Complex Numbers Multiplication

In this tutorial, I'll be your guide as we navigate the multiplication of complex numbers. We'll explore this concept with the aid of a practical example, providing you with a clear understanding of how it works.

Imagine you have two complex numbers - let's call them (a,b) and (c,d). When you multiply these, you'll end up with another pair, which looks like this: (ac-bd,ad+bc). So the formula for our exploration today is $$ (a,b) \cdot (c,d) = (ac-bd,ad+bc) $$

Here's what you do: first, calculate the difference between the product of the initial components (colored in yellow) and the second components (colored in orange).

Next, it's time to add up the products of the first components (those in yellow) and the second components (those in orange).

Let's bring this concept to life with a practical example:

$$ (a,b) \cdot (c,d) = (ac-bd,ad+bc) $$

 

Let's say we have two complex numbers z₁ and z₂.

$$ z_1 = (3,4) $$ $$ z_2 = (2,5) $$

Our task is to multiply these numbers using the formula we discussed above.

$$ z_1 \cdot z_2 = (ac-bd, ad+bc) $$

In this equation, our parameters are a=3, b=4 (because z₁=(a,b)=(3,4)) and c=2, d=5 (because z₂=(c,d)=(2,5)).

So, let's proceed with our computation:

$$ z_1 \cdot z_2 = (3 \cdot 2-4 \cdot 5, 3 \cdot 5+4 \cdot 2) $$

$$ z_1 \cdot z_2 = (6-20, 15+8) $$

$$ z_1 \cdot z_2 = (-14, 23) $$

So, we discover that the product of our two complex numbers is (-14,23).

If the formula isn't at your fingertips, worry not! You can also work out the multiplication of complex numbers algebraically.

First, let's convert our complex numbers into their algebraic forms:

$$ z_1 = (3,4) = 3+4i $$ $$ z_2 = (2,5) = 2+5i $$

A quick note: to translate a complex number (a,b) into its algebraic form, combine the real and imaginary parts of the number. Don't forget that the first component of a pair (a,b) is the real part, while the second component is the imaginary part.$$ (a,b)=a+b \cdot i $$

Now, let's compute the product of the binomials, sticking to our trusty algebraic rules.

We then proceed to:

$$ z_1 \cdot z_2 = (3+4i) \cdot (2+5i) $$

$$ z_1 \cdot z_2 = 3 \cdot (2+5i) + 4i \cdot (2+5i) $$

$$ z_1 \cdot z_2 = 6+15i + 8i+20i^2 $$

$$ z_1 \cdot z_2 = 6+23i+20i^2 $$

At this stage, remember that the imaginary unit "i" is the square root of -1 by definition. $$ i = \sqrt{-1} $$ As a result, the square of the imaginary unit i² is -1. $$ i^2 = -1 $$

With that in mind, we replace i² with -1 in our previous expression.

$$ z_1 \cdot z_2 = 6+23i+20 \cdot (-1) $$

$$ z_1 \cdot z_2 = 6+23i-20 $$

$$ z_1 \cdot z_2 = -14+23i $$

Whether you use the formula or algebra, you'll find that the final multiplication result remains the same: -14+23i = (-14,23).

Multiplication properties of complex numbers

The multiplication of complex numbers adheres to:

Commutative property

$$ z_1 \cdot z_2 = z_2 \cdot z_1 $$

Associative property

$$ z_1 \cdot (z_2 \cdot z_3) = (z_1 \cdot z_2) \cdot z_3 $$

Distributive property over addition

$$ z_1 \cdot ( z_2 + z_3 ) = z_1 \cdot z_2 + z_1 \cdot z_3 $$

The neutral element of multiplication is the number (1,0). Multiply any complex number (a,b) by (1,0), and you'll get (a,b) back.

$$ (a,b) \cdot (1,0) = (a,b) $$

The absorbing element of multiplication is the number (0,0). When any complex number (a,b) is multiplied by (0,0), you'll get (0,0).

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

If you've found this complex numbers tutorial by Nigiara valuable, do keep your eyes peeled for more.