Multiplication of Complex Numbers

Learn how to multiply complex numbers using the FOIL method and algebraic properties

Introduction to Multiplication

As stated in the introduction to complex numbers, the principle of permanence ensures that all calculation rules valid for real numbers also apply to complex numbers.

When multiplying complex numbers, we follow the same algebraic rules as with binomials. The key difference is handling the term \(i^2\), which equals \(-1\) by definition.

Key Principle:

Multiply complex numbers as if they were algebraic expressions with variable \(i\), then replace \(i^2\) with \(-1\) and simplify.

Multiplication Formula

To multiply two complex numbers, we use the distributive property (FOIL method) to expand the product, then simplify using \(i^2 = -1\).

General Multiplication Formula:

For complex numbers \(z_1 = a + bi\) and \(z_2 = c + di\):

Expansion using FOIL:
\(\displaystyle (a + bi)(c + di) = ac + adi + bci + bdi^2\)
Simplification (using \(i^2 = -1\)):
\(\displaystyle = ac + adi + bci - bd\)

\(\displaystyle = (ac - bd) + (ad + bc)i\)

The result is another complex number with:

  • Real part: \(ac - bd\)
  • Imaginary part: \(ad + bc\)

The FOIL Method for Complex Numbers

The FOIL method (First, Outer, Inner, Last) is a systematic way to multiply two binomials, which applies perfectly to complex numbers.

FOIL Steps Explained

First
Multiply the first terms of each binomial:
\(\displaystyle a \cdot c\)
Outer
Multiply the outer terms:
\(\displaystyle a \cdot di\)
Inner
Multiply the inner terms:
\(\displaystyle bi \cdot c\)
Last
Multiply the last terms:
\(\displaystyle bi \cdot di = bdi^2\)

Step-by-Step Example

Example 1: Basic Multiplication

Multiply \((3 + i)(1 - 2i)\)
1Apply FOIL - First
Multiply the first terms: \(3 \cdot 1 = 3\)
2Apply FOIL - Outer
Multiply the outer terms: \(3 \cdot (-2i) = -6i\)
3Apply FOIL - Inner
Multiply the inner terms: \(i \cdot 1 = i\)
4Apply FOIL - Last
Multiply the last terms: \(i \cdot (-2i) = -2i^2 = -2(-1) = 2\)

Combine all four parts:

\(\displaystyle 3 + (-6i) + i + 2 = 3 + 2 + (-6 + 1)i\)

Simplify:

\(\displaystyle = 5 - 5i\)

Result: \((3 + i)(1 - 2i) = 5 - 5i\)

Example 2: Complex Multiplication

Multiply \((2 + 3i)(4 - i)\)

Using FOIL:

  • First: \(2 \cdot 4 = 8\)
  • Outer: \(2 \cdot (-i) = -2i\)
  • Inner: \(3i \cdot 4 = 12i\)
  • Last: \(3i \cdot (-i) = -3i^2 = 3\)

Combine:

\(\displaystyle 8 + (-2i) + 12i + 3 = (8 + 3) + (-2 + 12)i = 11 + 10i\)

Result: \((2 + 3i)(4 - i) = 11 + 10i\)

Example 3: Purely Imaginary Multiplication

Multiply \(2i \cdot 3i\)
\(\displaystyle 2i \cdot 3i = 6i^2 = 6(-1) = -6\)

Result: The product is real! \(2i \cdot 3i = -6\)

Special Cases in Multiplication

Multiplying by Pure Imaginary Numbers

Example: \((5 + 2i) \cdot i\)
\(\displaystyle (5 + 2i) \cdot i = 5i + 2i^2 = 5i - 2 = -2 + 5i\)

Multiplying a Complex Number by its Conjugate

When a complex number is multiplied by its conjugate, the result is always a real number.

Conjugate:

The conjugate of \(z = a + bi\) is \(\overline{z} = a - bi\)

Multiply \((3 + 4i)(3 - 4i)\)

Using the formula: \((a+bi)(a-bi) = a^2 + b^2\)

\(\displaystyle (3 + 4i)(3 - 4i) = 9 - 12i + 12i - 16i^2 = 9 + 16 = 25\)

Perfect Square of a Complex Number

Calculate \((2 + i)^2\)
\(\displaystyle (2 + i)^2 = (2+i)(2+i) = 4 + 2i + 2i + i^2 \) \(\displaystyle = 4 + 4i - 1 = 3 + 4i\)

Properties of Complex Number Multiplication

Commutative
\(\displaystyle z_1 \cdot z_2 = z_2 \cdot z_1\)

Order doesn't matter
Associative
\(\displaystyle (z_1 \cdot z_2) \cdot z_3 = z_1 \cdot (z_2 \cdot z_3)\)

Grouping doesn't matter
Identity
\(\displaystyle z \cdot 1 = z\)

Multiplying by 1 unchanged
Distributive
\(\displaystyle z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3\)

Multiplication distributes

Common Mistakes to Avoid

Mistake 1: Forgetting to Replace \(i^2\) with \(-1\)

WRONG: \((2+i)(2+i) = 4 + 4i + i^2\) (left as \(i^2\)) ✗
RIGHT: \((2+i)(2+i) = 4 + 4i - 1 = 3 + 4i\) ✓

Mistake 2: Incorrect FOIL Application

WRONG: \((3+2i)(1-i) = 3 + 2i + 3(-i) + 2i(-i)\) (wrong grouping) ✗
RIGHT: \(= 3 - 3i + 2i - 2i^2 = 3 - i + 2 = 5 - i\) ✓

Mistake 3: Sign Errors with Negative Coefficients

WRONG: \((2 - 3i)(1 + i) = 2 + 2i - 3i + 3i^2 = 2 - i + 3\) (wrong sign) ✗
RIGHT: \(= 2 + 2i - 3i - 3i^2 = 2 - i + 3 = 5 - i\) ✓

Multiplication Summary

Case Formula/Example Result Form
General \((a+bi)(c+di)\) \((ac-bd)+(ad+bc)i\)
By Conjugate \((a+bi)(a-bi)\) \(a^2+b^2\) (always real)
Perfect Square \((a+bi)^2\) \((a^2-b^2)+2abi\)
By Pure Imaginary \((a+bi) \cdot ki\) \(-kb+(ka)i\

Practice Problems

Try These Multiplication Problems
  • \((1+2i)(3+i) = ?\) → Answer: \(1+7i\)
  • \((4-i)(2+3i) = ?\) → Answer: \(11+10i\)
  • \((5+2i)(5-2i) = ?\) → Answer: \(29\)
  • \((1+i)^2 = ?\) → Answer: \(2i\)
  • \(3i \cdot 2i = ?\) → Answer: \(-6\)
  • \((2-3i)(1-i) = ?\) → Answer: \(-1-5i\)


























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