In this section we will learn how to multiply and divide complex numbers, and in the process, we'll have to learn a technique for simplifying complex numbers we've divided. First let's look at multiplication.

Multiplying Complex Numbers

Multiplying complex numbers is almost as easy as multiplying two binomials together. I say "almost" because after we multiply the complex numbers, we have a little bit of simplifying work. Here's an example:

Example One
Multiply (3 + 2i)(2 - i).

Solution
Use the distributive property to write this as

3(2 - i) + 2i(2 - i)
6 - 3i + 4i - 2i²

Now we need to remember that i² = -1, so this becomes

6 - 3i + 4i + 2
8 + i

Example Two
Multiply (3 + 4i)(3 - 4i)

Solution
3(3 - 4i) + 4i(3 - 4i)

9 - 12i + 12i - 16i²

Conveniently, the imaginary parts cancel out, and -16i² = -16(-1) = 16, so we have:

9 + 16 = 25

This is very interesting; we multiplied two complex numbers, and the result was a real number! Would you like to see another example where this happens?

Example Three
Multiply (2 + 7i)(2 - 7i)

Solution
2(2 - 7i) + 7i(2 - 7i)
4 - 14i + 14i - 49i²
4 + 49
53

It turns out that whenever we have a complex number x + yi, and we multiply it by x - yi, the imaginary parts cancel out, and the result is a real number. We have a fancy name for x - yi; we call it the conjugate of x + yi. Every complex number has a conjugate, which we obtain by switching the sign of the imaginary part. Thus, the conjugate of 3 + 2i is 3 - 2i, and the conjugate of 5 - 7i is 5 + 7i.

We'll use this concept of conjugates when it comes to dividing and simplifying complex numbers.

Dividing Complex Numbers

Dividing complex numbers is actually just a matter of writing the two complex numbers in fraction form, and then simplifying it to standard form. In other words, there's nothing difficult about dividing - it's the simplifying that takes some work. Let's look at an example.

Suppose I want to divide 1 + i by 2 - i. I write it as follows:

. But it's not in simplest form, and that's a problem. Here's our rule for simplifying a complex fraction like this:

To simplify a complex fraction, multiply both the numerator and the denominator of the fraction by the conjugate of the denominator.

Why? Because doing this will result in the denominator becoming a real number.

So in the previous example, we would multiply the numerator and denomator by the conjugate of 2 - i, which is 2 + i:

• =

Now we need to multiply out the numerator, and we need to multiply out the denominator:

(1 + i)(2 + i) = 1(2 + i) + i(2 + i) = 2 + i +2i +i² = 1 + 3i

(2 - i)(2 + i) = 2(2 + i) - i(2 + i) = 4 + 2i - 2i - i² = 5

Thus, we have the result .

But this is still not in a + bi form, so we need to split the fraction up:

+ i.

Example Four
Divide 3 + 4i by 3 - 4i.

Solution
This problem comes from a previous page; I promised that we would show that = - + i. So here we go...

Multiply the numerator and the denominator by the conjugate of 3 - 4i:

Now we multiply out the numerator and the denominator:

(3 + 4i)(3 + 4i) = 3(3 + 4i) + 4i(3 + 4i) = 9 + 12i + 12i + 16i² = -7 + 24i

(3 - 4i)(3 + 4i) = 3(3 + 4i) - 4i(3 + 4i) = 9 + 12i - 12i - 16i² = 25

= = - + i

Example Five
Divide 3 + 4i by 2i

Solution
This one is a little different, because we're dividing by a pure imaginary number. We could do it the regular way by remembering that if we write 2i in standard form it's 0 + 2i, and its conjugate is 0 - 2i, so we multiply numerator and denominator by that.

But there's an easier way. First, we break it up into two fractions:

= + = + 2

Now we just need to get the i out of the denominator in , which we can do by multiplying the numerator and denominator by i:

= = -i

Thus, the answer is 2 - i

1.

Multiply 1 + 2i times 2 + i

2.

Multiply 3 + 5i times 3 - 5i

3.

What is the conjugate of 4 - 2i?

4.

What is the conjugate of 3 + i

5.

Multiply 2 + 3i times its conjugate.

6.

Multiply x + yi times its conjugate. Your answer will be in terms of x and y.

7.

Divide 1 + i by 2 + i

8.

Divide 4 - i by 2 + 3i

9.

Divide 5 + 7i by i

10.

Divide 1 by 1 + i