Conjugates and division | Year 11

01 / LEARN

The idea, step by step

Conjugates and a real product

The conjugate of $z = a + b i$ is $\bar{z} = a - b i$ : it changes the sign of the imaginary component, not the real component. On an Argand diagram this is reflection in the real axis.

The product $z \bar{z} = (a + b i) (a - b i) = a^{2} + b^{2}$ is real and positive when $z \neq 0$ . This identity is the reason the conjugate is useful in division.

Dividing without changing the number

Multiply numerator and denominator by the conjugate of the denominator. This multiplies the fraction by one, so its value is unchanged. Expand the numerator and simplify the real denominator before separating components.

Division by zero is still undefined. If $c + d i \neq 0$ , then $c^{2} + d^{2} > 0$ , and $\frac{a + b i}{c + d i} = \frac{(a + b i) (c - d i)}{c^{2} + d^{2}}$ .

A common trap

Changing both signs when conjugating $- 3 + 2 i$ .

Instead: Its conjugate is $- 3 - 2 i$ . Only the imaginary component changes sign.

02 / WORKED EXAMPLES

See the thinking

Example 1

Divide two non-real numbers

Write $\frac{3 + 4 i}{1 - 2 i}$ in Cartesian form.

Multiply top and bottom by $1 + 2 i$ .
The numerator is $3 + 6 i + 4 i + 8 i^{2} = - 5 + 10 i$ .
The denominator is $1^{2} + 2^{2} = 5$ .
Divide each numerator component by $5$ .

- 1 + 2 i

Example 2

Find a reciprocal

Find $\frac{1}{2 + i}$ exactly.

Multiply numerator and denominator by $2 - i$ .
The denominator is $(2 + i) (2 - i) = 5$ .
Thus the reciprocal is $(2 - i) / 5$ .
Check by multiplying it by $2 + i$ to obtain $1$ .

\frac{2}{5} - \frac{1}{5} i

Check your understanding

Find $1 / (- i)$ .

Show answer

$i$ .

Because $(- i) i = - i^{2} = 1$ , the reciprocal is $i$ .

03 / WORKSHEET

Your turn

6 questions · 17 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find the conjugate of $- 5 - 2 i$ .
2 marks
Find $z \bar{z}$ for $z = 4 - 3 i$ .
2 marks
Simplify $\frac{2 - i}{3 + i}$ .
3 marks
Simplify $\frac{5}{1 + 2 i}$ .
3 marks
Solve $(2 - i) z = 7 + i$ .
3 marks
Show that $\frac{z}{\bar{z}}$ has real value only when $a b = 0$ , where $z = a + b i \neq 0$ and $a, b$ are real.
4 marks

Read the worked solutions

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

Find the conjugate of $- 5 - 2 i$ .
2 marks
Keep the real part and reverse the imaginary component.
$- 5 + 2 i$ .
Find $z \bar{z}$ for $z = 4 - 3 i$ .
2 marks
$z \bar{z} = 4^{2} + {(- 3)}^{2} = 25$ .
$25$ .
Simplify $\frac{2 - i}{3 + i}$ .
3 marks
Multiply by $(3 - i) / (3 - i)$ .
The numerator is $6 - 2 i - 3 i + i^{2} = 5 - 5 i$ and denominator is $10$ .
$\frac{1}{2} - \frac{1}{2} i$ .
Simplify $\frac{5}{1 + 2 i}$ .
3 marks
Multiply by $(1 - 2 i) / (1 - 2 i)$ .
The denominator is $5$ , giving $5 (1 - 2 i) / 5$ .
$1 - 2 i$ .
Solve $(2 - i) z = 7 + i$ .
3 marks
$z = (7 + i) / (2 - i)$ .
Multiply by $(2 + i) / (2 + i)$ : the numerator is $13 + 9 i$ and denominator $5$ .
$z = \frac{13}{5} + \frac{9}{5} i$ .
Show that $\frac{z}{\bar{z}}$ has real value only when $a b = 0$ , where $z = a + b i \neq 0$ and $a, b$ are real.
4 marks
$z / \bar{z} = {(a + b i)}^{2} / (a^{2} + b^{2})$ .
The numerator is $a^{2} - b^{2} + 2 a b i$ .
The denominator is positive, so its imaginary part is zero exactly when $2 a b = 0$ .
The imaginary part is $\frac{2 a b}{a^{2} + b^{2}}$ , so it vanishes exactly when $a b = 0$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, real and complex numbers: Cartesian arithmetic, conjugates, quadratic roots and Argand geometry. Polar form and roots of unity are developed in Year 12.

Official curriculum source ↗

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