modulus $r = ∣ z ∣$ r = |z| and argument $θ = ar g z$ θ = arg z, the angle from the positive real axis.
Polar form: $z = r (cos θ + i sin θ) = r cis θ$ z = r(cos θ + isin θ) = r cis θ. Euler form (also called exponential form): $z = r e^{i θ}$ z = re^iθ.
Converting: $a = r cos θ$ a = rcos θ, $b = r sin θ$ b = rsin θ, and $r = a^{2} + b^{2}$ r = √(a² + b²).
Find $θ$ θ from a sketch: work out the acute angle from $tan α = \frac{b}{a}$ tan α = |b/a|, then place it in the right quadrant. The principal argument is $- π < θ \leq π$ −π < θ ≤ π.

Worked example Number and algebra · complex numbers

Converting to polar and Euler form

Question

Write

z = - 1 + 3 i

in the form

r e^{i θ}

, where

r > 0

and

- π < θ \leq π

1
Modulus: $r = (- 1)^{2} + (3)^{2} = 4 = 2$ r = √((−1)² + (√3)²) = √4 = 2
2
Sketch: real part negative, imaginary part positive, so $z$ z is in the second quadrant
3
Acute angle with the real axis: $tan α = \frac{3}{1}$ tan α = (√3)/1, so $α = \frac{π}{3}$ α = π/3
4
Second quadrant: $θ = π - \frac{π}{3} = \frac{2 π}{3}$ θ = π − π/3 = 2π/3

Answer

z = 2 e^{\frac{2 π}{3} i} = 2 cis \frac{2 π}{3}

⚠️ Watch out

arctan (\frac{3}{- 1}) = - \frac{π}{3}

puts the point in the fourth quadrant. Always check the quadrant with a sketch.

Quick check

?What is the principal argument of

- 2 - 2 i

Number and algebra · complex numbers

Multiplying and dividing in Euler form

Multiply: $r_{1} e^{i θ_{1}} \times r_{2} e^{i θ_{2}} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$ r₁e^iθ₁ × r₂e^iθ₂ = r₁ r₂ e^(i(θ₁ + θ₂)): multiply the moduli, add the arguments.
Divide: $\frac{r _{1} e ^{i θ_{1}}}{r _{2} e ^{i θ_{2}}} = \frac{r _{1}}{r _{2}} e^{i (θ_{1} - θ_{2})}$ (r₁e^iθ₁)/(r₂e^iθ₂) = r₁/r₂ e^(i(θ₁ − θ₂)): divide the moduli, subtract the arguments.
Geometrically, multiplying by $w$ w stretches by $∣ w ∣$ |w| and rotates by $ar g w$ arg w about the origin.
Adjust the final argument by $\pm 2 π$ ± 2π to bring it into $- π < θ \leq π$ −π < θ ≤ π.

Calculate

z = 6 e^{\frac{5 π}{6} i}

and

w = 2 e^{\frac{π}{3} i}

. Find the principal argument of

z w

, in radians, to 3 significant figures.

Worked example Number and algebra · complex numbers

A power using polar form

Question

Find

(1 + i)^{6}

in the form

a + bi

1
Convert: $∣1 + i ∣ = 2$ |1 + i| = √2 and $ar g (1 + i) = \frac{π}{4}$ arg(1 + i) = π/4, so $1 + i = 2 e^{\frac{π}{4} i}$ 1 + i = √2 e^(π/4 i)
2
Raise the modulus to the power 6 and multiply the argument by 6 (De Moivre's theorem): $(2)^{6} e^{\frac{6 π}{4} i} = 8 e^{\frac{3 π}{2} i}$ (√2)⁶ e^(6π/4 i) = 8e^(3π/2 i)
3
Bring the argument into range: $\frac{3 π}{2} - 2 π = - \frac{π}{2}$ 3π/2 − 2π = −π/2
4
Convert back: $8 (cos (- \frac{π}{2}) + i sin (- \frac{π}{2})) = - 8 i$ 8(cos (−π/2) + isin (−π/2)) = −8i

Answer

(1 + i)^{6} = - 8 i

⚠️ Watch out

the modulus is raised to the power (

(2)^{6} = 8

), not multiplied by 6.

Calculate

?Let

z = 1 - 3 i

. Find the real part of

z^{5}

Number and algebra · complex numbers

Adding waves with complex numbers

$A cos (ω t + α)$ Acos (ω t + α) is the real part of $A e^{i (ω t + α)} = A e^{i α} \times e^{iω t}$ Ae^(i(ω t + α)) = Ae^iα × e^(iω t).
To add two waves with the same frequency $ω$ ω, add their complex amplitudes $A e^{i α} + B e^{i β}$ Ae^iα + Be^iβ, then write the sum as $R e^{i ϕ}$ Re^iφ.
The total is $R cos (ω t + ϕ)$ Rcos (ω t + φ).
Example: $3 cos 2 t + 4 cos (2 t + \frac{π}{2})$ 3cos 2t + 4cos (2t + π/2): $3 + 4 i = 5 e^{0.927 i}$ 3 + 4i = 5e^0.927i, so the total is $5 cos (2 t + 0.927)$ 5cos (2t + 0.927).

Calculate

?Two AC voltage sources in a circuit give

V_{1} = 5 cos (50 t)

and

V_{2} = 5 cos (50 t + \frac{π}{3})

. The total voltage is

V = A cos (50 t + B)

. Use your GDC to find

A

to 3 significant figures.

Mission check

?Why does multiplying by

i

rotate a point by 90°?

Recap

Big ideas

Standard form is $a \times 1 0^{k}$ a × 10^k with $1 \leq a < 10$ 1 ≤ a < 10 and $k$ k an integer; never write calculator notation like 5.2E30.
To multiply, add the indices; to divide, subtract them; then adjust so $1 \leq a < 10$ 1 ≤ a < 10.
Rounded values have bounds ( $4.1$ 4.1 to 1 dp means $4.05 \leq x < 4.15$ 4.05 ≤ x < 4.15), and percentage error is $\frac{v _{A} - v _{E}}{v _{E}} \times 100%$ |(v_A − v_E)/v_E| × 100%.
Arithmetic sequence: $u_{n} = u_{1} + (n - 1) d$ uₙ = u₁ + (n−1)d, where $d$ d is the common difference.
Arithmetic series: $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d) = \frac{n}{2} (u_{1} + u_{n})$ Sₙ = n/2(2u₁ + (n−1)d) = n/2(u₁ + uₙ); sigma notation lists the terms to add.
Simple interest grows arithmetically; for real data, approximate $d$ d and judge how far a prediction can be trusted.
Geometric sequence: $u_{n} = u_{1} r^{n - 1}$ uₙ = u₁ rⁿ⁻¹, where $r$ r is the common ratio (next term ÷ previous term).
Geometric series: $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ Sₙ = (u₁(rⁿ − 1))/(r − 1) = (u₁(1 − rⁿ))/(1 − r), $r \neq = 1$ r ≠ 1.
If $∣ r ∣ < 1$ |r| < 1 the series converges and $S_{\infty} = \frac{u _{1}}{1 - r}$ S_(∞) = u₁/(1 − r).
Compound interest: $F V = P V (1 + \frac{r}{100 k})^{kn}$ FV = PV(1 + r/100k)^kn, with $k$ k compounding periods a year; depreciation multiplies by $1 - \frac{r}{100}$ 1 − r/100 each year.
For real value, use the interest rate minus the inflation rate (the approximation IB uses); to find a time, solve for $n$ n and round up to a whole period.
Loans and annuities: use the finance solver (N, I%, PV, PMT, FV, P/Y, C/Y), with money paid out entered as negative.
Laws of exponents: add indices to multiply, subtract to divide, multiply for a power; $a^{- n} = \frac{1}{a ^{n}}$ a⁻ⁿ = 1/aⁿ.
$a^{x} = b ⟺ lo g_{a} b = x$ a^x = b ⇔ log_a b = x; $lo g$ log means base 10 and $ln$ ln means base e. On a base-10 log scale, +1 means ×10.
Laws of logarithms turn products into sums and powers into multipliers; take logs to solve $a^{f (x)} = b$ a^(f(x)) = b.
$z = a + bi$ z = a + bi with $i^{2} = - 1$ i² = −1: $Re (z) = a$ Re(z) = a, $Im (z) = b$ Im(z) = b, conjugate $z^{*} = a - bi$ z^(*) = a − bi, modulus $∣ z ∣ = a^{2} + b^{2}$ |z| = √(a² + b²).
Multiply by expanding and using $i^{2} = - 1$ i² = −1; divide by multiplying top and bottom by the conjugate of the denominator.
On the Argand diagram $z$ z is the point $(a, b)$ (a, b); a real quadratic with $b^{2} - 4 a c < 0$ b² − 4ac < 0 has a conjugate pair of complex roots.
$z = r (cos θ + i sin θ) = r cis θ = r e^{i θ}$ z = r(cos θ + isin θ) = r cis θ = re^iθ: find $r$ r by Pythagoras and $θ$ θ from a sketch, with $- π < θ \leq π$ −π < θ ≤ π.
Multiply moduli and add arguments; divide moduli and subtract arguments; for powers, $(r e^{i θ})^{n} = r^{n} e^{in θ}$ (re^iθ)ⁿ = rⁿe^inθ. Multiplying by $w$ w is a stretch by $∣ w ∣$ |w| and a rotation by $ar g w$ arg w.
Waves with the same frequency add by adding complex amplitudes $A e^{i α}$ Ae^iα.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Number & Algebra (AI HL) for IB Diploma Mathematics: Applications & Interpretation HL. 🎉

Mission 1How many cells laid end to end would cross the Milky Way?
Mission 220 seats in row 1, two more in each row: how many seats in 30 rows?
Mission 3Double the rice on each chessboard square. How much on all 64?
Mission 4£10,000 at 4% for 10 years: what does monthly compounding add?
Mission 5A magnitude 8 earthquake versus a magnitude 6: how much bigger?
Mission 6 $x^{2} = - 1$ x² = −1 has no real solution. Is inventing one just cheating?
Mission 7Why does multiplying by $i$ i rotate a point by 90°?

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Specification points and sources for this lesson are listed at the bottom of the page.

Sources (11)

The specification points and sources behind this lesson.

1 IB Mathematics: applications and interpretation guide (first assessment 2021), SL 1.1, SL 1.6
2 IB Mathematics: applications and interpretation guide (first assessment 2021), SL 1.2
3 IB Mathematics: applications and interpretation guide (first assessment 2021), SL 1.3; AHL 1.11 (HL)
4 IB Mathematics: applications and interpretation guide (first assessment 2021), SL 1.4, SL 1.7
5 IB Mathematics: applications and interpretation guide (first assessment 2021), SL 1.5; AHL 1.9, AHL 1.10 (HL)
6 IB Mathematics: applications and interpretation guide (first assessment 2021), AHL 1.12, 1.13
7 IB Mathematics: applications and interpretation guide (first assessment 2021), AHL 1.13
8 AAT Bioquest, "What are the size ranges of mammalian cells?" (most mammalian cells are 10–100 µm in diameter), https://aatbio.com/resources/faq-frequently-asked-questions/What-are-the-size-ranges-of-mammalian-cells
9 NASA Imagine the Universe, "The Milky Way Galaxy" (approximately 100,000 light-years in diameter), https://imagine.gsfc.nasa.gov/science/featured_science/milkyway
10 IAU definition of the light-year: distance light travels in a Julian year, 9,460,730,472,580.8 km (about $9.46 \times 1 0^{15}$ 9.46 × 10¹⁵ m), IAU "Measuring the Universe", https://iauarchive.eso.org/public/themes/measuring/
11 USGS, "The Severity of an Earthquake" (each whole number increase in magnitude represents a tenfold increase in measured amplitude; about 31 times more energy per step), https://pubs.usgs.gov/gip/earthq4/severitygip.html

Number & Algebra (AI HL)

How many cells laid end to end would cross the Milky Way?

Standard form

Calculating in standard form

Dividing in standard form

Bounds and percentage error

Maximum percentage error in an area

20 seats in row 1, two more in each row: how many seats in 30 rows?

Arithmetic sequences

Finding a term and a term number

Arithmetic series and sigma notation

Evaluating a sum in sigma notation

When data are nearly arithmetic

Double the rice on each chessboard square. How much on all 64?

Geometric sequences

A decreasing geometric sequence

Geometric series

Sum to infinity

A bouncing ball

£10,000 at 4% for 10 years: what does monthly compounding add?

Compound interest

Compounding quarterly

Depreciation, inflation and time

Loans and annuities with a finance solver

Monthly loan repayments

A magnitude 8 earthquake versus a magnitude 6: how much bigger?

Laws of exponents

Logarithms

Solving with loglog and lnln

Laws of logarithms

An exponential equation

x2=−1x² = −1 has no real solution. Is inventing one just cheating?

The number ii and Cartesian form

Arithmetic and the conjugate

A product and a quotient

The Argand diagram and modulus

A quadratic with a negative discriminant

Why does multiplying by ii rotate a point by 90°?

Modulus–argument form

Converting to polar and Euler form

Multiplying and dividing in Euler form

A power using polar form

Adding waves with complex numbers

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Solving with $lo g$ log and $ln$ ln

$x^{2} = - 1$ x² = −1 has no real solution. Is inventing one just cheating?

The number $i$ i and Cartesian form

Why does multiplying by $i$ i rotate a point by 90°?