Complex arithmetic in both forms | Year 12

01 / LEARN

The idea, step by step

Addition follows coordinates

Add or subtract real parts and imaginary parts separately. A polar expression is still a number, but its modulus and argument cannot usually be added separately. If two vectors point in different directions, their resultant length is not the sum of their lengths.

For multiplication in Cartesian form, expand the brackets and replace $i^{2}$ by $- 1$ . For division by nonzero $w$ , multiply the numerator and denominator by $\bar{w}$ ; the denominator becomes the positive real number $∣ w ∣^{2}$ .

Multiplication follows length and angle

Trigonometric addition identities give $(r cis α) (s cis β) = r s cis (α + β)$ . For division, divide the moduli and subtract the arguments. Reduce the final angle only if a principal value is requested.

Choose the form that makes the operation transparent. When adding polar numbers, convert to coordinates; when multiplying many factors, polar form often avoids a large expansion.

\frac{r cis α}{s cis β} = \frac{r}{s} cis (α - β), s > 0

A common trap

Adding moduli and arguments when adding complex numbers.

Instead: Convert to Cartesian coordinates for addition; polar multiplication rules do not apply to sums.

02 / WORKED EXAMPLES

See the thinking

Example 1

Divide with a conjugate

Simplify $\frac{3 + i}{1 - 2 i}$ .

Multiply top and bottom by $1 + 2 i$ .
The numerator is $(3 + i) (1 + 2 i) = 1 + 7 i$ ; the denominator is $1 + 4 = 5$ .
Divide each numerator component by 5.

\frac{1}{5} + \frac{7}{5} i

Example 2

Use exact polar multiplication

Calculate $(2 cis (π / 3)) (3 cis (- π / 6))$ .

Multiply moduli: $2 \times 3 = 6$ .
Add angles: $π / 3 - π / 6 = π / 6$ .
Convert with exact sine and cosine values.

6 cis (π / 6) = 3 \sqrt{3} + 3 i

Check your understanding

Simplify $\frac{4 cis (π / 2)}{2 cis (π / 6)}$ .

Show answer

$2 cis (π / 3)$ .

Divide moduli and subtract angles.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Calculate $(2 - 3 i) + (1 + 5 i)$ .
2 marks
Expand $(2 + i) (3 - 2 i)$ .
2 marks
Simplify $\frac{2 - i}{1 + i}$ .
2 marks
Calculate $(4 cis (2 π / 3)) / (2 cis (- π / 6))$ .
2 marks
Find $2 cis (π / 3) + 2 cis (- π / 3)$ .
2 marks
If $z w = 6 i$ and $z = 1 + i$ , find $w$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Calculate $(2 - 3 i) + (1 + 5 i)$ .
2 marks
Add real parts $2 + 1 = 3$ and imaginary parts $- 3 + 5 = 2$ .
$3 + 2 i$ .
Expand $(2 + i) (3 - 2 i)$ .
2 marks
$6 - 4 i + 3 i - 2 i^{2} = 6 - i + 2 = 8 - i$ .
$8 - i$ .
Simplify $\frac{2 - i}{1 + i}$ .
2 marks
Multiply by $(1 - i) / (1 - i)$ : numerator $1 - 3 i$ , denominator 2.
$\frac{1}{2} - \frac{3}{2} i$ .
Calculate $(4 cis (2 π / 3)) / (2 cis (- π / 6))$ .
2 marks
Divide moduli to get 2; subtract angles to get $5 π / 6$ .
$2 cis (5 π / 6) = - \sqrt{3} + i$ .
Find $2 cis (π / 3) + 2 cis (- π / 3)$ .
2 marks
The terms are $1 + \sqrt{3} i$ and $1 - \sqrt{3} i$ , so the imaginary parts cancel.
$2$ .
If $z w = 6 i$ and $z = 1 + i$ , find $w$ .
3 marks
$w = 6 i / (1 + i) = 6 i (1 - i) / 2 = 3 + 3 i$ .
$3 + 3 i$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗

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