### Video Transcript

In this video, weβre going to learn
how to draw and interpret loci in the complex plane expressed in terms of the
modulus. Weβll begin by looking at geometry
in the complex plane before considering loci and the equations of a number of
different loci. These will include circular loci,
perpendicular bisectors, and elliptical loci.

We begin by recalling some
definitions. The locus of a point said in the
complex plane is the set of all points said, which satisfy a particular
condition. We also know that the modulus of a
complex number is the distance of the point representing that complex number plotted
on the Argand plane from the origin. For a complex number in algebraic
form, π§ equals π plus ππ, its modulus is the square root of π squared plus π
squared. In our first example, weβll simply
consider the geometry of the complex plane.

A complex number π€ lies at a
distance of five root two from π§ one, which is nine over two plus seven over two
π, and at a distance of four root five from π§ two, which is negative nine over two
minus seven over two π. Does the point π€ lie on the circle
centred at the origin that passes through π§ one and π§ two?

To answer this question, weβll
begin by considering the properties of circles. The complex numbers π§ one and π§
two can be plotted on an Argand diagram as shown. π§ one is represented by the point
whose cartesian coordinates are nine over two and seven over two. And similarly, π§ two has cartesian
coordinates negative nine over two and negative seven over two.

Notice how π§ one is equal to
negative π§ two. So they are diametrically
opposite. And the line segment that joins π§
one to π§ two must form the diameter of the circle. We can use circle theorems to
deduce that this means that π€ will lie on the circle if the triangle formed by π§
one, π§ two, and π€ is a right-angled triangle.

To check whether this is a
right-angled triangle, we can use the Pythagorean theorem to check whether the
lengths of the sides given as π, π, and π satisfy the formula π squared plus π
squared equals π squared, where π is of course the longest side, the hypotenuse,
in that triangle.

We already know that π€ is five
root two units away from π§ one and that itβs four root five units away from π§
two. So weβll need to find the length of
the third side of the triangle. Thatβs the length of the side
joining π§ one and π§ two. The distance of the length of this
side will be the modulus of the difference between these two complex numbers. Thatβs the modulus of π§ one minus
π§ two.

To find their difference, we simply
find the difference between their real parts and their imaginary parts. The real parts, thatβs nine over
two minus negative nine over two. And for the imaginary parts, thatβs
seven over two minus negative seven over two. And this means that the difference
between π§ one and π§ two is nine plus seven π.

Remember, weβre looking to find the
modulus of the difference between these two numbers. So thatβs the square root of nine
squared plus seven squared, which is the square root of 130.

Now that we know the lengths of
these three sides, we can check whether they satisfy the Pythagorean theorem. Root 130 is longer than five root
two and four root five. So weβre going to find the sum of
the squares of the two shorter lengths. Four root five squared is 50, and
five root two squared is 80. 50 plus 80 is 130, which is indeed
equal to the square of root 130.

Weβve seen that these three sides
satisfy the Pythagorean theorem and therefore form the sides of a right-angled
triangle. This in turn means that the line
joining π§ one and π§ two form the diameter of a circle for which π€ lies on the
circumference as required.

Letβs look to generalise this
idea. We saw that the distance between
two points is given by the modulus of their difference. We can therefore say that, for a
given constant complex number π§ one, the locus of a point π§, which satisfies the
equation the modulus of π§ minus π§ one is equal to π, is a circle centred at π§
one of radius π. Letβs now consider an example of
the application of this definition.

A complex number π§ satisfies the
equation the modulus of π§ minus two plus three π is equal to two. 1) Describe the locus of π§ and
give it a cartesian equation. 2) Find the range of the argument
of π§ in the interval negative π to π. 3) Find the range of the modulus of
π§.

Remember, for a constant complex
number π§ one, the locus of a point π§, which satisfies the equation given, is a
circle centred at π§ one with a radius π. To answer part one then, weβll
rewrite π§ minus two plus three π slightly by factoring negative one. And we get π§ minus two minus three
π. This means that since our complex
number π§ satisfies this equation, its locus is a circle with a radius of two, whose
centre is at two minus three π.

And there are two ways we can give
this a cartesian equation. We could substitute π§ equals π₯
plus π¦π into the given equation. Alternatively, we recall the
cartesian equation for a circle centred at ππ with a radius of π. Itβs π₯ minus π all squared plus
π¦ minus π all squared equals π squared.

Taking π₯ to be the real part and
π¦ to be the imaginary part, we know that, for our circle, the radius is two, π is
equal to two, and π is equal to negative three. Substituting these into the
formula, and we get π₯ minus two all squared plus π¦ minus negative three all
squared equals two squared. So the cartesian equation
simplifies to π₯ minus two squared plus π¦ plus three squared equals four.

And now we consider part two. Find the range of the argument of
π§ in the interval negative π to π.

Weβll begin by drawing the locus
given on an Argand diagram. Remember, itβs a circle whose
centre lies at two, negative three. And its radius is two units. This means that the imaginary axis
must be a tangent to this circle. Itβs quite clear then that the
smallest possible value for the argument must be negative π by two radians. But what about its maximum
value?

Well, letβs call that π minus π
by two. And weβll add another tangent to
the circle. Weβll call this at point π. We know that triangles ππ΅πΆ and
π΄ππΆ are congruent. They are right-angled triangles who
share a hypotenuse of equal length. They also both have the radius of
the circle as one of their sides. So they must be congruent. This means these acute angles at π
must be equal. Weβll call them π divided by
two.

Using right angle trig, we can see
that tan of π divided by two must be equal to π΄πΆ divided by π΄π. π΄πΆ is two units, and π΄π is
three units. So π divided by two must be equal
to arc tan of two-thirds. And we can therefore say that π is
equal to two times arctan of two-thirds. And the maximum value is therefore
two arctan of two-thirds minus π by two. And we have the range for the
argument.

So letβs look at part three. Find the range of the modulus of
π§.

We know that the minimum value of
π§ will lie at this point π·. And the maximum will lie at the
point πΈ. And thatβs because the modulus is
the distance between the point representing the complex number on the Argand diagram
and the origin. And we actually know the radius of
the circle. So we can define the minimum as the
length of ππΆ minus the radius and the maximum as the length of ππΆ plus the
radius.

Now of course we saw that the
radius is two. So the minimum is ππΆ minus two,
and the maximum is the length of ππΆ plus two. And we can use the distance formula
or the definition of the modulus to find the length of ππΆ. πΆ is at the point two, negative
three. So the length of ππΆ is the square
root of two squared plus negative three squared, which is root 13. And we can see that the range of
the modulus of π§ is values between root 13 minus two and root 13 plus two. We can also reverse this process
and apply standard coordinate geometry to allow us to describe a given locus as an
equation in terms of the modulus.

Letβs see what that might look
like.

The figure shows a locus of a point
π§ in the complex plane. Write an equation for the locus in
the form the modulus of π§ minus π equals π, where π is a complex number and π
is greater than zero, and theyβre constants to be found.

Remember, the locus of a point π§,
which satisfies the equation the modulus of π§ one is π§ one equals π, is a circle
centred at π§ one of radius π. We therefore need to find the
centre and radius of the circle represented on our diagram.

We see that the circumference of
the circle passes through three points. These are the points that represent
the complex numbers zero, four π, and negative 10. We see that the vertex at π§ one is
right. And then we therefore know that the
line joining π§ one and π§ two passes through the centre of the circle. Itβs the diameter of the
circle.

We can use the Pythagorean theorem
to calculate its length. And thus weβll be able to find the
radius. Denoting the diameter as π, we see
that π is equal to the square root of 10 squared plus four squared. And thatβs equal to two root
29. Its radius is half of this. So the radius of our circle is root
29 units.

We also know that the centre of the
circle must lie at the midpoint of the diameter. We could apply standard coordinate
geometry here. Or we could recall the fact that
the midpoint of two complex numbers is half of their sum. And we see that the centre of the
circle lies at the point which represents a complex number a half of negative 10
plus four π. Thatβs negative five plus two
π. So we have a circle whose radius is
root 29 and whose centre lies at negative five plus two π. This means that the equation of our
circle and therefore the equation of the locus given is the modulus of π§ minus
negative five plus two π equals root 29.

The next example, we use the fact
that an equation given by the modulus of π§ minus π§ one equals the modulus of π§
minus π§ two represents a perpendicular bisector of the line segment which joins the
points π§ one to π§ two. For example, the modulus of π§
equals the modulus of π§ minus six π represents the locus of all points equidistant
from the points zero, zero and zero, six. Letβs look at a slightly more
complicated example.

A complex number π§ satisfies the
modulus of π§ plus one plus π equals the modulus of π§ minus two minus six π. Describe the locus of π§ and give
its cartesian equation. Weβll begin by factoring the terms
inside each modulus to ensure that this locus looks like the general form.

The modulus of π§ minus π§ one
equals the modulus of π§ one minus π§ two. Thatβs the modulus of π§ minus
negative one minus π equals the modulus of π§ minus two plus six π. This means that the locus of π§ is
given by all points equidistant from negative one minus π and two plus six π. Itβs the perpendicular bisector of
the line segment between the two points on the Argand plane.

We can find its cartesian equation
as we would the cartesian equation of any line, by first finding its gradient. The gradient of the line segment
between these points representing our two complex numbers is six minus negative one
over two minus negative one, which is seven-thirds. Since the locus of the points
representing π§ form the perpendicular line bisector of this line segment, the
gradient is found by using the fact that the product of the gradients of two lines,
which are perpendicular, is negative one. So its gradient is negative
three-sevenths.

We also know that this line passes
through the midpoint of the points representing our two complex numbers. And this midpoint is half the sum
of these two complex numbers. Itβs a half of negative one minus
π plus two plus six π. Thatβs a half plus five over two
π. Using the formula π¦ minus π¦ one
equals π times π₯ minus π₯ one, with the cartesian coordinates a half, five over
two, we get π¦ minus five over two equals negative three-sevenths times π₯ minus
one-half.

And we can rearrange this. And we see that the equation of our
line is π¦ equals negative three-sevenths π₯ plus 19 over seven. For our two final examples, weβll
consider the locus of a point π§ that forms a circle different to our first example
and that which forms an ellipse. The first definition we need to
know is that the locus of a point π§, which satisfies the equation the modulus of π§
minus π§ one is equal to π times the modulus of π§ minus π§ two when π is greater
than zero and not equal to one, is a circle. We also need to know that the locus
of a point π§, which satisfies the equation the modulus of π§ minus π§ one plus the
modulus of π§ minus π§ two is equal to π, where the modulus of π§ one minus π§ two
is less than π, is an ellipse, with foci π§ one and π§ two and a major axis of
length π.

A complex number π§ satisfies the
equation the modulus of π§ plus one minus 13π equals three times the modulus of π§
minus seven minus five π. Find the cartesian equation of the
locus of π§.

We know that the locus of the
points π§ which satisfy this equation form a circle. We can find its centre and radius
by substituting the general algebraic form of the complex number into this
equation. Weβll let π§ be equal to π₯ plus
π¦π. On the left-hand side, we get π₯
plus π¦π plus one minus 13π. And on the right-hand side, it
becomes π₯ plus π¦π minus seven minus five π. We can gather real and imaginary
parts.

And next we consider the definition
of the modulus. The modulus of a complex number is
the square root of the sum of the squares of the real and imaginary parts. In our case, thatβs as shown. Weβre going to square both sides of
this equation. And next, we distribute inside our
parentheses and gather all the terms. We can then divide through by
eight. And we have π₯ squared minus 16π₯
plus π¦ squared minus eight π¦ plus 62 equals zero.

We can then complete the
square. And we see that this simplifies to
π₯ minus eight all squared plus π¦ minus four all squared equals 18. We found the cartesian equation of
the locus of π§. And in fact, we can describe this
locus as a circle, which we said earlier. However, we now know it has a
centre at the point whose cartesian coordinates are eight, four. And its radius is root 18, which
simplifies to three root two units.

A complex number satisfies the
modulus of π§ plus the modulus of π§ minus five minus three π equals eight. Describe the locus of π§.

Remember, the locus of a point π§
which satisfies the modulus of π§ minus π§ one plus the modulus of π§ minus π§ two
equals π, where the modulus of π§ one minus π§ two is less than π, is an ellipse,
with a foci at π§ one and π§ two and a major axis of length π. We can rewrite our equation
slightly to be the modulus of π§ minus zero plus the modulus of π§ minus five plus
three π equals eight. This means the locus of π§ is an
ellipse. It has foci at the origin and five
plus three π. And it has a major axis of length
eight units.

In this lesson, weβve seen that we
can use our understanding of geometry and the geometry of the complex plane to
interpret the loci of points which satisfy certain equations. Weβve seen that the locus of a
point π§, which satisfies this equation, is a circle of radius π. The locus of the points which
satisfy these equations are the perpendicular bisector of the line segment between
π§ one and π§ two. Weβve also seen the alternative
form for the locus of the points which are a circle, as well as the locus of points
which are an ellipse.