### Video Transcript

In this video, we are going to look at how to perform division using complex numbers. So these are some of the types of questions that we might want to look at. So the first example, we are dividing a complex number by the complex number π β so something with a purely imaginary complex number. In the second example, we are looking at dividing two complex numbers, where both have nonzero real and imaginary parts. And in the third example, we might want to look at how to solve an equation. So we want to work out the value of π§, where π§ represents some complex number. And in order to do that weβre going to need to use division of complex numbers. So within the course of this video, weβll look at how to cover all three of these types of questions.

Before we dive into actually doing division with complex numbers, we just need a quick reminder about something called the complex conjugate of a complex number. And what this is is the complex number that is found by just changing the sign of the imaginary part of a complex number. So for example, the complex number three plus four π, if I wanted the complex conjugate of that, I would just change the sign of the imaginary part. So it would change from positive four π to negative four π. And the complex conjugate of that number would be three minus four π.

For the second example that we have on the screen, seven. This complex number is actually just a real number; its imaginary part is zero. So the complex conjugate of seven is also just seven because if I change the sign of plus zero π, well Iβve still just got plus zero π. For the third example, we have the complex number negative two π β so purely an imaginary number. If I change the sign of the imaginary part, I will have the complex number positive two π although I wouldnβt write the plus in this instance. For the fourth example, negative four take away six π, if I change the sign just of the imaginary part, remember I would now have the complex number negative four plus six π. And if we want to generalize, if we have the complex number π plus ππ, then the complex conjugate of that will be the complex number π minus ππ β so just changing the sign. So we need to know about complex conjugates in order to be able to do division.

Now letβs look at why these complex conjugates are helpful when we do division. And that actually comes from really useful property that arises when we multiply a complex number by its own conjugate. So what Iβve got on the screen here- Iβve got the complex number three plus four π and I want to multiply it by its conjugate three minus four π. So Iβll change the sign of the imaginary part. Now this is just a pair of double brackets. So I need to expand them using FOIL or whatever method you are most comfortable with. So if I go ahead and do that, I will get the following four terms: nine plus twelve π minus twelve π minus sixteen π squared. Now thereβre various different things that Iβm going to do with this in order to simplify it. But the key point is here. And Iβve got plus twelve π take away twelve π and those two parts are going to cancel each other out, leaving me with an imaginary part of zero. The other key fact that I need to remember is that π squared is equal to negative one. And so this is gonna be useful for simplifying this final term here.

So if π squared is equal to negative one, then negative sixteen times π squared becomes negative sixteen times negative one, and therefore itβs equal to positive sixteen. So I can replace that final term. Instead of negative sixteen π squared, I can replace it with positive sixteen. And then the key point that we mentioned, this positive twelve π and negative twelve π are just gonna cancel each other out directly. And so what am I left with? Iβm left with nine plus sixteen which gives an answer of twenty-five for this multiplication sum. And hereβs the key bit; that is a purely real number. It has no imaginary component whatsoever; itβs just twenty-five.

Now this isnβt a coincidence. It doesnβt just work for this particular pair of complex conjugates; this will always be the case. If you take a complex number and you multiply it by its conjugate, you will always end up with a purely real result. And in fact, we can go a step further and we can generalize this. So if I take the complex number π plus ππ and I multiply it by its own conjugate π minus ππ, I will always end up with the answer π squared plus π squared, which is a purely real number. And this will be true for any complex number that I do this with.

But now letβs get going with dividing some complex numbers. So the example weβre going to look at first of all, four plus π and Iβm dividing it by π. Now the key steps that we need is that we are going to multiply both the numerator and the denominator of this fraction by the complex conjugate of the denominator. So in the denominator, I have the complex number π; its complex conjugate is negative π. So Iβm gonna multiply by negative π over negative π. Now because the numerator and the denominator of that fraction are the same β itβs equivalent to one β and therefore Iβm not changing the sum at all, Iβm just multiplying by something thatβs equivalent to one. So now I need to go ahead and actually multiply these two fractions together. So in the numerator, I would have four plus π multiplied by negative π. So four times negative π is negative four π. And then positive π times negative π is negative π squared. And then the denominator π multiplied by negative π is also negative π squared.

Now in order to simplify this, I need to remember that key fact that π squared is equal to negative one. And so there are various different things that weβll simplify here. That negative four π will stay as it is for now. And negative π squared, well if π squared is negative one, then negative π squared will be positive one. So that becomes positive one there. And the same in the denominator, that will also become positive one. Now if weβre dividing by one, weβre not going to write it like that. So we would just simplify this to a final answer of one subtract four π. And Iβve written it that way round, so that itβs in the usual way that we used to see in complex numbers.

So to recap on what we did here, the key step is this first stage here, where we multiplied both the numerator and the denominator of the fraction by the complex conjugate of the denominator. And thatβs how weβre always going to start any question that involves division of complex numbers. We then work through the arithmetic in terms of multiplying out the numerator and multiplying the denominators and then simplifying the result. But that key step, remembering to multiply the numerator and the denominator by the complex conjugate of the denominator.

Now letβs look at an example, where the denominator has both a real and an imaginary part. So the process is gonna be exactly the same; weβre going to start off by multiplying both the numerator and the denominator by the complex conjugate of the denominator. So in this case, the denominator is three plus four π. The complex conjugate is therefore three minus four π. So there we are, multiplying by three minus four π over three minus four π. Now it may help to picture some brackets around each of these different parts because what we need to do now is multiply the numerators together and multiply the denominators together. And thatβs really just a case of expanding two pairs of double brackets.

So if I expand the brackets in the numerator, I will have four terms: six minus eight π plus three π minus four π squared. And if I do the same in the denominator, again four terms: nine plus twelve i minus twelve π minus sixteen π squared. Now weβve got various bits of simplification to do here. Remembering of course π squared is equal to negative one; thatβs going to be important in terms of simplifying. But also the fact that in the denominator, this plus twelve π and negative twelve π are going to cancel out. And remember thatβs not by accident; thatβs by design so that we end up with a real number in the denominator. So theyβre going to cancel each other out. And then all of the terms that involve an π squared can be replaced with negative one for that part.

So in the numerator, the negative four π squared is negative four times negative one. So it becomes plus four. And in the denominator, negative sixteen π squared, again negative sixteen times negative one it becomes plus sixteen. So we end up with this result here. The next stage is just to simplify each part. So in the numerator, I have six plus four giving a real part of ten. I have negative eight π plus three π giving an imaginary part of negative five π. And in the denominator, remember itβs all real: nine plus sixteen giving a real part of twenty-five.

Finally, to the numbers involved that are multiples of five, so I can divide three by five in order to simplify, which will give me two minus π in the numerator and five in the denominator. And that would be absolutely fine as a final answer or I could alternate it to be given as two-fifths minus one-fifth of π or two-fifths minus π over five, if there was a particular format requested. But either of those would be absolutely fine. Again the key step was this stage here where I multiplied both the numerator and the denominator of the fraction by the complex conjugate of the denominator.

Okay, these examples here Iβm not going to do in full, but weβll just do the first step. So the first one, weβre dividing by two π. So the first step would be to multiply by the complex conjugate of that denominator. So we would be multiplying by negative two π over negative two π. And then weβd go through the process as we did in the previous question. For example four again, I would be multiplying by the complex conjugate of that denominator. So changing the sign of the imaginary part, I would be multiplying by three plus two π over three plus two π. So in each case, itβs the complex conjugate that we need to be using.

One final application that we would like to consider is how you can use the division of complex numbers to solve an equation such as the one we have here. So I have π§ multiplied by one plus π equals four minus three π, and I want to work out what π§ is. So the first step to solving this equation will be to divide both sides of the equation by one plus π. So if I go ahead and do that, I will have the next stage of working out: π§ is equal to four minus three π divided by one plus π. And now this looks very similar to the example that weβve already looked at. So the next step in order to perform this division is going to be multiplying numerator and denominator by the complex conjugate of the denominator, so thatβs multiplying them both by one minus π. As discussed before, you may want to picture the brackets around each of these different parts so that youβll remember youβre multiplying out pairs of double brackets.

But if I go ahead and expand all of those double brackets, I will get to the stage here: four minus four π minus three π plus three π squared in the numerator and one plus π minus π minus π squared in the denominator. Now as usual, weβre gonna simplify it by using this key fact that π squared is equal to negative one. And as before, the plus π and the minus π in the denominator are gonna directly cancel out so that weβre left with a purely real part.

If I replace π squared with negative one everywhere, then in this last term in the numerator three π squared will become negative three because itβs three times negative one. And this term in the denominator negative π squared will become plus one because itβs minus minus one. Final step then is just to tidy up the result. So in the numerator, four take three is one for the real part. Negative four π take three π is negative seven π for the imaginary part. And in the denominator, one plus one is equal to two. And I can either leave it like that or I could write it as a half minus seven over two π, if I wanted to separate the real and imaginary parts out a bit more. But either of those two would be perfectly fine as final answers here.

So there you have it. The key to dividing complex numbers is about remembering complex conjugates and in particular multiplying both the numerator and denominator by the complex conjugate of whatever you have in the denominator of the division.