Video: Division of Complex Numbers

In this lesson, we will learn how to divide complex numbers.

17:41

Video Transcript

In this video, we’ll learn how to divide complex numbers. We’ll learn first how to divide a complex number by a real and then a purely imaginary number, before generalizing these techniques to allow us to divide a complex number by another complex number. We’ll then learn how to use these processes to solve equations involving the division of complex numbers.

Let’s begin by learning how to divide a complex number by a real number. This is very much an extension of the process we use to multiply complex numbers by real numbers. We multiply a complex number by a real number by using the distributive property and multiplying each part by the real number. We can think of division by a real number 𝑐 as the same as multiplying by the reciprocal of 𝑐, multiplying by one over 𝑐. One over 𝑐 multiplied by 𝑧 is 𝑧 divided by 𝑐. We multiply one over 𝑐 by π‘Ž and we get π‘Ž over 𝑐. And we multiply one over 𝑐 by 𝑏𝑖 and we get 𝑏 over 𝑐𝑖. We can see then that, to divide a complex number by a real number 𝑐, we simply divide the real part and then divide the imaginary part.

Let’s look at an example. Given that 𝑧 is equal to five plus three 𝑖, express 𝑧 over two in the form π‘Ž plus 𝑏𝑖.

We’ve been given a complex number, five plus three 𝑖, and we’re looking to work out the value or the complex number given by 𝑧 divided by two. 𝑧 divided by two is five plus three 𝑖 divided by two. To divide a complex number by a real number, we need to divide both the real and the imaginary part of the complex number by that real number. We divide five β€” that’s the real part β€” by two. And we divide the imaginary part, three, by two. So we see that, for our complex number, 𝑧 over two is the same as five over two plus three over two 𝑖.

And what about dividing a complex number by a purely imaginary number? Simplify two plus four 𝑖 over 𝑖.

To work out how to divide two plus four 𝑖 by 𝑖, we recall the definition of 𝑖. It’s a solution to the equation π‘₯ squared equals negative one. And we say that 𝑖 squared is equal to negative one. Or often 𝑖 is equal to the square root of negative one.

If we consider this fraction as two plus four 𝑖 divided by the square root of negative one, we can see that, to simplify, we’d need to perform the same process as rationalizing the denominator when we’re dealing with any other radical. We multiply both the numerator and the denominator of our fraction by the square root of negative one.

In fact, we know that the square root of negative one is 𝑖. So we’re going to be multiplying both the numerator and the denominator of this fraction by 𝑖. And we’re allowed to do that because multiplying by 𝑖 over 𝑖 is the same as multiplying by one. Essentially, we’re creating an equivalent fraction.

Let’s apply the distributive property for 𝑖 multiplied by two plus four 𝑖. 𝑖 multiplied by two is two 𝑖, and 𝑖 multiplied by four 𝑖 is four 𝑖 squared. Now of course 𝑖 squared is equal to negative one. So our expression becomes two 𝑖 plus four multiplied by negative one, which is negative four plus two 𝑖.

On the denominator, we have 𝑖 multiplied by 𝑖, which is of course 𝑖 squared, which is negative one. So we can rewrite two plus four 𝑖 over 𝑖 as negative four plus two one [𝑖] over negative one. And now we’re simply dividing by a real number. And to divide a complex number by a real number, we divide the real part and then separately divide the imaginary part. Negative four divided by negative one is four, and two 𝑖 divided by negative one is negative two 𝑖. And we fully simplify two plus four 𝑖 over 𝑖. It’s four minus two 𝑖.

In fact, we can use a similar technique to help us divide two complex numbers. Just as we’ve used the rules for rationalizing the denominator when that denominator is a radical, we can also apply the rules for rationalizing a denominator when that denominator is an expression involving a rational number and a radical.

Let’s consider an example. Simplify three minus six 𝑖 over one minus five 𝑖.

To simplify this fraction or divide three minus six 𝑖 by one minus five 𝑖, we need to find a way to get a real number as our denominator. So what can we do to get that real number? Well, remember, if we multiply a complex number by its complex conjugate β€” that’s found by changing the sign of the imaginary part β€” we end up with a real number. So for a complex number of the form π‘Ž plus 𝑏𝑖, its conjugate is π‘Ž minus 𝑏𝑖.

So if we multiply the numerator and denominator of this fraction by the complex conjugate of one minus five 𝑖, we’ll end up with a real denominator. And the conjugate of one minus five 𝑖 is one plus five 𝑖. We then distribute these parentheses as normal. There’s a number of different methods we could use.

Let’s look at the FOIL method. β€œF” stands for β€œfirst.” We multiply the first term in the first bracket by the first term in the second bracket. That’s three. β€œO” stands for β€œouter.” We multiply the outer terms. That’s 15𝑖. β€œI” stands for β€œinner.” We multiply the inner terms. That’s negative six 𝑖. And β€œL” stands for β€œlast.” We multiply the last term in each bracket. And this time, we get negative 30𝑖 squared.

Since we know that 𝑖 squared is equal to negative one, our last term becomes negative 30 multiplied by negative one, which is 30. And we can simplify our numerator to 33 plus nine 𝑖.

And we’ll repeat this process with the denominator. We get one plus five 𝑖 minus five 𝑖 minus 25𝑖 squared. And of course, 𝑖 squared is negative one. So we end up with our final term being positive 25. And then five 𝑖 minus five 𝑖 is zero. So we’re left with 26. And we have that real denominator we were looking for. So we’ve partly simplified our fraction. We now have 33 plus nine 𝑖 divided by 26.

To divide a complex number by a real number though, we can divide the real part and then separately divide the imaginary part. The real part of our answer is 33 over 26, and the imaginary part is nine over 26. So our fraction in its simplest form is 33 over 26 plus nine over 26𝑖.

And in fact, we could’ve saved ourselves a little bit of time by recalling the general formula for the product of a complex number with its conjugate. If the complex number is of the form π‘Ž plus 𝑏𝑖, we say that that can be found by π‘Ž squared plus 𝑏 squared. So in general, to divide a complex number by another complex number, we write it as a fraction. And we then multiply both the numerator and the denominator of this fraction by the conjugate of the denominator. We can then distribute and simplify as far as possible.

Let’s see what this looks like in general. One: Expand and simplify 𝑝 plus π‘žπ‘– multiplied by 𝑝 minus π‘žπ‘–. Two: Expand π‘Ž plus 𝑏𝑖 multiplied by 𝑝 minus π‘žπ‘–. Three: Hence, find a fraction which is equivalent to π‘Ž plus 𝑏𝑖 over 𝑝 plus π‘žπ‘– and whose denominator is real.

For part one of this question, we’re looking to multiply two complex numbers. We could absolutely apply any method for distributing brackets, such as the FOIL method or the grid method. However, if we look really carefully, we can see that these two complex numbers are conjugates of one another.

For a complex number of the form π‘Ž plus 𝑏𝑖, where π‘Ž is the real part and 𝑏 is the imaginary part, its conjugate is found by changing the sign of the imaginary part. And this is really useful. It allows us to use a formula for the product of a complex number with its conjugate. It’s π‘Ž squared plus 𝑏 squared. We square the real part and add it to the square of the imaginary part. The real part of our complex number is 𝑝, and the imaginary part is π‘ž. So the product of 𝑝 plus π‘žπ‘– with its conjugate 𝑝 minus π‘žπ‘– is 𝑝 squared plus π‘ž squared.

Unfortunately, there are no nice tricks that will allow us to multiply π‘Ž plus 𝑏𝑖 with 𝑝 minus π‘žπ‘–. We’ll use the FOIL method instead. We multiply the first terms. π‘Ž multiplied by 𝑝 is π‘Žπ‘. We multiply the outer terms, and we get π‘Žπ‘žπ‘–. Multiplying the inner terms gives us 𝑏𝑝𝑖. And clearing a little space to multiply the last terms, we get negative π‘π‘žπ‘– squared.

Now in fact, we should recall that 𝑖 squared is equal to negative one. And so this last term becomes positive π‘π‘ž. We’re going to rearrange this a little so it looks like a complex number. We add the real parts and we get π‘Žπ‘ plus π‘π‘ž. And we separately add the imaginary parts. And when we do, we find that the imaginary part of the distribution of these brackets is 𝑏𝑝 minus π‘Žπ‘ž. So the answer to part two is π‘Žπ‘ plus π‘π‘ž plus 𝑏𝑝 minus π‘Žπ‘žπ‘–.

And the final part is to find an equivalent fraction to π‘Ž plus 𝑏𝑖 over 𝑝 plus π‘žπ‘–. And of course, it’s no coincidence that we’ve been asked to do the working out that we already have. We want to create an equivalent fraction which has a real denominator. To achieve this, we multiply both the numerator and the denominator of our fraction by the complex conjugate of the denominator. And of course, we already evaluated these.

So we see that an equivalent fraction to π‘Ž plus 𝑏𝑖 over 𝑝 plus π‘žπ‘– whose denominator is real β€” and in fact the general form of π‘Ž plus 𝑏𝑖 divided by another complex number 𝑝 plus π‘žπ‘– β€” is π‘Žπ‘ plus π‘π‘ž plus 𝑏𝑝 minus π‘Žπ‘žπ‘– all over 𝑝 squared plus π‘ž squared. Now remember, whilst it’s all fine and well to derive this formula, it’s important to focus on applying the processes each time.

If π‘Ž plus 𝑏𝑖 equals negative three minus five 𝑖 over negative three plus five 𝑖, is it true that π‘Ž squared plus 𝑏 squared equals one?

We’ve been given the quotient of two complex numbers. And we’re told that this can be expressed as a single complex number π‘Ž plus 𝑏𝑖. In order to evaluate the expression π‘Ž squared plus 𝑏 squared, we’re going to need to work out what this single complex number actually is. To do so, we apply the processes for dividing complex numbers.

We need to multiply both the numerator and denominator of this fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the imaginary part. And if we do that, we see that the conjugate of our denominator is negative three minus five 𝑖.

We’re then going to multiply by applying the distributive property. Let’s begin with the numerator. We’re going to use the FOIL method. Negative three multiplied by negative three is nine. Multiplying the outer terms, we get 15𝑖. And indeed the inner terms gives us 15𝑖. And when we multiply the last two terms, we get 25𝑖 squared. But of course, 𝑖 squared is equal to negative one. So this last term is 25 multiplied by negative one, which is negative 25. We then add the real parts and separately add the imaginary parts, although we can think about this like collecting like terms. And we get the numerator of our fraction to be negative 16 plus 30𝑖.

Now we could repeat this process for the denominator. However, if we recall, for a general complex number of the form π‘Ž plus 𝑏𝑖 whose conjugate is of course π‘Ž minus 𝑏𝑖, the product of these two numbers is found by π‘Ž squared plus 𝑏 squared. The real part of our complex number is negative three, and the imaginary part is five. So when we distribute these parentheses, we’re going to get negative three squared plus five squared. That’s nine plus 25, which is 34. So negative three minus five 𝑖 over negative three plus five 𝑖 is negative 16 plus 30𝑖 over 34.

And of course, if we’re dividing a complex number by a real number, we divide the real parts and then the imaginary parts separately. Negative 16 over 34 simplifies to negative eight over 17. And 30 over 34 simplifies to 15 over 17. And so we can see that π‘Ž must be equal to negative eight over 17 and 𝑏 must be equal to 15 over 17.

All that’s left is to consider the sum of their squares, π‘Ž squared plus 𝑏 squared. That’s negative eight over 17 squared plus 15 over 17 squared. They’ll have the same denominator. So that’s 64 plus 225 over 289. But actually, 64 plus 225 is 289. So π‘Ž squared plus 𝑏 squared equals 289 over 289, which is of course one. And it is indeed true that π‘Ž squared plus 𝑏 squared equals one.

Now in fact, in this example, the two complex numbers we were dividing were conjugates of one another. It is actually a general rule that if we divide a complex number by its conjugate, the sum of the squares of the real and imaginary part of that new complex number will be one.

We’ll now consider how to solve equations which involve the division of complex numbers. Solve the equation 𝑧 multiplied by two plus 𝑖 equals three minus 𝑖 for 𝑧.

To solve this equation for 𝑧, we’ll need to apply inverse operations. We’ll begin by dividing both sides of this equation by two plus 𝑖. And we see that 𝑧 is equal to three minus 𝑖 divided by two plus 𝑖. To divide three minus 𝑖 by two plus 𝑖, we’re going to need to multiply both the numerator and the denominator of the fraction by the conjugate of two plus 𝑖. To find the conjugate, we change the sign of the imaginary part. And we see that the conjugate of two plus 𝑖 is two minus 𝑖.

We’ll distribute the parentheses at the top of this fraction by using the FOIL method. Three multiplied by two is six. Three multiplied by negative 𝑖 is negative three 𝑖. We then get negative two 𝑖. And our last term gives us 𝑖 squared. 𝑖 squared is of course negative one. So our last term is negative one.

And we can collect like terms or add the real parts and separately add the imaginary parts. And we see that three minus 𝑖 multiplied by two minus 𝑖 is five minus five 𝑖. And we could repeat this process for the denominator.

However, there is a special rule we can use to multiply a complex number by its conjugate. We can find the sum of the squares of the real and imaginary parts. The real part is two, and the imaginary part, the coefficient of 𝑖, is one. So the product of these two complex numbers is four plus one, which is five. And we see that 𝑧 is equal to five minus five 𝑖 over five.

We can then divide the real parts by this real number. We get five divided by five, which is one. And separately we divide the imaginary part by this real number. Five divided by five is one. So we get one minus 𝑖. And we’ve solved our equation for 𝑧. We’ve seen that division of complex numbers can be quite time-consuming.

We’ll look at one final example of where we can simplify our working somewhat. Simplify three minus four 𝑖 over two plus two 𝑖 plus three minus four 𝑖 over two minus two 𝑖.

In this question, we’re looking to find the sum of two fractions whose denominators and numerators are both complex numbers. We could apply the rules for dividing complex numbers and work from there. However, that’s quite a lengthy process, especially for two fractions. Instead, we notice that the numerator of each fraction is the same. And we can therefore rewrite this expression by taking out a factor of three minus four 𝑖. And we have three minus four 𝑖 multiplied by one over two plus two 𝑖 plus one over two minus two 𝑖.

Next, we’ll add these fractions by finding a common denominator. The common denominator is the product of these two numbers. It’s two plus two 𝑖 multiplied by two minus two 𝑖. And when we multiply the numerator of the first fraction by two minus two 𝑖, we get two minus two 𝑖. And for the numerator of the second fraction, we get two plus two 𝑖. So we’ll simplify this next.

For the numerator, negative two 𝑖 plus two 𝑖 is zero. So we’re simply left with four. And we won’t actually expand the brackets on the denominator. Instead, we use the fact that they are complex conjugates of one another. And we can find their product by finding the sum of the squares of the real parts and the imaginary parts. That’s two squared plus two squared, which is eight.

Now four over eight simplifies to one-half. So we need to find one-half of three minus four 𝑖. A half of the real part is three over two, and a half of the imaginary part is negative two. So our solution is three over two minus two 𝑖.

In this video, we’ve learned that we can divide complex numbers by using the same techniques as we do when rationalizing the denominator. And that’s by multiplying both the numerator and the denominator of the fraction by the conjugate of the denominator, expanding the brackets, and then simplifying. We’ve also seen that it can be useful to look for common factors to help simplify any more complicated expressions.

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