Why the imaginary unit is needed | Year 11

01 / LEARN

The idea, step by step

Extending the number system

For a real number $x$ , the square $x^{2}$ is never negative. An equation such as $x^{2} = - 9$ therefore has no real solution. That does not make the equation meaningless: it tells us the real number system does not contain the required values.

We introduce a new number $i$ by the rule $i^{2} = - 1$ . Multiples of $i$ can square to negative real numbers. The real numbers remain part of the larger complex number system; their usual calculation rules still work.

Square roots and equation solutions

Because ${(3 i)}^{2} = 9 i^{2} = - 9$ , both $3 i$ and $- 3 i$ solve $x^{2} = - 9$ . We write the two solutions as $x = \pm 3 i$ .

The symbol $\sqrt{- 9}$ denotes the principal square root, $3 i$ , rather than both signs. Solve an equation by considering both roots, but do not attach an unexplained $\pm$ to every square-root expression. For a positive real $a$ , $\sqrt{- a} = i \sqrt{a}$ .

A common trap

Calling $i$ an ordinary variable whose value can be chosen.

Instead: The symbol $i$ is a fixed number defined by its square. Unknown complex values should use another symbol such as z.

02 / WORKED EXAMPLES

See the thinking

Example 1

Solve a previously impossible equation

Solve $x^{2} + 16 = 0$ over the complex numbers.

Rearrange to $x^{2} = - 16$ .
Use $i^{2} = - 1$ : ${(4 i)}^{2} = 16 i^{2} = - 16$ .
The negative value $- 4 i$ has the same square, so retain both signs.

x = 4 i

x = - 4 i

Example 2

Simplify a radical

Write $\sqrt{- 12}$ in exact form.

Use the principal-root rule $\sqrt{- 12} = i \sqrt{12}$ .
Factor $12 = 4 \times 3$ and take out the perfect square.
Thus $i \sqrt{12} = 2 i \sqrt{3}$ .

2 i \sqrt{3}

Check your understanding

What are the complex solutions of $x^{2} = - 49$ ?

Show answer

$x = \pm 7 i$ .

Each value squares to $49 i^{2} = - 49$ ; both signs must be included.

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

State why $x^{2} = - 5$ has no real solution.
2 marks
Simplify the principal square root $\sqrt{- 36}$ .
2 marks
Solve $z^{2} + 81 = 0$ .
2 marks
Solve $4 z^{2} + 64 = 0$ .
2 marks
Evaluate ${(3 i)}^{2}$ and explain why the answer is real.
2 marks
Simplify $\sqrt{- 20}$ exactly, then check its square.
3 marks

Read the worked solutions

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

State why $x^{2} = - 5$ has no real solution.
2 marks
For $x \geq 0$ or $x < 0$ , multiplying $x$ by itself gives a non-negative result.
The square of every real number is non-negative.
Simplify the principal square root $\sqrt{- 36}$ .
2 marks
$\sqrt{- 36} = i \sqrt{36} = 6 i$ .
$6 i$ .
Solve $z^{2} + 81 = 0$ .
2 marks
$z^{2} = - 81$ .
Both $9 i$ and $- 9 i$ square to $- 81$ .
$z = \pm 9 i$ .
Solve $4 z^{2} + 64 = 0$ .
2 marks
$4 z^{2} = - 64$ , hence $z^{2} = - 16$ .
Take both complex roots.
$z = \pm 4 i$ .
Evaluate ${(3 i)}^{2}$ and explain why the answer is real.
2 marks
${(3 i)}^{2} = 9 i^{2} = 9 (- 1) = - 9$ .
The factor $i^{2}$ is real.
$- 9$ .
Simplify $\sqrt{- 20}$ exactly, then check its square.
3 marks
$\sqrt{- 20} = i \sqrt{4 \times 5} = 2 i \sqrt{5}$ .
${(2 i \sqrt{5})}^{2} = 4 (- 1) 5 = - 20$ .
$2 i \sqrt{5}$ ; its square is $- 20$ .

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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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