Explainer: Multiplying Complex Numbers

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 𝑧=𝑎+𝑏𝑖. If we multiply both sides of this equation by a real number 𝑐, we get 𝑐𝑧=𝑐(𝑎+𝑏𝑖). Using the distributive property, we can rewrite this as 𝑐𝑧=𝑐𝑎+𝑐𝑏𝑖, much as we would expect.

Example 1: Multiplying Complex Numbers by Real Numbers

If 𝑟=5+2𝑖 and 𝑠=82𝑖, find 2𝑟+3𝑠.

Answer

Substituting in the values of 𝑟 and 𝑠 into the expression, we get 2𝑟+3𝑠=2(5+2𝑖)+3(82𝑖).

Expanding the brackets using the distributive property, we get 2𝑟+3𝑠=10+4𝑖+(246𝑖).

Finally, we can gather our like terms to find 2𝑟+3𝑠=342𝑖.

Having considered the simplest case of multiplying a complex number by a real numbers, we can now consider multiplying a complex number by a purely imaginary number. Taking 𝑧=𝑎+𝑏𝑖, we can multiply both sides of the equation by a purely imaginary number 𝑐𝑖 to get (𝑐𝑖)𝑧=𝑐𝑖(𝑎+𝑏𝑖). Once again, we can use the distributive property to rewrite this as (𝑐𝑖)𝑧=𝑐𝑎𝑖+𝑐𝑏𝑖.

Since 𝑖=1, we can simplify this to (𝑐𝑖)𝑧=𝑐𝑏+𝑐𝑎𝑖.

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 7𝑖(5+5𝑖)?

Answer

Expanding the brackets using the distributive property, we can write 7𝑖(5+5𝑖)=35𝑖35𝑖.

Since 𝑖=1, we can simplify this to 7𝑖(5+5𝑖)=35+35𝑖.

Let us now consider the product of two general complex numbers 𝑧=𝑎+𝑏𝑖 and 𝑧=𝑐+𝑑𝑖: 𝑧𝑧=(𝑎+𝑏𝑖)(𝑐+𝑑𝑖).

Using any technique for multiplying two binomials (FOIL, grid/area, or long multiplication), we can express this as 𝑧𝑧=𝑎𝑐+𝑎𝑑𝑖+𝑏𝑐𝑖+𝑏𝑑𝑖.

Using the fact that 𝑖=1, and gathering our like terms, we can rewrite this as 𝑧𝑧=𝑎𝑐𝑏𝑑+(𝑎𝑑+𝑏𝑐)𝑖.

We summarize this as follows.

The Product of Two Complex Numbers

For two complex numbers 𝑧=𝑎+𝑏𝑖 and 𝑧=𝑐+𝑑𝑖, we define the product 𝑧𝑧=𝑎𝑐𝑏𝑑+(𝑎𝑑+𝑏𝑐)𝑖.

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.

Example 3: Multiplying Complex Numbers

Multiply (3+𝑖) by (2+5𝑖).

Answer

Using FOIL, or any other technique, we can expand the brackets as follows: (3+𝑖)(2+5𝑖)=(3)×2+(3)×5𝑖+2𝑖+5𝑖=615𝑖+2𝑖+5𝑖.

Since 𝑖=1, we can rewrite this as (3+𝑖)(2+5𝑖)=615𝑖+2𝑖5.

Finally, gathering our like terms, we have (3+𝑖)(2+5𝑖)=1113𝑖.

Example 4: Squares of Complex Numbers

If 𝑟=2+4𝑖 and 𝑠=8𝑖, find (𝑟𝑠).

Answer

For a question like this, we have a choice: do we multiply out (𝑟𝑠) and then substitute in the values of 𝑟 and 𝑠, or do we do it the other way around? If we try the first method, we will find we have to calculate 𝑟, 𝑠, and 𝑟𝑠. This is a lot of computation. However, if we first calculate 𝑟𝑠, 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 𝑟𝑠 as follows: 𝑟𝑠=2+4𝑖(8𝑖)=10+5𝑖.

To find the square of this number, we can express it as a product and use the technique for multiplying complex numbers. Hence, (𝑟𝑠)=(10+5𝑖)=(10+5𝑖)(10+5𝑖).

Using FOIL or another technique for expanding brackets, we have (𝑟𝑠)=(10)+(10)5𝑖+(10)5𝑖+(5𝑖)=10050𝑖50𝑖+25𝑖.

Using the fact that 𝑖=1, we can gather our like terms and rewrite this as (𝑟𝑠)=75100𝑖.

This example raises the question of what the general form of the square of a complex number 𝑧=𝑎+𝑏𝑖 is. Using the techniques we have been developing to multiply complex numbers, we can derive a formula for it as follows: 𝑧=(𝑎+𝑏𝑖)(𝑎+𝑏𝑖).

Expanding the brackets, we get 𝑧=𝑎+𝑎𝑏𝑖+𝑎𝑏𝑖+𝑏𝑖.

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

The Square of a Complex Number

For a complex number 𝑧=𝑎+𝑏𝑖, 𝑧=𝑎𝑏+2𝑎𝑏𝑖.

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.

Example 5: Squares of Complex Numbers

Find Re(72𝑖).

Answer

We begin by calculating the square of 72𝑖 as follows: (72𝑖)=(72𝑖)(72𝑖).

Expanding the brackets, we have (72𝑖)=49+7(2𝑖)+7(2𝑖)+(2)𝑖=4914𝑖14𝑖+4𝑖.

Using the fact that 𝑖=1, and gathering like terms, we have (72𝑖)=4528𝑖.

Finally, taking the real part, we get Re(72𝑖)=45.

Alternatively, we could save ourselves some calculation by using the fact that, for a complex number 𝑧=𝑎+𝑏𝑖, 𝑧=𝑎𝑏+2𝑎𝑏𝑖. Taking the real part, we have Re𝑧=𝑎𝑏.

Hence, Re(72𝑖)=7(2)=494=45.

Example 6: Powers of Complex Numbers

If 𝑟=2+𝑖, express 𝑟 in the form 𝑎+𝑏𝑖.

Answer

We start by calculating 𝑟: 𝑟=(2+𝑖)=(2+𝑖)(2+𝑖).

Expanding the brackets, we have 𝑟=4+2𝑖+2𝑖+𝑖=3+4𝑖.

To calculate 𝑟, we can now multiply both sides of the equation by 𝑟 to get 𝑟=(3+4𝑖)(2+𝑖). Multiplying out the brackets, we get 𝑟=6+3𝑖+8𝑖+4𝑖=2+11𝑖.

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 𝑖𝑧=4+3𝑖.

Answer

On first impression, we might think that we need to divide both sides of the equation by 𝑖 to isolate 𝑧. However, by remembering that 𝑖=1, we could simply multiply the whole equation by 𝑖 or, better yet, by 𝑖.

Hence, by multiplying both sides of the equation by 𝑖, we get (𝑖)𝑖𝑧=(𝑖)(4+3𝑖).

Expanding the brackets, and simplifying, we arrive at the solution: 𝑧=3+4𝑖.

It is always good practice to check our answer. To do this, we can multiply our solution by 𝑖 and check that we get back our original equation: 𝑖𝑧=𝑖(3+4𝑖).

Expanding the brackets and simplifying, we find 𝑖𝑧=3𝑖+4𝑖=4+3𝑖 as required.

Key Points

  1. To multiply complex numbers, we use the same techniques that we use to multiply binomials.
  2. For a complex number 𝑧=𝑎+𝑏𝑖, 𝑧=𝑎𝑏+2𝑎𝑏𝑖.
  3. 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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