MathematicsChapter 1: Complex Numbers

Multiplication of Complex Numbers

Use FOIL (or the distributive law) and replace i^2 by -1.

Learning objectives

Explain multiplication of complex numbers using the correct definitions, conditions, and notation.
Apply i^2 -> -1, ac - bd correctly in routine and exam-style questions.
Show each important step, simplify the result, and check that it satisfies the original conditions.

Definition

Use FOIL (or the distributive law) and replace i^2 by -1. (a+bi)(c+di) = (ac - bd) + (ad + bc)i. A number times its conjugate is a real square sum: (a+bi)(a-bi) = a^2 + b^2.

Formula

i = -1 + 5i

Where / symbols

MULTIPLICATION OF COMPLEX NUMBERS: Use FOIL (or the distributive law) and replace i^2 by -1. (a+bi)(c+di) = (ac - bd) + (ad + bc)i. A number times its conjugate is a real square sum: (a+bi)(a-bi) = a^2 + b^2.
TWO COMPLEX FACTORS: (2+3i)(1+i) = 2*1 + 2*i + 3i*1 + 3i*i = 2 + 2i + 3i + 3i^2 = 2 + 5i - 3 = -1 + 5i. Collect real leftover after i^2, then i terms.
SQUARES AND CONJUGATES: (1+i)^2 = 1 + 2i + i^2 = 2i. Product with conjugate: (3+2i)(3-2i) = 9 - (2i)^2 = 9 - 4(-1) = 13.

Overview

Expand, then tidy i^2:

Two Complex Factors

(2+3i)(1+i) = 21 + 2i + 3i1 + 3ii = 2 + 2i + 3i + 3i^2 = 2 + 5i - 3 = -1 + 5i. Collect real leftover after i^2, then i terms.

Examples (2i)(3i)=6i^2=-6 · (1+i)(1-i)=2 · i(4+2i)=-2+4i

Squares and Conjugates

(1+i)^2 = 1 + 2i + i^2 = 2i. Product with conjugate: (3+2i)(3-2i) = 9 - (2i)^2 = 9 - 4(-1) = 13.

Examples (1+i)^2 = 2i · (3+2i)(3-2i)=13 · Not 9+4i

WHAT — MULTIPLICATION OF COMPLEX NUMBERS

Use FOIL (or the distributive law) and replace i^2 by -1. (a+bi)(c+di) = (ac - bd) + (ad + bc)i. A number times its conjugate is a real square sum: (a+bi)(a-bi) = a^2 + b^2. The symbols and conditions are part of the mathematics, so include them when stating the rule.

WHY — Why this topic matters

Multiplication Of Complex Numbers matters because it extends the real number system and makes equations with negative square roots solvable. The same rules are used later in algebra, quadratics, and graphical interpretation.

HOW — How to solve and explain it

Start by identifying what is given and what must be found. Work through TWO COMPLEX FACTORS, then SQUARES AND CONJUGATES. Worked: (2 + 3i)(1 + i). (2 + 3i)(1 + i) = 2(1) + 2(i) + 3i(1) + 3i(i) = 2 + 2i + 3i + 3i^2 i^2 = -1 so 3i^2 = -3 = (2 - 3) + (2 + 3)i = -1 + 5i Finish by checking signs, restrictions, units, and whether the answer matches the question.

Worked example

(2 + 3i)(1 + i).
(2 + 3i)(1 + i) = 2(1) + 2(i) + 3i(1) + 3i(i)
= 2 + 2i + 3i + 3i²
i² = -1 so 3i² = -3
= (2 - 3) + (2 + 3)i
= -1 + 5i

In simple words

The real part is ac - bd, not ac + bd. The minus comes from i^2 = -1.

Key points

i^2 -> -1
ac - bd
FOIL then i^2
conjugate product is real

Exam preparation

Exam Focus

Quick revision

Revision Zone

Review the key facts, formulas and checks before you move on.