Conjugates and division | Year 11

01 / LEARN

Understand the idea.

Conjugates and a real product

The conjugate of $z = a + b i$ is $\bar{z} = a - b i$ . Only the imaginary coefficient changes sign. Geometrically the point reflects in the real axis, so the real coordinate stays fixed.

Conjugate multiplication removes the imaginary cross terms: $(a + b i) (a - b i) = a^{2} + b^{2}$ . This is a positive real number for $z \neq 0$ , and zero only when both components vanish. The identity turns a complex denominator into a real one.

Dividing without changing the number

To divide by $c + d i \neq 0$ , multiply the numerator and denominator by $c - d i$ . The quotient stays unchanged because the extra factor is $(c - d i) / (c - d i) = 1$ . Expand the numerator carefully and divide both resulting components by the real denominator.

The denominator becomes $c^{2} + d^{2} > 0$ , so it cannot vanish unless the original divisor was zero. Complex numbers do not permit division by zero. Check a calculated reciprocal by multiplying it by the original number and obtaining one.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Divide two non-real numbers

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

Worked solution

1
Multiply by the denominator conjugate.
$\frac{3 + 4 i}{1 - 2 i} = \frac{(3 + 4 i) (1 + 2 i)}{(1 - 2 i) (1 + 2 i)}$
Multiplying the whole fraction by one preserves its value.
2
Expand the numerator.
$\begin{aligned} (3 + 4 i) (1 + 2 i) & = 3 + 6 i + 4 i + 8 i^{2} \\ = - 5 + 10 i \end{aligned}$
The imaginary product $8 i^{2}$ contributes negative eight to the real part.
3
Calculate the real denominator.
$(1 - 2 i) (1 + 2 i) = 1^{2} + 2^{2} = 5$
A number times its conjugate is the sum of its squared components.
4
Divide each component by five.
$\frac{- 5 + 10 i}{5} = - 1 + 2 i$
The result is now in Cartesian form with a real denominator removed.

Answer

$- 1 + 2 i$ .

EXAMPLE 02

Find a reciprocal

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

Worked solution

1
Use the conjugate of the divisor.
$\frac{1}{2 + i} = \frac{2 - i}{(2 + i) (2 - i)}$
Multiplying numerator and denominator by $2 - i$ leaves the quotient unchanged.
2
Simplify the denominator and separate components.
$\frac{2 - i}{2^{2} + 1^{2}} = \frac{2}{5} - \frac{1}{5} i$
The conjugate product is five, which divides each numerator component.
3
Check the reciprocal condition.
$(2 + i) (\frac{2}{5} - \frac{1}{5} i) = \frac{(2 + i) (2 - i)}{5} = 1$
A reciprocal must multiply the original nonzero number to one.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

Find $1 / (- i)$ .

Show answer & reasoning

$i$ .

The number $i$ is the reciprocal because $(- i) i = - i^{2} = 1$ . Hence $1 / (- i) = i$ .

Keep practising

6 worksheet questions · 17 marks

Open worksheet 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 $\frac{3 - i}{3 - i}$ .
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 $\frac{1 - 2 i}{1 - 2 i}$ .
The denominator is $5$ , giving $\frac{5 (1 - 2 i)}{5}$ .
$1 - 2 i$ .
Solve $(2 - i) z = 7 + i$ .
3 marks
$z = \frac{7 + i}{2 - i}$ .
Multiply by $\frac{2 + i}{2 + i}$ : 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
$\frac{z}{\bar{z}} = \frac{{(a + b i)}^{2}}{a^{2} + b^{2}}$ .
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 ↗

In this lesson

Understand the idea.

Conjugates and a real product

Dividing without changing the number

See the method in action.

Divide two non-real numbers

Multiply by the denominator conjugate.

Expand the numerator.

Calculate the real denominator.

Divide each component by five.

Find a reciprocal

Use the conjugate of the divisor.

Simplify the denominator and separate components.

Check the reciprocal condition.

Try it yourself.

Check your understanding

Keep practising