### Video Transcript

In this video, weβre going to learn
about distances and midpoints on the complex plane. And in doing so, weβll see some
basic examples of how complex numbers can help solve geometric problems. Letβs 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 whatβs the distance
between these two numbers on the complex plane. Well, negative two and six arenβt
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 itβs exactly the same as the
distance between the real numbers negative two and six on the real number line.

Letβs now see an example involving
imaginary numbers.

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

Letβs draw an Argand diagram to
help us and mark negative three π and seven π. 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 π to
zero, you have to move three units up. And continuing from zero to seven
π, you have to move a further seven units. So the total distance is three plus
seven, which is 10.

Letβs now look at an example where
the two numbers donβt lie on the same axis.

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

First, letβs identify π§ one and π§
two. The real part of π§ one is negative
two. And its imaginary part is
seven. So this is the complex number
negative two plus seven π as itβs represented by the point negative two, seven. We do the same thing for π§
two. It turns out to be six minus three
π, which is represented by the point six, negative three. And weβre looking for the distance
between these two numbers on the complex plane. Recall that the distance between
the points π₯ one, π¦ one and π₯ two, π¦ two on a coordinate plane is the square
root of π₯ one minus π₯ two squared plus π¦ one minus π¦ two squared. We can substitute the coordinate to
the points that corresponds to our complex numbers into this formula to find our
distance.

Weβre looking for the distance
between negative two, seven and six, negative three. So π₯ one is negative two. And π¦ one is seven. π₯ two is six. And π¦ 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 didnβt 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, π§ one equals π₯ one plus π¦ one π and π§ two equals π₯ two plus π¦ two
π, is the square root of π₯ one minus π₯ two squared plus π¦ one minus π¦ two
squared. The only difference here is that
weβre talking about the complex numbers π₯ one plus π¦ one π and π₯ two plus π¦ two
π, instead of the points π₯ one, π¦ one and π₯ two, π¦ 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. Letβs see what that approach leads
to.

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

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

Letβs finish off the problem using
this method then. We know that π§ one is negative two
plus seven π. And π§ two is six minus three
π. Subtracting their real and
imaginary parts, we get the modulus of negative eight plus 10π. 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.

Itβs worth writing down our
conclusions again. Pause and take a look if youβd 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.

Letβs solve a final problem.

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

So we have π§ one equals three plus
five π. And π§ two equals negative six
minus two π, which we can mark accurately on our Argand diagram or complex
plane. However, itβs hard to guess where
the complex number π€ should go. All we know is that it lies at the
distance of five root two from π§ one and four root five from π§ 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 weβll call π. π is the distance between the
complex numbers π§ one and π§ two. And so itβs the modulus of π§ one
minus π§ two. We substitute the known values of
π§ one and π§ two, subtract the complex numbers to get nine plus seven π. 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 π. We just need to check then whether
π squared equals the sum of squares of the other two side lengths. As π is root 130, π 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 itβs 130 equals 50 plus 80. Yes, it is. And so our triangle is a right
triangle, with the right angle at π€.

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

Weβre 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. Weβre looking for the midpoint of
this line segment. Thatβs 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 π₯ one, π¦ one and π₯ two, π¦ two is the point
π₯ one plus π₯ two over two, π¦ one plus ~~π¦ one~~ [π¦ 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 π. 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 π.

We can take our fact from
coordinate geometry and write it in the notation of complex numbers. So the point π₯ one, π¦ one becomes
a complex number π§ one equals π₯ one plus π¦ one π. And the point π₯ two, π¦ two
becomes the complex number π§ two equals π₯ two plus π¦ two π. And the midpoint becomes π₯ one
plus π₯ two over two plus π¦ one plus π¦ two over two π. 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 π§ one plus π§ 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 donβt 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.

Letβs see a quick application.

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

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

Before we see our final example,
letβs consider a generalisation of the midpoint.

The midpoint of two complex numbers
π§ one and π§ two divides the line segment connecting them into two equal parts. But what if we didnβt 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 π€ 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
π§ one and π§ two. And we also think of the line
segment connecting π§ one and π§ two as a vector. What is this vector? Well, we can get from its tail to
its tip by going opposite to π§ one and then along the vector π§ two. This is therefore the vector
negative π§ one plus π§ two or, equivalently, π§ two minus π§ one.

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

We can divide the line segment in
any other ratio we like, the arbitrary ratio π to π, for example. This ratio can be rewritten so that
the sum of the numbers is one. And so if π over π plus π is π,
then π over π plus π is one minus π. Writing this way, we find that π€
is π§ one plus π times π§ two minus π§ one or one minus π times π§ one plus π
times π§ two. Setting π 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 youβve studied it. In that context, π is not
constrained to be between zero and one but can take any real value. Letβs now use this generalisation
of the midpoints to solve a problem in geometry.

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

Letβs draw an arbitrary triangle in
the complex plane with vertices π, π, and π. Weβre 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? Itβs the line segment between a
vertex of the triangle and the midpoint of the side opposite that vertex. So if we take the vertex π, then
we need to find the midpoint of the side opposite. So I think itβs 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. Itβs 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 π, π, and π. How weβre going to do that? Well, we know that the midpoint of
the complex numbers π and π is just their arithmetic mean, π plus π over
two. Letβs call this π for
simplicity. And the centroid divides the line
segment from π to π in the ratio two-thirds to one-third. So we can get to it by going to π
and then going a third of the way from π to π. Using what we know about π, we can
write the centroid in terms of π, π, and π. 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 π plus π plus π over three. Using complex numbers then with
simple methods, we get this elegant result.

To summarise, the distance π
between two complex numbers π§ one equals π₯ one plus π¦ one π and π§ two equals π₯
two plus π¦ two π can be expressed in terms of the modulus of a complex number as
π equals the modulus of π§ one minus π§ two, which is equivalent to the square root
of π₯ one minus π₯ two squared plus π¦ one minus π¦ two squared. The midpoint π of two complex
numbers π§ one and π§ two lies at their arithmetic mean, that is, π equals π§ one
plus π§ two over two. And we can generalise this to find
the point π€ that is some fraction between zero and one along the line segment from
π§ one to π§ two. Itβs given by π€ equals one minus
π times π§ one plus π times π§ two. And if π is greater than one, then
π€ lies on the line extended past π§ two. And if π is less than zero, π€
lies on the line extended past π§ 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.