Video: Addition and Subtraction of Complex Numbers

In this video, we will learn how to equate, add, and subtract complex numbers.

15:16

Video Transcript

In this lesson, we will learn how to add and subtract complex numbers. We’ll begin by recapping what we mean by a complex number and what it means for two complex numbers to be equal. We will then learn how to add and subtract these numbers, extending this idea into solving simple equations involving complex numbers.

Remember, a complex number 𝑧 is a number of the form π‘Ž plus 𝑏𝑖. It is important that π‘Ž and 𝑏 are both real numbers. And 𝑖 is defined as the solution to the equation π‘₯ squared equals negative one. We say that 𝑖 squared is equal to negative one and sometimes 𝑖 is equal to the squared root of negative one.

And for our complex number π‘Ž plus 𝑏𝑖, we say that the real part of 𝑧 is π‘Ž and the imaginary part is 𝑏. It’s important that the imaginary part is 𝑏, not 𝑏𝑖. It’s essentially the coefficient of 𝑖. And just as the set of real numbers is denoted by the letter ℝ, the set of complex numbers is denoted by the letter β„‚, as shown.

Before we can perform addition and subtraction and indeed solve equations with complex numbers, we should define what it means for two complex numbers to be equal. We already saw that a complex number is made up of two parts: a real part and an imaginary part. π‘Ž and 𝑏 are the real and imaginary parts, respectively. And they are both part of the real number set.

Let’s say then we have two complex numbers π‘Ž plus 𝑏𝑖 and 𝑐 plus 𝑑𝑖. We want to know what it means for these two complex numbers to be equal. Well, in fact, it follows that their real parts must be equal and their imaginary parts must separately be equal. We can say then that π‘Ž plus 𝑏𝑖 equals 𝑐 plus 𝑑𝑖 if π‘Ž is equal to 𝑐 and 𝑏 is equal to 𝑑. In other words, two complex numbers are equal if their real parts are equal and separately their imaginary parts are equal. And of course, the equivalent is also true. If π‘Ž is equal to 𝑐 and 𝑏 is equal to 𝑑, for two complex numbers π‘Ž plus 𝑏𝑖 and 𝑐 plus 𝑑𝑖, then π‘Ž plus 𝑏𝑖 must be equal to 𝑐 plus 𝑑𝑖. Let’s consider our problem for which this definition might be useful.

If the complex numbers seven plus π‘Žπ‘– and 𝑏 minus three 𝑖 are equal, what are the values of π‘Ž and 𝑏?

Remember, for two complex numbers to be equal, their real parts must be equal and their imaginary parts must also be equal. And the beauty of this fact is it takes a problem about complex numbers and makes it purely about real numbers, since both the real parts and imaginary parts of each complex number must be real numbers.

Let’s have a look at the complex numbers seven plus π‘Žπ‘– and 𝑏 minus three 𝑖 then. The real part of the first complex number is seven, and the real part of our second complex number is 𝑏. The imaginary part of our first complex number is π‘Ž, and the imaginary part of our second complex number is negative three. It follows then that seven must be equal to 𝑏 and π‘Ž must be equal to negative three. Both negative three and seven are real numbers, which satisfies our criteria for the real and imaginary parts of a complex number. So for the complex numbers seven plus π‘Žπ‘– and 𝑏 minus three 𝑖 to be equal, π‘Ž must be equal to negative three and 𝑏 must be equal to seven.

And what about adding and subtracting complex numbers? Remember, a complex number is the result of adding a real number and an imaginary number. We can compare this a little to the idea of an algebraic expression like four plus seven π‘₯. This is the result of adding a number and a term in π‘₯. We could add, for example, four plus seven π‘₯ and another expression such as two plus five π‘₯ by individually adding the numbers to get six and adding the terms in π‘₯. That’s seven π‘₯ and five π‘₯, which is 12π‘₯.

We can add complex numbers in exactly the same way, remembering that the letter 𝑖 doesn’t actually represent a variable. But it’s the solution to the equation π‘₯ squared equals negative one. Let’s generalise this for complex numbers π‘Ž plus 𝑏𝑖 and 𝑐 plus 𝑑𝑖. Their sum is π‘Ž plus 𝑏𝑖 plus 𝑐 plus 𝑑𝑖. And we can add the real parts π‘Ž and 𝑐, and we get π‘Ž plus 𝑐, and then add their imaginary parts. That’s 𝑏 plus 𝑑. So we see that the sum of these two complex numbers is π‘Ž plus 𝑐 plus 𝑏 plus 𝑑 𝑖.

Their difference is π‘Ž plus 𝑏𝑖 minus 𝑐 plus 𝑑𝑖. This time, we add their real parts and we get π‘Ž minus 𝑐. And we add their imaginary parts. We get 𝑏. And then we distribute the brackets and we get negative 𝑑. So the difference is π‘Ž minus 𝑐 plus 𝑏 minus 𝑑 𝑖. So to add and subtract complex numbers, we add or subtract their real parts and separately add or subtract their imaginary parts.

In fact, what we’ve seen so far is that we can take a problem about complex numbers and turn it into a problem about real numbers by considering the real and imaginary parts. This is great because it means we can take the skills we already have for working with real numbers and extend them into working with complex numbers. Let’s see what this might look like.

What is negative nine plus seven plus four 𝑖 plus negative four minus four 𝑖 minus one plus three 𝑖?

Remember, we can add or subtract complex numbers by adding their real parts and separately adding their imaginary parts. Here we have four complex numbers. Now it might not look like it, but we could say that negative nine is actually a complex number. It’s negative nine plus zero 𝑖.

So to solve this problem, we’re going to work out the real part first. That’s negative nine plus seven minus four minus one, which is negative seven. Similarly, the imaginary parts are zero, four, negative four, and negative three.

Remember, we distribute this final set of parentheses. And a negative multiplied by a positive is a negative. That gives us negative three. So the real part of our solution is negative seven, and the imaginary part is negative three. So negative nine plus seven plus four 𝑖 plus negative four minus four 𝑖 minus one plus three 𝑖 is negative seven minus three 𝑖.

Now we also know that many algebraic properties can be extended into the concept of negative numbers. We could actually have collected like terms. Distributing that final set of parentheses and then rearranging slightly, we get negative nine plus seven plus negative four minus one plus four 𝑖 plus negative four 𝑖 minus three 𝑖, which once again gives us negative seven minus three 𝑖. This latter method, the one of collecting like terms, is generally the one that we use when adding and subtracting complex numbers. It’s useful to remember though that there are other methods that can work. Let’s see why.

If π‘Ÿ equals five plus two 𝑖 and 𝑠 equals nine minus 𝑖, find the real part of π‘Ÿ minus 𝑠.

Here we have two complex numbers to find as five plus two 𝑖 and nine minus 𝑖. We can see that the real part of π‘Ÿ is five and the real part of 𝑠 is nine. The imaginary part of π‘Ÿ is two and the imaginary part of 𝑠 is negative one. We’re being asked to find the real part of the difference between π‘Ÿ and 𝑠. And we could absolutely work out the entire answer to π‘Ÿ minus 𝑠 by collecting like terms. That’s five plus two 𝑖 minus nine minus 𝑖.

It’s important that we use these parentheses here because it reminds us that we’re subtracting everything inside these brackets, nine minus 𝑖. If we distribute these parentheses, we get five plus two 𝑖 minus nine plus 𝑖, since subtracting a negative is the same as adding a positive. Then we would simplify by collecting like terms. However, that’s probably a little more work than we really need to do.

In fact, we recall that, to subtract complex numbers, we simply subtract the real parts and then subtract the imaginary parts separately. We’re being asked to work out the real parts of the complex number π‘Ÿ minus 𝑠. So actually, we just need to subtract the real part of 𝑠 from the real part of π‘Ÿ. We can formalise this and say that the real part of π‘Ÿ minus 𝑠 equals the real part of π‘Ÿ minus the real part of 𝑠. We already saw that the real part of π‘Ÿ is five and the real part of 𝑠 is nine. Five minus nine is negative four. So the real part of π‘Ÿ minus 𝑠 in this case is negative four.

Now that we’ve established what it means for two complex numbers to be equal and learned how to add and subtract complex numbers, this will allow us to solve simple equations involving these types of numbers.

Determine the real numbers π‘₯ and 𝑦 that satisfy the equation five π‘₯ plus two plus three 𝑦 minus five 𝑖 equals negative three plus four 𝑖.

Let’s look carefully at what we’ve been given. We have been given two complex numbers that we’re told are equal to each other. Now I know it doesn’t look like it, but that expression to the left of the equal sign is indeed a complex number. Remember, a complex number is one of the form π‘Ž plus 𝑏𝑖, where π‘Ž and 𝑏 are real numbers. And we’re told that π‘₯ and 𝑖 are real numbers. And this means that the expression five π‘₯ plus two must be real and three 𝑦 minus five must be real. So five π‘₯ plus two plus three 𝑦 minus five 𝑖 is a complex number. It has a real part of five π‘₯ plus two and an imaginary part of three 𝑦 minus five.

Next, we’ll recall what it actually means for two complex numbers to be equal. We see that two complex numbers π‘Ž plus 𝑏𝑖 and 𝑐 plus 𝑑𝑖 are equal if π‘Ž is equal to 𝑐 and 𝑏 is equal to 𝑑. In other words, their real parts must be equal and their imaginary parts must separately be equal.

Let’s begin with the real parts in our question. We saw that the real part of the complex number on the left is five π‘₯ plus two. And on the right, it’s negative three. This means that five π‘₯ plus two must be equal to negative three. We’ll solve this as normal by applying a series of inverse operations. We’ll subtract two from both sides, and then we’ll divide through by five. And we see that π‘₯ is equal to negative one.

Let’s repeat this process for the imaginary parts. We said that the imaginary part for our number on the left is three 𝑦 minus five. And on the right, we can see it’s four. This means that three 𝑦 minus five must be equal to four. We can add five to both sides of this equation. And then we’ll divide through by three. And we see that 𝑦 must be equal to three. And we’ve solved the equation for π‘₯ and 𝑦. π‘₯ equals negative one and 𝑦 equals three.

In fact, it’s always sensible to check our answers by substituting them back into the equation and making sure that it makes sense. If we do, we get five multiplied by negative one plus two plus three multiplied by three minus five 𝑖. This does indeed give us negative three plus four 𝑖 as required. Our final example uses everything we’ve looked at in this video, with just a little more complexity.

Let 𝑧 one equal four π‘₯ plus two 𝑦𝑖 and 𝑧 two equal four 𝑦 plus π‘₯𝑖, where π‘₯ and 𝑦 are real numbers. Given that 𝑧 one minus 𝑧 two is equal to five plus two 𝑖, find 𝑧 one and 𝑧 two.

Let’s look carefully at what we’ve been given. We’ve been given two complex numbers in terms of π‘₯ and 𝑦. And we know these are complex numbers because we’re told that π‘₯ and 𝑦 are real numbers. That’s an important definition of a complex number. Both the real and imaginary parts of the complex numbers must be made up by real numbers. We’re also told that the difference between these two numbers is five plus two 𝑖.

Let’s recall: to subtract complex numbers, we simply subtract the real parts and then subtract the imaginary parts separately. This means that the real part of 𝑧 one minus 𝑧 two must be equal to the difference between the real parts of 𝑧 one and 𝑧 two. The real part of 𝑧 one minus 𝑧 two is five. The real part of 𝑧 one is four π‘₯, and the real part of our second complex number is four 𝑦. So five is equal to four π‘₯ minus four 𝑦.

Let’s repeat this process for our imaginary numbers. The imaginary part of the difference is two. The imaginary part of 𝑧 one is two 𝑦, and the imaginary part of 𝑧 two is π‘₯. So two equals two 𝑦 minus π‘₯. And now we see we have a pair of simultaneous equations in π‘₯ and 𝑦. We can use any method we’re comfortable with to solve these.

Now I think substitution lends itself quite nicely to these equations. Let’s rearrange this second equation to make π‘₯ the subject. We add π‘₯ to both sides and then subtract two. And we get π‘₯ equals two 𝑦 minus two. We then substitute this into our first equation. And we see that five is equal to four lots of our value of π‘₯, which is two 𝑦 minus two. And then we subtract that four 𝑦.

We distribute these parentheses by multiplying each term by four. And we see that five is equal to eight 𝑦 minus eight minus four 𝑦. Eight 𝑦 minus four 𝑦 is four 𝑦. We’ll solve this equation by adding eight to both sides to get 13 equals four 𝑦. And then we’ll divide through by four. And we see that 𝑦 is equal to 13 over four.

We could substitute this value back into any of our original equations. But it’s sensible to choose the rearranged form of the second equation. π‘₯ is equal to two multiplied by 13 over four minus two. Two multiplied by 13 over four is the same as 13 over two. And two is the same as four over two. 13 over two minus four over two is nine over two. And this is usually where we would stop.

But we’ve been asked to find the complex numbers 𝑧 one and 𝑧 two. So we need to substitute our values for π‘₯ and 𝑦 into each of these. We get 𝑧 one equals four multiplied by nine over two plus two multiplied by 13 over four 𝑖. That’s 18 plus 13 over two 𝑖. 𝑧 two is four multiplied by 13 over four plus nine over two 𝑖. This is 13 plus nine over two 𝑖.

And it’s sensible to check our answer by subtracting 𝑧 two from 𝑧 one and checking we do indeed get five plus two 𝑖. We subtract their real parts. 18 minus 13 is five, as required. And we subtract their imaginary parts. 13 over two minus nine over two is four over two, which simplifies to two. And the imaginary part is two as required.

In this video, we’ve seen that we can turn a problem about complex numbers into one involving real numbers by considering their real and imaginary parts. And we’ve seen that this can be useful because we know how to add, subtract, and equate with real numbers already. We’ve also learned that we can extend ideas about rules for algebraic expressions to help us work with complex numbers.

We’ve seen that two complex numbers are equal if individually their real components are equal and their imaginary parts are equal. And finally, we’ve learned that we can add and subtract complex numbers by adding or subtracting their real parts and adding or subtracting their imaginary parts.

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