Year 12 Complex Numbers: Multiplication

Multiplication is where complex numbers start to feel different from ordinary algebra, because the symbol $i$ brings the relation $i^{2} = - 1$ into play. Even so, the core method is still familiar: expand carefully, simplify powers of $i$ , and rewrite the result in standard form. Students often find this easier once they realise that complex multiplication is really just bracket expansion plus one new rule.

Expand first, then replace every

i^{2}

with

- 1

, then combine real and imaginary parts.

Subtopic 1: Expanding two complex numbers

The safest method is to treat the multiplication exactly like a binomial product. Multiply each term in the first bracket by each term in the second. Once the full expansion is visible, simplify any $i^{2}$ term and then collect like parts. Keeping the steps separate prevents a lot of avoidable sign errors.

Worked example 1

Problem: Simplify

(2 + 3 i) (4 - i)

Expand: $2 (4) + 2 (- i) + 3 i (4) + 3 i (- i)$ .
This gives $8 - 2 i + 12 i - 3 i^{2}$ .
Use $i^{2} = - 1$ : $- 3 i^{2} = 3$ .
Combine real parts and imaginary parts: $11 + 10 i$ .

Answer:

(2 + 3 i) (4 - i) = 11 + 10 i

Subtopic 2: Squaring a complex number

Squaring is a very common special case. You can either expand directly or use the algebraic identity for a square. The important point is that the middle term stays imaginary, while the final $i^{2}$ contribution becomes real after simplification. This often produces a result where both the real and imaginary parts change, which is worth checking carefully.

Worked example 2

Problem: Simplify

{(1 - 2 i)}^{2}

Expand: $(1 - 2 i) (1 - 2 i)$ .
This gives $1 - 2 i - 2 i + 4 i^{2}$ .
Combine the imaginary terms: $1 - 4 i + 4 i^{2}$ .
Use $i^{2} = - 1$ : $4 i^{2} = - 4$ .
So the result is $- 3 - 4 i$ .

Answer:

{(1 - 2 i)}^{2} = - 3 - 4 i

Why multiplication changes both parts

In addition and subtraction, the real and imaginary parts mostly stay in their own lanes. In multiplication, they interact. A real part can contribute to the imaginary output, and two imaginary terms can combine to create a real term through $i^{2} = - 1$ . This is one reason the working should stay line by line. If you try to compress too many steps into one jump, it becomes much harder to see where each contribution went.

Worked example 3

Problem: Simplify

(3 + i) (3 - i)

Expand: $9 - 3 i + 3 i - i^{2}$ .
The imaginary terms cancel.
Use $i^{2} = - 1$ , so $- i^{2} = 1$ .
The result is $10$ .

Answer:

(3 + i) (3 - i) = 10

Why conjugate-style products matter

The previous example shows an important pattern: when a complex number is multiplied by its conjugate, the result is purely real. This becomes especially useful when dividing complex numbers, because it lets you clear imaginary parts from a denominator. Even if division is the later topic, multiplication is where that idea begins.

Common traps

Replacing $i^{2}$ with $1$ instead of $- 1$ .
Forgetting one of the four products when expanding.
Combining real and imaginary terms before simplifying the power of $i$ .
Leaving the result not fully simplified into $a + b i$ form.

Revision focus

In revision, separate the expansion step from the simplification step every time. First write all the products. Only then simplify $i^{2}$ and collect terms. This is slower at first, but it is much more reliable and usually becomes fast with practice anyway.

It also helps to look for conjugate patterns deliberately. Products like $(a + b i) (a - b i)$ are worth recognising instantly because they simplify cleanly and prepare you for the logic of rationalising denominators later.

Practice links

Basic operations on complex numbers