### Video Transcript

In this video, we will learn how to use the conjugate of a complex number to evaluate expressions. We will begin by defining what we mean by the complex conjugate, before considering their properties and how these can be exploited to help us solve equations involving complex numbers. Throughout this video, we will look to, where possible, derive general results that can be used in more complicated complex number problems.

Every single complex number has associated with it another complex number, known as its conjugate. The definition of the word conjugate is having features in common but opposite or inverse in some particular. Outside of mathematics, it can mean to juxtapose or join together, indicating that a complex number and its conjugate have a special relationship. Letβs look at the definition. For a complex number of the form π plus ππ, its conjugate denoted by π§ bar or π§ star is π minus ππ. In laymanβs terms, the conjugate of a complex number is found by changing the sign for the imaginary part of the number.

For example, a complex number given as three plus two π β weβll have a complex conjugate of three minus two π. Similarly, a complex number, four minus six π β weβll have a conjugate of four plus six π. And in fact, the complex conjugate of the conjugate is four minus six π. And we can see that we can generalize this and say that the complex conjugate of the conjugate is simply the original number. Itβs π§.

And what about a purely real number? Will this have a conjugate? Well yes. In fact, we can say that a real number is of the form π. This is actually a complex number. But itβs one of the form π plus zero π. Its imaginary part is zero. Since we change the sign of the imaginary part to find the conjugate, the conjugate of this number will be π minus zero π. But of course, this is still π. So the conjugate of the real number is simply that number. Another beauty of the conjugate is that it shares all of the same properties as any other complex number. Itβs distributive over addition and multiplication. And weβre going to see now what that might look like.

If π equals eight plus two π, what is π plus π star?

Remember, π star is the conjugate of the complex number π , given by eight plus two π. We can say that the complex conjugate of a number π§, given by π plus ππ, is π§ star equals π minus ππ. Our complex number has a real part of eight and an imaginary part of two. This means its complex conjugate is eight minus two π. And this also means that the sum of the two numbers is eight plus two π plus eight minus two π. And we add these by adding their real parts and separately adding their imaginary parts, which we often think of like collecting like terms. Eight plus eight is 16. And two π minus two π is zero. And we see that π plus π star is 16.

Notice how the sum of a complex number and its conjugate is just a real number. And this makes a lot of sense if we consider the general form. π§ plus π§ star is π plus ππ plus π minus ππ. ππ minus π π is zero. And we therefore see that the sum of a complex number with its complex conjugate is simply two π. Or alternatively, we might say that the sum of a complex number and its conjugate is two lots of the real part of that number.

Similarly, their difference is π plus ππ minus π minus ππ. We distribute the second bracket by multiplying each part by negative one. And we get π plus ππ minus π plus ππ. This time, π minus π is zero, which is equal to two ππ. We can therefore say that the difference between a complex number and its conjugate is two π multiplied by the imaginary part of that complex number. Weβll now look at a detailed example of an equation involving the sum and difference of a complex number and its conjugate.

Find the complex number π§ which satisfies the following equations. π§ plus π§ star is equal to negative five. π§ star minus π§ is equal to three π.

Remember, π§ star represents the conjugate of the complex number π§. π§ will be of the form π plus ππ, where π and π are real numbers. And π§ star will be of the form π minus ππ. We changed the sign of the imaginary part. The first equation tells us the sum of these two numbers. We can form an expression for their sum by using the general form of the complex number. Itβs π plus ππ plus π minus ππ. This simplifies to two π or two multiplied by the real part of π§.

Now in fact, we know that the sum of these numbers is negative five. So we can say that negative five is equal to two multiplied by the real part of π§. And we solve by dividing through by two. And we see that the real part of π§ is equal to negative five over two.

The second equation tells us the difference of these two numbers. Thatβs π minus ππ minus π plus ππ. We distribute the second bracket by multiplying each term by negative one. And we get π minus ππ minus π minus ππ, which is negative two ππ. This is equal to negative two π multiplied by the imaginary part of π§. And of course, we know that this is actually equal to three π. So we see that three π is equal to negative two π multiplied by the imaginary part of π§. To solve this equation, we divide through by negative two π. And since π divided by π is one, we see that the imaginary part of π§ is negative three over two.

Itβs useful to spot that we could alternatively have multiplied through by negative one. That wouldβve given us π§ minus π§ star is equal to negative three π. But this wouldβve resulted in the same solution. So we know that the complex number π§, which satisfies the two equations given, has a real part of negative five over two and an imaginary part of negative three over two. So this solution is negative five over two minus three over two π. You should now be able to see why learning the definition of the sum and difference of a complex number and its conjugate can be a real time saver. Next, weβll consider the product of a complex number and its conjugate.

Find the complex conjugate of one plus π and the product of this number with its complex conjugate.

Remember, the complex conjugate is found by changing the sign of the imaginary part of the complex number. This means that the complex conjugate of one plus π is one minus π. And we want to find the product of one plus π and one minus π. We find the product of these two numbers just as we would with any two binomials.

The FOIL method can be a nice way to do this. F stands for first. We multiply the first term in the first bracket by the first term in the second bracket. One multiplied by one is one. O stands for outer. We multiply the outer two terms. One multiplied by negative π is negative π. I stands for inner. We multiply the inner terms. And π multiplied by one is π. And finally, L stands for last. We multiply the last term in each bracket. π multiplied by negative π is negative π squared. And of course, π squared is equal to negative one. Since negative π plus π is zero, this becomes one minus negative one, which is two. And the product of this number with its complex conjugate is two. We can generalize this result. And weβll soon see that there are a number of applications for the complex conjugate.

Let π plus ππ be a complex number whose conjugate is π minus ππ. Their product is π plus ππ multiplied by π minus ππ. If we expand these brackets as before, we get π squared minus πππ plus πππ minus π squared π squared. And of course, negative πππ plus πππ is zero. And you might notice this is just like the difference of two squares. Weβll substitute negative one for π squared. And we see that the product of this complex number with its conjugate is π squared plus π squared. So π§ multiplied by π§ star is π squared plus π squared. For our last two examples, weβll consider some more complicated questions involving sums and products of complex numbers and their conjugates.

Consider π§ equals five minus π root three and π€ equals root two plus π root five. Part one, calculate π§ star and π€ star. Part two, find π§ star plus π€ star and π§ plus π€ star. Part three, find π§ star π€ star and π§π€ star.

In this question, weβve been given two complex numbers. And we need to find their conjugates. Remember, to find the conjugate of a complex number, we change the sign of its imaginary part. This means that the complex conjugate of five minus π root three is five plus π root three. Now, donβt worry that the π is in front of the root three here. Yes, thatβs not of the general form. But itβs a sensible way to write it when weβre dealing with roots.

If we, instead, chose to write root three multiplied by π, it could look a little bit like weβre finding the square root of three π rather than the square root of three multiplied by π. Next, we see since π€ is root two plus π root five, its conjugate is root two minus π root five.

For part two, we need to calculate two numbers. We need to find the sum of the conjugates. And we need to find the conjugate of the sum of the original complex numbers. Letβs begin by finding the sum of their conjugates. Thatβs five plus π root three plus root two minus π root five. We can find their sum by collecting like terms. And when we do, we see that π§ star plus π€ star is five plus root two plus π multiplied by root three minus root five. And we can also work out π§ plus π€ star.

This time, we add π§ and π€ first before finding the conjugate. Thatβs the conjugate of five minus π root three plus root two plus π root five. Once again, we do this by adding the real parts and then the imaginary parts. We get five plus root two plus π multiplied by negative root three plus root five. So we change the sign of the imaginary part to find the conjugate. And we get five plus root two minus π multiplied by negative root three plus root five. And in fact, if we multiply the imaginary part by negative one, we see that this is equal to five plus root two plus π multiplied by root three minus root five. Notice that π§ star plus π€ star is actually the same as π§ plus π€ star.

For this third part, we need to find the product of the conjugate of π§ and π€. Thatβs five plus π root three multiplied by root two minus π root five. Multiplying the first term in each bracket, we get five root two. Multiplying the outer terms, we get negative five π root five. Multiplying the inner terms, thatβs π root three multiplied by root two, gives us π root six. And multiplying the last terms gives us π root three multiplied by negative π root five, which is negative π squared multiplied by root 15. And since π squared is equal to negative one, this last term becomes positive root 15. Collecting like terms, we see that the product of the conjugate of π§ and π€ is five root two plus root 15 plus π multiplied by root six minus five root five.

Next, we find the product of π§ and π€ and then find their conjugate. This time, their product is five minus π root three multiplied by root two plus π root five. Expanding these brackets, and we get five root two plus root 15 minus π multiplied by root six minus five root five. And it follows that the conjugate of this number is five root two plus root 15 plus π multiplied by root six minus five root five, once again.

Weβve seen in this example that, for complex numbers π§ and π€, the sum of their conjugates is equal to the conjugate of their sum. And weβve also seen that the product of their conjugates is equal to the conjugate of their product. And this in fact is a general rule, which holds for all complex numbers.

Solve two π§ minus π§ bar equals five in β.

Here we have a complex number. And we can say that that could be of the form π plus ππ, where π and π are real numbers. π§ bar is its conjugate. Thatβs π minus ππ. And β is used to denote the set of complex numbers. Letβs substitute π§ and π§ bar into our equation.

When we do, we see that two multiplied by π plus ππ minus π minus ππ equals five. Then, we distribute the brackets by multiplying the real and imaginary part by the number on the outside. For the first bracket, thatβs two multiplied by π and two multiplied by ππ. And for the second bracket, thatβs negative one multiplied by π plus negative one multiplied by negative ππ. So we get two π plus two ππ minus π plus ππ equals five. And of course, we can collect like terms. And we see that π plus three ππ is equal to five.

Now, this number is purely real. Or we can say itβs a complex number whose imaginary part is zero. And once weβve identified that, we can equate the real and imaginary parts. We see that π must be equal to five. And three π must be equal to zero. In fact, if three π is equal to zero, π must also be equal to zero. Weβre solving for π§. And weβve established that π β its real part is equal to five. And π β its imaginary part is equal to zero. So we could say that π§ is equal to five plus zero π though we donβt need to write the imaginary part. So we say that π§ is simply equal to five.

In this video, weβve learned that a complex number of the form π plus ππ has a complex conjugate π minus ππ. And this is often denoted by π§ star or sometimes π§ bar. Weβve also seen that, for two complex numbers π§ one and π§ two, there exist a whole set of rules that relate the complex numbers with their conjugates. And finally, weβve learned that a complex number is equal to its conjugate if and only if its imaginary part is zero, in other words, if itβs a real number.