Video: Distances And Midpoints on the Complex Plane

In this video, we will learn how to find the distance and midpoint of two complex numbers in the complex plane.

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Video Transcript

In this video, we’re going to learn about distances and midpoints on the complex plane. And in doing so, we’ll see some basic examples of how complex numbers can help solve geometric problems. Let’s jump straight in with an example.

What is the distance between the numbers negative two and six on the complex plane?

We see that we already have a complex plane or Argand diagram drawn for us, with the numbers negative two and six marked. Our question is what’s the distance between these two numbers on the complex plane. Well, negative two and six aren’t just any complex numbers. They are also real numbers. And so they lie on the real axis of the complex plane, which we can just think of as the normal real number line.

The distance is measured along this real number line. And we see that to get from negative two to six, we have to move two units to get to the zero and then a further six units to get to six, making a total of eight units. This is the distance between negative two and six on the complex plane. And it’s exactly the same as the distance between the real numbers negative two and six on the real number line.

Let’s now see an example involving imaginary numbers.

What is the distance between the numbers negative three 𝑖 and seven 𝑖 in the complex plane?

Let’s draw an Argand diagram to help us and mark negative three 𝑖 and seven 𝑖. Both numbers are purely imaginary and so lie on the imaginary axis of the complex plane. And so the distance between them is measured along this axis. To get from negative three 𝑖 to zero, you have to move three units up. And continuing from zero to seven 𝑖, you have to move a further seven units. So the total distance is three plus seven, which is 10.

Let’s now look at an example where the two numbers don’t lie on the same axis.

Find the distance between the complex numbers 𝑧 one and 𝑧 two shown on the complex plane. Give your answer in an exact simplified form.

First, let’s identify 𝑧 one and 𝑧 two. The real part of 𝑧 one is negative two. And its imaginary part is seven. So this is the complex number negative two plus seven 𝑖 as it’s represented by the point negative two, seven. We do the same thing for 𝑧 two. It turns out to be six minus three 𝑖, which is represented by the point six, negative three. And we’re looking for the distance between these two numbers on the complex plane. Recall that the distance between the points 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two on a coordinate plane is the square root of 𝑥 one minus 𝑥 two squared plus 𝑦 one minus 𝑦 two squared. We can substitute the coordinate to the points that corresponds to our complex numbers into this formula to find our distance.

We’re looking for the distance between negative two, seven and six, negative three. So 𝑥 one is negative two. And 𝑦 one is seven. 𝑥 two is six. And 𝑦 two is negative three. Substituting, we get the square root of negative two minus six squared plus seven minus negative three squared. Negative two minus six is negative eight. And seven minus negative three is 10. And negative eight squared is just eight squared. So the distance is the square root of eight squared plus 10 squared, which is the square root of 64 plus 100, in other words, the square root of 164. And 164 is two squared times 41. So in simplified surd form, this is two root 41.

We didn’t have to use the distance formula. We could use the Pythagorean theorem as well, drawing a right triangle on our diagram, counting squares to see that we have side lengths of eight and 10. These are the differences of the real and imaginary parts of our complex numbers, respectively. The Pythagorean theorem would then tell us that the length of the hypotenuse, which is the distance between the two complex numbers, is the square root of eight squared plus 10 squared, which is exactly what we got in this line of working here. The Pythagorean theorem is of course how the distance formula for points on a coordinate grid is proved.

In the context of the complex plane, these points represent complex numbers. And so we can rewrite our formula with this in mind. The distance between the complex numbers, 𝑧 one equals 𝑥 one plus 𝑦 one 𝑖 and 𝑧 two equals 𝑥 two plus 𝑦 two 𝑖, is the square root of 𝑥 one minus 𝑥 two squared plus 𝑦 one minus 𝑦 two squared. The only difference here is that we’re talking about the complex numbers 𝑥 one plus 𝑦 one 𝑖 and 𝑥 two plus 𝑦 two 𝑖, instead of the points 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two. This is what you get when you think about complex numbers as points on the complex plane. But we can also think of complex numbers as vectors. Let’s see what that approach leads to.

We’re now thinking about the complex numbers 𝑧 one, which is negative two plus seven 𝑖, and 𝑧 two, which is six minus three 𝑖, as vectors. And instead of just thinking of the distance between 𝑧 one and 𝑧 two, we consider this vector here, which I’ll call 𝑉. To go from the tail or initial point of 𝑉 to the tip or terminal point, you can travel by negative 𝑧 two to the origin. And then, 𝑧 one takes you where you want to go. 𝑉 is therefore negative 𝑧 two plus 𝑧 one or 𝑧 one minus 𝑧 two.

And of course, as a vector on the complex plane, it also represents a complex number, that’s being the complex number 𝑧 one minus 𝑧 two. The distance between the two complex numbers is the magnitude of the vector 𝑉, which is the modulus of the complex number 𝑉. And of course, 𝑉 as a complex number is just 𝑧 one minus 𝑧 two. We get another way of thinking about the distance between the complex numbers 𝑧 one and 𝑧 two then. This distance is the modulus of their difference.

Let’s finish off the problem using this method then. We know that 𝑧 one is negative two plus seven 𝑖. And 𝑧 two is six minus three 𝑖. Subtracting their real and imaginary parts, we get the modulus of negative eight plus 10𝑖. And using the formula for the modulus, we get the square root of negative eight squared plus 10 squared, which after simplification becomes two root 41.

It’s worth writing down our conclusions again. Pause and take a look if you’d like to. And we can see here how the modulus really does too for complex numbers what the absolute value function does for real numbers. The distance between two real numbers is the absolute value of their difference. The distance between two complex numbers is the modulus of their difference.

Let’s solve a final problem.

A complex number 𝑤 lies at a distance of five root two from 𝑧 one equals three plus five 𝑖 and at a distance of four root five from 𝑧 two equals negative six minus two 𝑖. Is the triangle formed by the points 𝑤, 𝑧 one, and 𝑧 two a right triangle?

So we have 𝑧 one equals three plus five 𝑖. And 𝑧 two equals negative six minus two 𝑖, which we can mark accurately on our Argand diagram or complex plane. However, it’s hard to guess where the complex number 𝑤 should go. All we know is that it lies at the distance of five root two from 𝑧 one and four root five from 𝑧 two. The question is whether the triangle with these vertices is a right triangle. And as we know, two of the lengths are suggested using the Pythagorean theorem.

If the square of the length of the longest side equals the sum of the squares of the other two sides, then this is a right triangle. But first, we need to find this longest side length, which we’ll call 𝑑. 𝑑 is the distance between the complex numbers 𝑧 one and 𝑧 two. And so it’s the modulus of 𝑧 one minus 𝑧 two. We substitute the known values of 𝑧 one and 𝑧 two, subtract the complex numbers to get nine plus seven 𝑖. Its modulus is the square root of nine squared plus seven squared, which is the square root of 130.

Now, we can apply the converse of the Pythagorean theorem. We need to identify the longest side then. Remember, our diagram might not be that accurate. We can unsimplify the other two side lengths to get root 50 and root 80, respectively. And so the length of the longest side really is 𝑑. We just need to check then whether 𝑑 squared equals the sum of squares of the other two side lengths. As 𝑑 is root 130, 𝑑 squared is 130. Five root two squared is five squared, which is 25 times two. And similarly, four root five squared is four squared, which is 16 times five. And it’s 130 equals 50 plus 80. Yes, it is. And so our triangle is a right triangle, with the right angle at 𝑤.

Find the midpoint of three plus five 𝑖 and seven minus 13𝑖.

We’re thinking of these complex numbers on Argand diagram or complex plane. We can mark these numbers on the Argand diagram and connect them with a line segment. We’re looking for the midpoint of this line segment. That’s the point on the line segment which divides the line segment into two equal halves. Now, I can use the fact that complex numbers on an Argand diagram behave like points. And we can recall from coordinate geometry that the midpoint of points 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two is the point 𝑥 one plus 𝑥 two over two, 𝑦 one plus 𝑦 one [𝑦 two] over two to see that the midpoint must have coordinates three plus seven over two, five minus 13 over two. And this corresponds to the complex number three plus seven over two plus five minus 13 over two 𝑖. We just need to simplify this number then. Three plus seven is 10. And five minus 13 is negative eight. So we see the midpoint represents the complex number five minus four 𝑖.

We can take our fact from coordinate geometry and write it in the notation of complex numbers. So the point 𝑥 one, 𝑦 one becomes a complex number 𝑧 one equals 𝑥 one plus 𝑦 one 𝑖. And the point 𝑥 two, 𝑦 two becomes the complex number 𝑧 two equals 𝑥 two plus 𝑦 two 𝑖. And the midpoint becomes 𝑥 one plus 𝑥 two over two plus 𝑦 one plus 𝑦 two over two 𝑖. Why is this interesting? It turns out we can rearrange this somewhat. Combining the fractions and rearranging two of the terms in the numerator, we see that we get 𝑧 one plus 𝑧 two over two. The midpoint of two complex numbers on an Argand diagram is simply their arithmetic mean. This generalises the fact that the midpoint of two real numbers on the real number line is represented by their arithmetic mean.

Moving to complex number notations makes the statement of the fact about midpoints cleaner. We don’t have to say that the midpoint is a point whose coordinates are the arithmetic means of the coordinates of the two points. We just have to say that the midpoint is the arithmetic mean of the two points.

Let’s see a quick application.

Let 𝑧 one, 𝑚, and 𝑧 two be complex numbers such that 𝑚 lies at the midpoint of the line segment connecting 𝑧 one to 𝑧 two. Given that 𝑧 two equals four plus five 𝑖 and 𝑚 equals negative 12 plus 20𝑖, find 𝑧 one.

Well, we could draw an Argand diagram and reason geometrically. But there’s another way. We know that the midpoint 𝑚 is the arithmetic mean of the complex numbers 𝑧 one and 𝑧 two. And we can rearrange this equation to find 𝑧 one in terms of 𝑚 and 𝑧 two. We multiply both sides by two, subtract 𝑧 two from both sides, and swap the sides to find that 𝑧 one is two 𝑚 minus 𝑧 two. We know the values of 𝑚 and 𝑧 two. And so we substitute them. We distribute the two and the minus sign and simplify to find that 𝑧 one is negative 28 plus 35𝑖.

Before we see our final example, let’s consider a generalisation of the midpoint.

The midpoint of two complex numbers 𝑧 one and 𝑧 two divides the line segment connecting them into two equal parts. But what if we didn’t want those parts to be equal. What if, instead, we wanted to divide the line segment in the ratio one to two. How would we find the complex number 𝑤 which corresponded to the point, which divides the line segment in this ratio. The trick is to use vectors. We think of the position vectors of 𝑧 one and 𝑧 two. And we also think of the line segment connecting 𝑧 one and 𝑧 two as a vector. What is this vector? Well, we can get from its tail to its tip by going opposite to 𝑧 one and then along the vector 𝑧 two. This is therefore the vector negative 𝑧 one plus 𝑧 two or, equivalently, 𝑧 two minus 𝑧 one.

We want to find the position vector of 𝑤. And we see that we can reach 𝑤 by going along the vector 𝑧 one and then part of the way along the vector 𝑧 two minus 𝑧 one. This vector that we add to 𝑧 one is some multiple of the long green vector 𝑧 two minus 𝑧 one. But which multiple? Well, we can rewrite the ratio as a third to two-thirds. And then, in total, we have one. It’s now not hard to see that we need to add a third of 𝑧 two minus 𝑧 one to 𝑧 one. And we can distribute the one-third to get two-thirds 𝑧 one plus one-third 𝑧 two.

We can divide the line segment in any other ratio we like, the arbitrary ratio 𝑎 to 𝑏, for example. This ratio can be rewritten so that the sum of the numbers is one. And so if 𝑎 over 𝑎 plus 𝑏 is 𝑘, then 𝑏 over 𝑎 plus 𝑏 is one minus 𝑘. Writing this way, we find that 𝑤 is 𝑧 one plus 𝑘 times 𝑧 two minus 𝑧 one or one minus 𝑘 times 𝑧 one plus 𝑘 times 𝑧 two. Setting 𝑘 equal to a half, we get the midpoint. And so we see that we have a generalisation of the midpoint. You may also recognise the equation from the vector equation of a straight line if you’ve studied it. In that context, 𝑘 is not constrained to be between zero and one but can take any real value. Let’s now use this generalisation of the midpoints to solve a problem in geometry.

A triangle has vertices at points 𝑎, 𝑏, and 𝑐 in the complex plane. Find an expression for the centroid of the triangle in terms of 𝑎, 𝑏, and 𝑐. You can use the fact that the centroid divides the median in the ratio two to one.

Let’s draw an arbitrary triangle in the complex plane with vertices 𝑎, 𝑏, and 𝑐. We’re looking for the centroid of that triangle. And we use the fact that it divides any median of the triangle in the ratio two to one. So what is a median of the triangle? It’s the line segment between a vertex of the triangle and the midpoint of the side opposite that vertex. So if we take the vertex 𝑎, then we need to find the midpoint of the side opposite. So I think it’s that here. And connecting the two points, we get a median. To find the centroid, we use the fact that it divides any median in the ratio two to one. So the centroid is about here. It’s twice as far from the vertex as it is from the midpoint of the opposite side.

Now, we want to find an expression for this centroid in terms of 𝑎, 𝑏, and 𝑐. How we’re going to do that? Well, we know that the midpoint of the complex numbers 𝑏 and 𝑐 is just their arithmetic mean, 𝑏 plus 𝑐 over two. Let’s call this 𝑚 for simplicity. And the centroid divides the line segment from 𝑎 to 𝑚 in the ratio two-thirds to one-third. So we can get to it by going to 𝑚 and then going a third of the way from 𝑚 to 𝑎. Using what we know about 𝑚, we can write the centroid in terms of 𝑎, 𝑏, and 𝑐. Now, we just need to simplify.

We multiply the numerator and denominator of the second fraction by two to get the fraction with denominator six. And so we write the first fraction over six as well. We can then combine the fractions. And in doing so, we collect some like terms in the numerator. Finally, we cancel the common factor of two in the numerator and denominator and rearrange some of the terms in the numerator to get that the centroid is 𝑎 plus 𝑏 plus 𝑐 over three. Using complex numbers then with simple methods, we get this elegant result.

To summarise, the distance 𝑑 between two complex numbers 𝑧 one equals 𝑥 one plus 𝑦 one 𝑖 and 𝑧 two equals 𝑥 two plus 𝑦 two 𝑖 can be expressed in terms of the modulus of a complex number as 𝑑 equals the modulus of 𝑧 one minus 𝑧 two, which is equivalent to the square root of 𝑥 one minus 𝑥 two squared plus 𝑦 one minus 𝑦 two squared. The midpoint 𝑚 of two complex numbers 𝑧 one and 𝑧 two lies at their arithmetic mean, that is, 𝑚 equals 𝑧 one plus 𝑧 two over two. And we can generalise this to find the point 𝑤 that is some fraction between zero and one along the line segment from 𝑧 one to 𝑧 two. It’s given by 𝑤 equals one minus 𝑘 times 𝑧 one plus 𝑘 times 𝑧 two. And if 𝑘 is greater than one, then 𝑤 lies on the line extended past 𝑧 two. And if 𝑘 is less than zero, 𝑤 lies on the line extended past 𝑧 one. These were the key points that we used in this video to solve geometric problems using complex numbers in a simple and elegant way.

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