# Video: Multiplying Complex Numbers

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

17:58

### Video Transcript

In this video, we’re going to learn how to extend operations on complex numbers into multiplication. We’ll begin by looking at how to perform multiplication of a complex number first by real numbers and then by another complex number. We’ll then extend this to include deriving a general rule for squaring complex numbers and consider how this might help us to raise a complex number to exponents higher than two. And finally, we’ll learn how to apply these processes to help us solve equations.

If you’ve been studying complex numbers for a little while now, you might be aware that operations on complex numbers are very similar if not at times identical to operations on algebraic expressions. In fact, multiplying complex numbers is just like multiplying algebraic expressions except that we remember that the letter 𝑖 is not a variable. 𝑖 is of course the solution to the equation 𝑥 squared equals negative one. This means that 𝑖 squared is equal to negative one and in fact, we often say that 𝑖 is equal to the square root of negative one.

So let’s begin by considering how we might multiply complex number of the form 𝑧 equals 𝑎 plus 𝑏𝑖 by a constant. Let’s call our constant 𝑐, where 𝑐 is a real number. 𝑐 multiplied by 𝑧 — written simply as 𝑐𝑧 — is exactly the same as 𝑐 multiplied by the whole complex number 𝑎 plus 𝑏𝑖. And we add parentheses to show that this is the case.

We recall the distributive property which allows us to multiply each part of the complex number by the real number 𝑐. In doing so, we see that 𝑐𝑧 is equal to 𝑐𝑎 plus 𝑐𝑏𝑖. And for clarity, this might sometimes be written as 𝑎𝑐 plus 𝑏𝑐𝑖. And this probably comes of no surprise. This is exactly what we would expect if we were to multiply any two-term algebraic expression by a real constant.

Let’s now have a look at an example of how this might work.

If 𝑟 equals negative five plus two 𝑖 and 𝑠 equals negative eight minus two 𝑖, find two 𝑟 plus three 𝑠.

Here, we’ve been given two complex numbers 𝑟 and 𝑠. We want to find the sum of two 𝑟 and three 𝑠. So we’ll split the problem up and begin by working out two 𝑟 and three 𝑠 separately. Two 𝑟 is two multiplied by the complex number negative five plus two 𝑖. We’ll distribute these brackets by multiplying each part of the complex number by our constant two. Two multiplied by negative five is negative 10 and two multiplied by two 𝑖 is four 𝑖. And we see that two 𝑟 is equal to negative 10 plus four 𝑖.

We’ll now repeat this process for three 𝑠. This time, we multiply each part of the complex number 𝑠 by the constant three. Three multiplied by negative eight is negative 24 and three multiplied by negative two 𝑖 is negative six 𝑖. And now that we know the complex numbers two 𝑟 and three 𝑠, we need to find their sum. That’s negative 10 plus four 𝑖 plus negative 24 minus six 𝑖.

And remember to add two complex numbers, we simply add their real parts and then individually add their imaginary parts. Negative 10 plus negative 24 is negative 34 and four 𝑖 plus negative six 𝑖 is negative two 𝑖. So for the complex numbers given, two 𝑟 plus three 𝑠 is equal to negative 34 minus three 𝑖.

Now, this is all fine and well. But what it if we’d actually been multiplying our complex numbers by a purely imaginary number? We’ve already seen that the distributive property is really helpful in multiplying complex numbers by a real constant. And in fact, we can use this property to multiply a complex number by a purely imaginary number. That’s a number of the form 𝑐𝑖, where 𝑐 is a real number and 𝑖 is an imaginary number, the solution to the equation 𝑥 squared equals negative one.

This time we’re going to multiply a complex number 𝑎 plus 𝑏𝑖 by 𝑐𝑖. When we do, we get 𝑐𝑖 multiplied by 𝑎 plus 𝑏𝑖. 𝑐𝑖 multiplied by 𝑎 is 𝑐𝑎𝑖 and 𝑐𝑖 multiplied by 𝑏𝑖 is 𝑐𝑏𝑖 squared. But since 𝑖 squared equals is equal to negative one, we can write this as negative 𝑐𝑏. And if we instead write this in complex number form, we see that our complex number multiplied by a purely imaginary number is negative 𝑐𝑏 plus 𝑐𝑎𝑖.

Now we have developed a formula for multiplying a complex number by a purely imaginary number. We should really focus on applying the processes each time rather than trying to learn these by heart. Here, we’re going to consider an example of where we can apply these processes to multiply a complex number by a purely imaginary number.

What is negative seven 𝑖 multiplied by negative five plus five 𝑖?

We have a complex number negative five plus five 𝑖 and we want to multiply it by a purely imaginary number negative seven 𝑖. And we know that multiplying complex numbers is just like multiplying algebraic expressions. Here, we can apply the distributive property for expanding brackets. We multiply each part inside the bracket by the number on the outside. That’s negative seven 𝑖 multiplied by negative five which is 35𝑖 and negative seven 𝑖 multiplied by five 𝑖 which is negative 35𝑖 squared.

And here, we recall the fact that 𝑖 is the solution to the equation 𝑥 squared equals negative one such that 𝑖 squared must be equal to negative one. So negative 35𝑖 squared is the same as negative 35 multiplied by negative one which is simply 35. And since we now have a complex number which is of course a result of adding a real and a purely imaginary number, we write it as 35 plus 35𝑖.

Now as we might expect, we can extend these ideas into multiplying two complex numbers. And we’ll begin by considering the general product of two complex numbers. Let’s say we have two complex numbers 𝑧 one and 𝑧 two such that 𝑧 one is equal to 𝑎 plus 𝑏𝑖 and 𝑧 two is equal to 𝑐 plus 𝑑𝑖. Their product 𝑧 one 𝑧 two is the product 𝑎 plus 𝑏𝑖 and 𝑐 plus 𝑑𝑖.

And we’ve already seen that we can apply algebraic techniques to complex numbers. Here we can use any technique we like for multiplying two binomials. FOIL method and the grid method are two common methods. We’re going to look at the FOIL method.

F stands for first. We multiply the first term in the first bracket by the first term in the second bracket. 𝑎 multiplied by 𝑐 is simply 𝑎𝑐. O stands for outer. We multiply the outer terms and we get 𝑎𝑑𝑖. I stands for inner. We multiply the inner terms and we get 𝑏𝑐𝑖. And L stands for last. We multiply the last term in each bracket which is 𝑏𝑑𝑖 squared.

And since 𝑖 squared is equal to negative one, we can write this last part as negative 𝑏𝑑. And we can rearrange this slightly and we see that the products of 𝑧 one and 𝑧 two is 𝑎𝑐 minus 𝑏𝑑 plus 𝑎𝑑 plus 𝑏𝑐𝑖. It’s a complex number with a real part 𝑎𝑐 minus 𝑏𝑑 and an imaginary part 𝑎𝑑 plus 𝑏𝑐. Now once again, we have developed a formula for multiplying a complex number by another complex number. But we should really focus on applying the processes each time.

Multiply negative three plus 𝑖 by two plus five 𝑖.

Multiplying two complex numbers is just like multiplying two binomials and we can use any technique we like. Let’s try the grid method. Two multiplied by negative three is negative six and two multiplied by 𝑖 is two 𝑖. Five 𝑖 multiplied by negative three is negative 15 𝑖 and five 𝑖 multiplied by 𝑖 is five 𝑖 squared. And of course, 𝑖 squared is equal to negative one. So five 𝑖 squared is five multiplied by negative one which is negative five.

We’re going to simplify in a moment by collecting like terms. But currently, adding each part, we get negative six minus five plus two 𝑖 minus 15𝑖. Negative six minus five is negative 11 and two 𝑖 minus 15𝑖 is negative 13𝑖. So when we multiply negative three plus 𝑖 by two plus five 𝑖, we get negative 11 minus 13𝑖.

In our next example, we’ll look at how we’ll extend these ideas into squaring complex numbers.

If 𝑟 is equal to negative two plus four 𝑖 and 𝑠 is equal to eight minus 𝑖, find 𝑟 minus 𝑠 all squared.

In this question, we’ve been given two complex numbers and we’re being asked to find the square of their difference, 𝑟 minus 𝑠. Now, we absolutely could write 𝑟 minus 𝑠 squared as 𝑟 minus 𝑠 multiplied by 𝑟 minus 𝑠 and expand these brackets as normal. When we do, we see that we have three unique parts: 𝑟 squared, 𝑠 squared, and negative two 𝑟𝑠. It’s rather a lot of work for us to evaluate each of these complex numbers.

Instead, we’ll find the difference between the terms first and then we’ll square them. 𝑟 minus 𝑠 is negative two plus four 𝑖 minus eight minus 𝑖. And to subtract complex numbers, we subtract their real parts and their imaginary parts. Alternatively, we can think of this and a little like collecting like terms. Before we do that though, let’s distribute the second lot of brackets by multiplying each part inside the bracket by negative one. That gives us negative eight plus 𝑖.

Negative two minus eight is negative 10 and four 𝑖 plus is five 𝑖. So 𝑟 minus 𝑠 is negative 10 plus five 𝑖. This means that 𝑟 minus 𝑠 squared is negative 10 plus five 𝑖 squared. But since squaring a number is just the same as multiplying it by itself, we write this as negative 10 plus five 𝑖 multiplied by negative 10 plus five 𝑖.

And multiplying two complex numbers is just like multiplying binomials. We can use any technique we like. Let’s look at the FOIL method. We’ll begin by multiplying the first term in the first bracket by the first term in the second bracket. Negative 10 multiplied by negative 10 is 100 we multiply the outer terms: negative 10 multiplied by five 𝑖 is negative 50𝑖. And we get the same if we multiply the inner two terms.

Finally, we multiply the last term in each bracket. And we get 25𝑖 squared. But since 𝑖 squared is equal to negative one, we can write this as 25 multiplied by negative one which is negative 25. 100 minus 25 is 75. And negative 50 minus 50 is negative 100. So we get negative 100𝑖. And we can see that 𝑟 minus 𝑠 all squared is 75 minus 100𝑖.

Now that we’ve seen an example of how to square a complex number, let’s extend this and derive the general form. Let’s say we have a complex number 𝑧 in the form 𝑎 plus 𝑏𝑖, where 𝑎 and 𝑏 are real numbers. 𝑧 squared is 𝑎 plus 𝑏𝑖 all squared. But we know that we can square a complex number by multiplying it by itself and applying the same techniques we use for expanding brackets.

We multiply the first term in each bracket and we get 𝑎 squared. When we multiply the outer terms, we get 𝑎𝑏𝑖. And when we multiply the inner terms, we once again get 𝑎𝑏𝑖. And when we multiply the last terms, we get 𝑏 squared 𝑖 squared. But of course, 𝑖 squared is equal to negative one. So this last term is negative 𝑏 squared. And we see that 𝑧 squared is equal to 𝑎 squared minus 𝑏 squared plus two 𝑎𝑏𝑖.

And we can say that in general when we square a complex number 𝑧 in the form 𝑎 plus 𝑏𝑖, the real part of 𝑧 squared is 𝑎 squared minus 𝑏 squared and the imaginary part is two 𝑎𝑏. We’ve seen throughout this video though that remembering the technique is more important than remembering the formula. In this case, learning the formula for a square of a complex number can actually be extremely useful.

Let’s see an example of where it might help us to simplify a calculation.

Find the real part of seven minus two 𝑖 all squared.

Here, we’ve been given a complex number seven minus two 𝑖 for which we’re being asked to find the real part of its square. We could begin by writing seven minus two 𝑖 squared as seven minus two 𝑖 multiplied by seven minus two 𝑖 and then expanding the brackets fully. And we could use any technique for multiplying binomials. For example, we could use the FOIL method.

We multiply the first term in the first bracket by the first term in the second bracket. That’s seven multiplied by seven which is 49. We multiply the outer terms. That’s seven multiplied by negative two 𝑖 which is negative 14𝑖. And we get the same number when we multiply the inner terms. We can multiply the last term in each bracket and we get positive four 𝑖 squared. But of course, 𝑖 squared is equal to negative one. So this simplifies somewhat to 49 minus 28𝑖 minus four which is 45 minus 28𝑖.

Now for a complex number of the form 𝑎 plus 𝑏𝑖, its real part is 𝑎 and its imaginary part is 𝑏. And in this case, we can say that the real part of our complex number is 45. Now while this method is perfectly valid, the amount of work is a little unnecessary. Instead, we’re going to recall the general form for the square of a complex number 𝑎 plus 𝑏𝑖. It’s given by the formula 𝑧 squared equals 𝑎 squared minus 𝑏 squared plus two 𝑎𝑏𝑖.

The real part is 𝑎 squared minus 𝑏 squared and the imaginary part is two 𝑎𝑏. Now in our complex number, the real part is seven; 𝑎 is equal to seven. And the imaginary part 𝑏 is equal to negative two. So we can say that the real part of 𝑧 squared is seven squared minus negative two squared or 49 minus four. 49 minus four is equal to five. And that’s the same answer we worked out earlier.

The rules we’ve learned for squaring complex numbers can actually also help us find higher powers of these numbers.

If 𝑟 is equal to two plus 𝑖, express 𝑟 cubed in the form 𝑎 plus 𝑏𝑖.

We’ll begin here by rewriting 𝑟 cubed slightly. 𝑟 cubed is the same as 𝑟 times 𝑟 times 𝑟 which is the same as two plus 𝑖 times two plus 𝑖 times two plus 𝑖. We’ll begin by multiplying two plus 𝑖 by two plus 𝑖. And we could expand these two brackets just like expanding binomials.

Alternatively, we recall that for complex number 𝑎 plus 𝑏𝑖, the square of that complex number is 𝑎 squared minus 𝑏 squared plus two 𝑎𝑏𝑖. The real part of our complex number 𝑎 is two. And the imaginary part is the coefficient of 𝑖; it’s one. We can square this number then by substituting 𝑎 is equal to two and B is equal to one into that formula. We get two squared minus one squared plus two times two times one 𝑖 and that’s all multiplied by two plus 𝑖.

Two squared is four and one squared is one. So two squared minus one squared is three and two times two times one is four. So we have three plus four 𝑖 multiplied by two plus 𝑖. And we can multiply these brackets by using the FOIL method. We multiply the first term in each bracket. Three times two is six. We multiply the outer terms to get three 𝑖. And for the inner terms, it’s four 𝑖 multiplied by two which is eight 𝑖. We then multiply the last terms four 𝑖 multiplied by 𝑖 is four 𝑖 squared.

And then since we know that 𝑖 squared is equal to negative one, four 𝑖 squared becomes negative four. And we also know that three 𝑖 plus eight 𝑖 is 11𝑖. So we see that our expression simplifies to two plus 11𝑖. So 𝑟 cubed in the form required is two plus 11𝑖.

Now there are techniques that exist to simplify this process for higher powers of 𝑧, which we’ll learn about as we become more confident working with complex numbers. Our last example though will demonstrate how to solve an equation using complex numbers.

Solve the equation 𝑖𝑧 is equal to negative four plus three 𝑖.

Usually, we look to apply the rules for solving algebraic expressions. Here that will be dividing by 𝑖. But there is another technique we can use. We know that 𝑖 squared is negative one. So we’re going to multiply both sides of this equation by 𝑖. We then distribute the brackets by multiplying each bit inside the bracket by 𝑖. And since 𝑖 squared is negative one, we get negative 𝑧 equals negative four 𝑖 minus three 𝑖. We’ll multiply through by negative one and we see that 𝑧 is equal to three plus four 𝑖.

And it’s always sensible to check our answers by substituting it back into the original equation: 𝑖 multiplied by three plus four 𝑖 is three 𝑖 plus four 𝑖 squared. And since four 𝑖 squared is four multiplied by negative one, we get negative four. And this is of course the same as negative four plus three 𝑖.

In this video, we’ve learned that we can multiply two complex numbers using standard methods such as the grid method or FOIL method. We’ve seen that the square of a complex number 𝑎 plus 𝑏𝑖 is 𝑎 squared minus 𝑏 squared plus two 𝑎𝑏𝑖. And we also saw how we can use these techniques for higher powers of 𝑧 though this isn’t necessarily the most efficient method.