Multiplying Imaginary Numbers

The product of two imaginary numbers is a real number: $$ (0,a) \cdot (0,b) = (- a \cdot b, 0) = - a \cdot b $$. Writing these numbers in algebraic form, where (0,b) = bi and (0,a) = ai, we find: $$ ai \cdot bi = - a \cdot b $$

The Proof

Let’s examine two imaginary numbers:

$$ z_1 = (0,a)=ai $$ $$ z_2 = (0,b)=bi $$

When multiplying these two imaginary numbers in Cartesian form, (0,a) and (0,b), we use the complex number multiplication rule:

$$ z_1 \cdot z_2 = (0,a) \cdot (0,b) $$

$$ z_1 \cdot z_2 = (0 \cdot 0 – a \cdot b, 0 \cdot b + a \cdot 0)$$

$$ z_1 \cdot z_2 = (0 – ab, 0 + 0)$$

$$ z_1 \cdot z_2 = ( – ab, 0)$$

Thus, the result is the real number -ab.

We get the same result when calculating the product of imaginary numbers in algebraic form:

$$ z_1 \cdot z_2 = ai \cdot bi $$

Using the rules of algebraic multiplication:

$$ z_1 \cdot z_2 = a \cdot b \cdot i \cdot i $$

$$ z_1 \cdot z_2 = a \cdot b \cdot i^2 $$

Since \( i^2 = -1 \):

$$ z_1 \cdot z_2 = a \cdot b \cdot (-1) $$

$$ z_1 \cdot z_2 = -a \cdot b $$

Note. I find it easy to remember the algebraic calculation because it follows standard algebraic rules. You just need to keep in mind that the square of the imaginary unit is \(-1\), i.e., \( i^2 = -1 \).

A Practical Example

Let’s take two imaginary numbers:

$$ (0,2) = 2i $$ $$ (0,3) = 3i $$

The product of these two imaginary numbers is the real number (-6,0):

$$ (0,2) \cdot (0,3) = (0 \cdot 0 – 2 \cdot 3, 0 \cdot 3 + 2 \cdot 0) = (-6,0) $$

The same result is achieved when the two imaginary numbers are written in algebraic form:

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

The result is always the real number (-6,0) = -6.

Note. When imaginary numbers are expressed in Cartesian form, (0,b), multiplication follows the complex number multiplication rule: $$ (a,b) \cdot (c,d) = (ac-bd, ad+bc) $$. Conversely, when imaginary numbers are expressed in algebraic form, you can simply apply the algebraic product of monomials: $$ ai \cdot bi = a \cdot b \cdot i^2 = a \cdot b \cdot (-1) = -ab.$$

Additional Notes

Here are a few additional points about multiplying imaginary numbers:

• Commutative Property
  The multiplication of imaginary numbers is commutative: swapping the order of the factors does not change the result. $$ ai \cdot bi = bi \cdot ai $$

  Example: $$ 2i \cdot 3i = 3i \cdot 2i $$ $$ 2 \cdot 3 \cdot i^2 = 3 \cdot 2 \cdot i^2 $$ $$ 6 \cdot (-1) = 6 \cdot (-1) $$ $$ -6 = -6 $$

• Associative Property
  The multiplication of imaginary numbers satisfies the associative property: grouping the factors in different ways does not affect the result. $$ (ai \cdot bi) \cdot ci = ai \cdot (bi \cdot ci) $$

  Example: $$ (2i \cdot 3i) \cdot 4i = 2i \cdot (3i \cdot 4i) $$ $$ (6i^2) \cdot 4i = 2i \cdot (12i^2) $$ $$ (6 \cdot (-1)) \cdot 4i = 2i \cdot (12 \cdot (-1)) $$ $$ (-6) \cdot 4i = 2i \cdot (-12) $$ $$ -24i = -24i $$

And so on.