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Lesson Explainer: Multiplying Complex Numbers Mathematics • First Year of Secondary School

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In this explainer, we will learn how to multiply two complex numbers.

The algebra of complex numbers is very similar to the algebra of binomials. Hence, applying the knowledge we have of working with binomials will take us a long way when working with complex numbers. Before we look at multiplication of complex numbers in full generality, we will consider the simpler cases of a complex number multiplied by a real number and a complex number multiplied by a purely imaginary number.

Consider a complex number ๐‘ง=๐‘Ž+๐‘๐‘–. If we multiply both sides of this equation by a real number ๐‘, we get ๐‘๐‘ง=๐‘(๐‘Ž+๐‘๐‘–). Using the distributive property, we can rewrite this as ๐‘๐‘ง=๐‘๐‘Ž+๐‘๐‘๐‘–, much as we would expect.

Example 1: Multiplying Complex Numbers by Real Numbers

If ๐‘Ÿ=โˆ’5+2๐‘– and ๐‘ =โˆ’8โˆ’2๐‘–, find 2๐‘Ÿ+3๐‘ .

Answer

Substituting in the values of ๐‘Ÿ and ๐‘  into the expression, we get 2๐‘Ÿ+3๐‘ =2(โˆ’5+2๐‘–)+3(โˆ’8โˆ’2๐‘–).

Expanding the parentheses using the distributive property, we get 2๐‘Ÿ+3๐‘ =โˆ’10+4๐‘–+(โˆ’24โˆ’6๐‘–).

Finally, we can gather our like terms to find 2๐‘Ÿ+3๐‘ =โˆ’34โˆ’2๐‘–.

Having considered the simplest case of multiplying a complex number by a real number, we can now consider multiplying a complex number by a purely imaginary number. Taking ๐‘ง=๐‘Ž+๐‘๐‘–, we can multiply both sides of the equation by a purely imaginary number ๐‘๐‘– to get (๐‘๐‘–)๐‘ง=๐‘๐‘–(๐‘Ž+๐‘๐‘–). Once again, we can use the distributive property to rewrite this as (๐‘๐‘–)๐‘ง=๐‘๐‘Ž๐‘–+๐‘๐‘๐‘–.๏Šจ

Since ๐‘–=โˆ’1๏Šจ, we can simplify this to (๐‘๐‘–)๐‘ง=โˆ’๐‘๐‘+๐‘๐‘Ž๐‘–.

The formulae for multiplying complex numbers by real and imaginary numbers are not formulae that you need to learn. Instead, we should focus on becoming familiar with the algebraic techniques required to work with complex numbers in general.

Example 2: Multiplying Complex Numbers by Imaginary Numbers

What is โˆ’7๐‘–(โˆ’5+5๐‘–)?

Answer

Expanding the parentheses using the distributive property, we can write โˆ’7๐‘–(โˆ’5+5๐‘–)=35๐‘–โˆ’35๐‘–.๏Šจ

Since ๐‘–=โˆ’1๏Šจ, we can simplify this to โˆ’7๐‘–(โˆ’5+5๐‘–)=35+35๐‘–.

Let us now consider the product of two general complex numbers ๐‘ง=๐‘Ž+๐‘๐‘–๏Šง and ๐‘ง=๐‘+๐‘‘๐‘–๏Šจ: ๐‘ง๐‘ง=(๐‘Ž+๐‘๐‘–)(๐‘+๐‘‘๐‘–).๏Šง๏Šจ

Using any technique for multiplying two binomials (FOIL, grid/area, or long multiplication), we can express this as ๐‘ง๐‘ง=๐‘Ž๐‘+๐‘Ž๐‘‘๐‘–+๐‘๐‘๐‘–+๐‘๐‘‘๐‘–.๏Šง๏Šจ๏Šจ

Using the fact that ๐‘–=โˆ’1๏Šจ, and gathering our like terms, we can rewrite this as ๐‘ง๐‘ง=๐‘Ž๐‘โˆ’๐‘๐‘‘+(๐‘Ž๐‘‘+๐‘๐‘)๐‘–.๏Šง๏Šจ

We summarize this as follows.

The Product of Two Complex Numbers

For two complex numbers ๐‘ง=๐‘Ž+๐‘๐‘–๏Šง and ๐‘ง=๐‘+๐‘‘๐‘–๏Šจ, we define the product ๐‘ง๐‘ง=๐‘Ž๐‘โˆ’๐‘๐‘‘+(๐‘Ž๐‘‘+๐‘๐‘)๐‘–.๏Šง๏Šจ

Although we have stated a general form, rather than simply memorizing it, it is more important to be familiar with the techniques of multiplying complex numbers.

Now, let us consider an example where we demonstrate the multiplication of two complex numbers.

Example 3: Multiplying Complex Numbers

Multiply (โˆ’3+๐‘–) by (2+5๐‘–).

Answer

Using FOIL, or any other technique, we can expand the parentheses as follows: (โˆ’3+๐‘–)(2+5๐‘–)=(โˆ’3)ร—2+(โˆ’3)ร—5๐‘–+2๐‘–+5๐‘–=โˆ’6โˆ’15๐‘–+2๐‘–+5๐‘–.๏Šจ๏Šจ

Since ๐‘–=โˆ’1๏Šจ, we can rewrite this as (โˆ’3+๐‘–)(2+5๐‘–)=โˆ’6โˆ’15๐‘–+2๐‘–โˆ’5.

Finally, gathering our like terms, we have (โˆ’3+๐‘–)(2+5๐‘–)=โˆ’11โˆ’13๐‘–.

In the next example, let us look at how to calculate the square of the difference of two complex numbers.

Example 4: Squares of Complex Numbers

If ๐‘Ÿ=โˆ’2+4๐‘– and ๐‘ =8โˆ’๐‘–, find (๐‘Ÿโˆ’๐‘ )๏Šจ.

Answer

For a question like this, we have a choice: do we multiply out (๐‘Ÿโˆ’๐‘ )๏Šจ and then substitute in the values of ๐‘Ÿ and ๐‘ , or do we do it the other way around? If we try the first method, we will find we have to calculate ๐‘Ÿ๏Šจ, ๐‘ ๏Šจ, and ๐‘Ÿ๐‘ . This is a lot of computation. However, if we first calculate ๐‘Ÿโˆ’๐‘ , we will only need to find the square of the result, which is much more efficient. This is the technique we will use here.

Firstly, we calculate ๐‘Ÿโˆ’๐‘  as follows: ๐‘Ÿโˆ’๐‘ =โˆ’2+4๐‘–โˆ’(8โˆ’๐‘–)=โˆ’10+5๐‘–.

To find the square of this number, we can express it as a product and use the technique for multiplying complex numbers. Hence, (๐‘Ÿโˆ’๐‘ )=(โˆ’10+5๐‘–)=(โˆ’10+5๐‘–)(โˆ’10+5๐‘–).๏Šจ๏Šจ

Using FOIL or another technique for expanding parentheses, we have (๐‘Ÿโˆ’๐‘ )=(โˆ’10)+(โˆ’10)5๐‘–+(โˆ’10)5๐‘–+(5๐‘–)=100โˆ’50๐‘–โˆ’50๐‘–+25๐‘–.๏Šจ๏Šจ๏Šจ๏Šจ

Using the fact that ๐‘–=โˆ’1๏Šจ, we can gather our like terms and rewrite this as (๐‘Ÿโˆ’๐‘ )=75โˆ’100๐‘–.๏Šจ

This example raises the question of what the general form of the square of a complex number ๐‘ง=๐‘Ž+๐‘๐‘– is. Using the techniques we have been developing to multiply complex numbers, we can derive a formula for it as follows: ๐‘ง=(๐‘Ž+๐‘๐‘–)(๐‘Ž+๐‘๐‘–).๏Šจ

Expanding the parentheses, we get ๐‘ง=๐‘Ž+๐‘Ž๐‘๐‘–+๐‘Ž๐‘๐‘–+๐‘๐‘–.๏Šจ๏Šจ๏Šจ๏Šจ

Gathering our like terms, we can state the general form as follows.

The Square of a Complex Number

For a complex number ๐‘ง=๐‘Ž+๐‘๐‘–, ๐‘ง=๐‘Žโˆ’๐‘+2๐‘Ž๐‘๐‘–.๏Šจ๏Šจ๏Šจ

Even though it is very important to know the techniques needed to derive equations like this, it can also be useful to commit the general form of the square of a complex number to memory. In the next example, we will see how remembering the formula can simplify calculations.

Now, let us consider an example where we have to calculate the real part of the square of a complex number.

Example 5: Squares of Complex Numbers

Find Re๏€บ(7โˆ’2๐‘–)๏†๏Šจ.

Answer

We begin by calculating the square of 7โˆ’2๐‘– as follows: (7โˆ’2๐‘–)=(7โˆ’2๐‘–)(7โˆ’2๐‘–).๏Šจ

Expanding the parentheses, we have (7โˆ’2๐‘–)=49+7(โˆ’2๐‘–)+7(โˆ’2๐‘–)+(โˆ’2)๐‘–=49โˆ’14๐‘–โˆ’14๐‘–+4๐‘–.๏Šจ๏Šจ๏Šจ๏Šจ

Using the fact that ๐‘–=โˆ’1๏Šจ, and gathering like terms, we have (7โˆ’2๐‘–)=45โˆ’28๐‘–.๏Šจ

Finally, taking the real part, we get Re๏€บ(7โˆ’2๐‘–)๏†=45.๏Šจ

Alternatively, we could save ourselves some calculation by using the fact that, for a complex number ๐‘ง=๐‘Ž+๐‘๐‘–, ๐‘ง=๐‘Žโˆ’๐‘+2๐‘Ž๐‘๐‘–๏Šจ๏Šจ๏Šจ. Taking the real part, we have Re๏€น๐‘ง๏…=๐‘Žโˆ’๐‘.๏Šจ๏Šจ๏Šจ

Hence, Re๏€บ(7โˆ’2๐‘–)๏†=7โˆ’(โˆ’2)=49โˆ’4=45.๏Šจ๏Šจ๏Šจ

Example 6: Powers of Complex Numbers

If ๐‘Ÿ=2+๐‘–, express ๐‘Ÿ๏Šฉ in the form ๐‘Ž+๐‘๐‘–.

Answer

We start by calculating ๐‘Ÿ๏Šจ: ๐‘Ÿ=(2+๐‘–)=(2+๐‘–)(2+๐‘–).๏Šจ๏Šจ

Expanding the parentheses, we have ๐‘Ÿ=4+2๐‘–+2๐‘–+๐‘–=3+4๐‘–.๏Šจ๏Šจ

To calculate ๐‘Ÿ๏Šฉ, we can now multiply both sides of the equation by ๐‘Ÿ to get ๐‘Ÿ=(3+4๐‘–)(2+๐‘–).๏Šฉ Multiplying out the parentheses, we get ๐‘Ÿ=6+3๐‘–+8๐‘–+4๐‘–=2+11๐‘–.๏Šฉ๏Šจ

Clearly, working like this to calculate higher and higher powers of complex numbers could be quite laborious. However, as we learn more about complex numbers we will learn alternative techniques which will significantly simplify the process. We finish by looking at one last example where we can apply what we know about complex numbers and multiplication to solve an equation involving complex numbers.

Example 7: Solving Equations Involving Complex Numbers

Solve the equation ๐‘–๐‘ง=โˆ’4+3๐‘–.

Answer

On first impression, we might think that we need to divide both sides of the equation by ๐‘– to isolate ๐‘ง. However, by remembering that ๐‘–=โˆ’1๏Šจ, we could simply multiply the whole equation by ๐‘– or, better yet, by โˆ’๐‘–.

Hence, by multiplying both sides of the equation by โˆ’๐‘–, we get (โˆ’๐‘–)๐‘–๐‘ง=(โˆ’๐‘–)(โˆ’4+3๐‘–).

Expanding the parentheses, and simplifying, we arrive at the solution: ๐‘ง=3+4๐‘–.

It is always good practice to check our answer. To do this, we can multiply our solution by ๐‘– and check that we get back our original equation: ๐‘–๐‘ง=๐‘–(3+4๐‘–).

Expanding the parentheses and simplifying, we find ๐‘–๐‘ง=3๐‘–+4๐‘–=โˆ’4+3๐‘–๏Šจ as required.

Key Points

  • To multiply complex numbers, we use the same techniques that we use to multiply binomials.
  • For two complex numbers ๐‘ง=๐‘Ž+๐‘๐‘–๏Šง and ๐‘ง=๐‘+๐‘‘๐‘–๏Šจ, we define the product as ๐‘ง๐‘ง=๐‘Ž๐‘โˆ’๐‘๐‘‘+(๐‘Ž๐‘‘+๐‘๐‘)๐‘–.๏Šง๏Šจ
  • For a complex number ๐‘ง=๐‘Ž+๐‘๐‘–, ๐‘ง=๐‘Žโˆ’๐‘+2๐‘Ž๐‘๐‘–.๏Šจ๏Šจ๏Šจ
  • Although, theoretically, using these techniques we can calculate arbitrarily large powers of complex numbers, it requires a large amount of calculation.

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