### Video Transcript

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.