Multiplication using Complex Numbers

06 Dec 2013

Maths

Here are three useful methods for multiplying complex terms:

• FOIL – First, Outer, Inner, Last
• Gathering Like Terms
• Expand Terms

Consider the rule:

(a + bi)(c + di)

Where i denotes an imaginary number.

The FOIL Method

The FOIL method expands the rule like this:

(a + bi)(c + di)
=> ac +adi + bic + bdi

Consider the following equation:

(2 + 3i) * (4 - i)

Isolating each term and assigning it to its counterpart variable:

a = 2
bi = 3i
c = 4
di = -i

Now we can use this information and plug it into the initial rule:

(a + bi)(c + di)
=> ac +adi + bic + bdi
=> (2 * 4) + (2 * -i) + (3i * 4) + (3i * -i)
=> (8) + (-2i) + (12i) + (-3i2)

Remember that:

i2 = -1

So exercising that rule:

=> 8 + 10i - 3i2
=> 8 + 10i - 3(-1)
=> 8 + 3 + 10i
=> 11 + 10i

The Gathering Like Terms Method

Using the gathering like terms method:

(a + bi)(c + di)
=> (ac – bd) + (ad + bc)i

Don't get caught out on this next step. Normally, a term with no coefficient is simply that term with a coefficient of 1, however in this method, we replace a missing coefficient with 0.

=> ((2 * 4) - (3 * 0)) + ((2 * 0) + (3 * 4)(i))
=> ((8) - (0)) + ((0) + (12)(i))
=> 8 + (-2 + 12)(i)
=> (11) + (-2 + 12)(i)
=> 11 + 10i

The Expand Terms Method

The expand terms method is the method I find the simplest to implement. The rule for this becomes:

(a + bi)(c + di)
=> a(c + di) + bi(c + di)

Using this rule to solve our problem:

(2 + 3i) * (4 - i)
=> a(c + di) + bi(c + di)
=> 2(4 - i) + 3i(4 - i)
=> 8 - 2i + 12i -3i2
=> 8 + 10i -3(-1)
=> 8 + 10i + 3
=> 11 + 10i