### Video Transcript

In this video, weβll learn how to
express regions in a complex plane in terms of inequalities and how to interpret
inequalities as regions in the complex plane. Itβs likely you all have worked
extensively with regions in the real Cartesian plane, considering them as
inequalities. And this video looks to extend
those concepts. Weβll begin by considering regions
defined by half lines, circles, and perpendicular line bisectors, before looking at
how set notation can help us to find composite regions.

Remember, there are two equations
that are used to define the locus given by a circle. The equations, the modulus of π§
minus π§ one equals π, represent a circle with a radius of π and a centre at π§
one. The equations, the modulus of π§
minus π§ one equals π times the modulus of π§ minus π§ two, also represent a circle
when π is greater than zero but not equal to one. In these situations, the radius and
centre will need to be found for each case.

We also need to know the equations
of the form the modulus of π§ minus π§ one equals the modulus of π§ minus π§ two
give a perpendicular bisector of the line segment that joins π§ one to π§ two, as
shown in this diagram. And finally, a half line is
described using the argument. The equations, the argument of π§
minus π§ one equals π, represent a half line from, but not including, π§ one. This half line makes an angle of π
with the horizontal half line that also extends from π§ one in the positive
π₯-direction.

Just as with regions in the
Cartesian plane, weβll also need to use the fact that a strict inequality, thatβs
greater than or less than, will be represented by a dashed line whereas a weak
inequality, greater than or equal to or less than or equal to, will be represented
by a solid line. Letβs begin by considering a simple
example that uses some of these definitions and look at how to decide which regions
are the ones required.

Sketch on an Argand diagram the
region represented by the argument of π§ plus three minus two π is greater than or
equal to negative π by two and less than π by four.

To sketch this region, weβll begin
by considering the boundaries. They are given by the argument of
π§ plus three minus two π is equal to negative π by two and the argument of π§
plus three minus two π is equal to π by four. Each of these represents a half
line. We can rewrite π§ plus three minus
two π by factoring negative one. And we get π§ minus negative three
plus two π. The point that represents this
complex number will have Cartesian coordinates negative three, two. And of course, we represent this
with an open circle since we know that the locus of points doesnβt actually include
this point.

The first boundary is going to make
an angle of negative π by two with the positive horizontal, measured in a
counterclockwise direction. This is the same as measuring an
angle of positive π by two in the clockwise direction. And this is a weak inequality. So we draw a solid line for this
one as shown. The half line for our next boundary
will make an angle of π by four radians with the positive horizontal, measured in a
counterclockwise direction. This time, itβs a weak
inequality. So we need to draw a dashed line as
shown.

Now that we have the boundaries for
our region, we need to decide which side of the region weβre going to shade. Weβre interested in all the complex
numbers such that the argument of π§ plus three minus two π is greater than or
equal to negative π by two and less than π by four. Thatβs going to be the region that
lies between these two half lines. So we shade this region. And weβre done. Weβve sketched the required region
on an Argand diagram.

In our next example, weβll consider
a circular region.

The figure shows a region in the
complex plane. Write an algebraic description of
the shaded region.

We can clearly see that this is a
circle. But there are two ways to describe
the locus that forms a circle. They are the modulus of π§ minus π§
one equals π and the modulus of π§ minus π§ one equals π times the modulus of π§
minus π§ two. In this example, it makes much more
sense to use the first form. In fact, we try to use this form
when describing regions as itβs much more simple to find the centre and the radius
then find two points whose distance to the circle are in constant ratio.

We can see that the centre of our
circle is represented by the complex number four plus π. The Cartesian coordinates of this
point are four, one. And we could use the distance
formula to calculate the radius with either zero, seven or zero, negative five as
one of the other points. Alternatively, we can find the
modulus of the difference between the complex number four plus π and either seven
π or negative five π. Letβs use seven π.

Seven π minus four plus π is the
same as six π minus four or negative four plus six π. So we need to find the modulus of
negative four plus six π. To find the modulus, we square the
real and imaginary parts, find their sum, and then find the square root of this
number. So thatβs the modulus of negative
four squared plus six squared, which is two root 13. So we know that the boundary for
our region, the circle, is described by the equation, the modulus of π§ minus four
plus π, cause thatβs the centre, is equal to two root 13 since thatβs the
radius. And we can distribute these
parentheses and write it as shown.

We do however need to consider the
region. Itβs the region outside of the
circle. Each point in the region is further
away from the centre of the circle than the distance of the radius. Itβs also a solid line which means
it represents a weak inequality. And we can therefore say that the
region is represented by the modulus of π§ minus four minus π is greater than or
equal to two root 13.

Weβve looked so far at simple
regions defined by half lines and circles. To define composite regions,
however, we need to use set notation. Weβll briefly recall the ones weβre
interested in.

π΄ union π΅ is the set of all
elements in π΄ or π΅ or both. π΄ intersection π΅ is the set of
all elements that are in both π΄ and π΅. Itβs the overlap. π΄ dash is the complement of
π΄. And thatβs the set of all elements
that are not in π΄, as shown by this third Venn diagram. Weβre going to now have a look at
how we can define composite regions using this notation.

Describe the shaded region in the
following figure algebraically in the form π΄ intersection π΅ intersection πΆ, where
π΄ is the set of complex numbers where the imaginary part of π§ is less than π΄. π΅ is the set of complex numbers
such that the modulus of π§ is less than or equal to the modulus of π§ minus π§
one. And πΆ is the set of complex
numbers such that the modulus of π§ is less than or equal to the modulus of π§ minus
π§ two. And π, which is a real number, and
π§ one and π§ two, which are complex numbers, are constants to be found.

We can see that the region is
bounded by three lines. We have one horizontal line, thatβs
L one, and two diagonal lines that Iβve labelled L two and L three. We can see that our horizontal line
has an equation of the imaginary part of π§ is equal to two. Now, weβre interested in the region
directly below this. And we can see that itβs
represented by a dashed line. So itβs going to be a strict
inequality. We can say then that π΄ is equal to
the set of complex numbers, where the imaginary part of this complex number must be
less than two.

Letβs now consider the diagonal
line that passes through the points three, zero and zero, three. Remember, we described diagonal
lines in the complex plane as perpendicular bisectors of a line segment joining a
point to the origin. And itβs possible here to do this
by inspection. Notice that the line itself passes
through two vertices of a square at three, zero and zero, three. This means the line it bisects must
pass through the other two vertices at zero, zero and three, negative three.

So our line is the perpendicular
bisector of the line segment joining the point three, negative three to the
origin. That represents the complex number
three minus three π. So we can say that π΅ is equal to
the set of complex numbers such that the modulus of this complex number is less than
or equal to the modulus of π§ minus three minus three π.

Weβll now consider the third
line. Itβs not quite so easy to find the
line it bisects by inspection. So we can look at a method that
generalises nicely. Weβll begin by finding the gradient
of our line. The formula here is change in π¦
divided by change in π₯ or π¦ two minus π¦ one over π₯ two minus π₯ one. The gradient is therefore negative
three minus zero over zero minus negative two, which is negative three over two.

Since this line passes through the
imaginary axis at negative three, we can say its equation is π¦ equals negative
three over two π₯ minus three. And we can also find the gradient
of the line it bisects. Since the line it bisects is
perpendicular to this line, its gradient is two-thirds. And of course, it passes through
the origin. So its equation is π¦ equals
two-thirds π₯.

Next, weβll find the point of
intersection of these lines. And weβll do that by equating
them. We add negative three over two π₯
to both sides and then divide through by three. And we see that the π₯-value of the
point where these two lines intersect is negative 18 over 13. Substituting this into either of
the equations we just used and we get a π¦-value of negative 12 over 13. So we now know the point at which
these two lines intersect. Our line is the perpendicular line
bisector of the line segment that joins a complex number with the origin. So this coordinate we just found
must be half the value of the complex number we need.

We can therefore double both the
π₯- and π¦-values. And we can see that the complex
number we need, the π§ two in part πΆ, is given by the point whose Cartesian
coordinates are negative 36 over 13, negative 24 over 13. And that tells us that πΆ is equal
to the set of complex numbers such that the modulus of π§ is less than or equal to
the modulus of π§ minus negative 36 over 13 minus 24 over 13 π. And of course, the question did say
that π, π§ one, and π§ two are constants to be found. So we add that π is equal to
two. π§ is equal to three minus three
π. And π§ two is equal to negative 36
over 13 minus 24 over 13.

Weβll now consider one further
example of representing composite regions in the complex plane.

The complex number π§ satisfies the
following conditions. The modulus of π§ is greater than
or equal to two times the modulus of π§ plus 12 minus nine π. The modulus of π§ minus two π is
greater than or equal to the modulus of π§ plus six plus four π. And the imaginary part of π§ is
less than 12. Represent the region on an Argand
diagram.

Weβll begin by considering the
first region. We can find the centre and radius
of the circle by substituting π§ equals π₯ plus π¦π into our equation. Remember at the moment, weβre just
finding the boundary for the region. We then square both sides of this
equation. We can instantly replace two
squared with four.

But for the other bit, weβre going
to need to use the definition of the modulus. We know that the modulus of π₯ plus
π¦π is equal to the square root of π₯ squared plus π¦ squared. So the left-hand side of our
equation becomes π₯ squared plus π¦ squared. We then gather the real and
imaginary parts on the right-hand side. And we get four times π₯ plus 12
all squared plus π¦ minus nine all squared. Expanding the parentheses and then
simplifying, on the right-hand side, we get four π₯ squared plus 96π₯ plus four π¦
squared minus 72π¦ plus 900. We subtract π₯ squared and π¦
squared from both sides of this equation. And then, we divide through by
three.

Now weβre looking to find the
Cartesian equation of a circle. So weβre going to complete the
square in π₯ and π¦. For π₯, we get π₯ plus 16 all
squared minus 256. And for π¦, we get π¦ minus 12 all
squared minus 144. And we add 300 of course. Negative 256 minus 144 plus 300 is
negative 100. So we add 100 to both sides. And we have the Cartesian equation
of a circle. It has a centre at negative 16, 12
and a radius of 10 units.

The boundary for our first region
is therefore this circle as shown. But how do we decide whether to
shade inside or outside of this circle? Well, letβs choose a point that we
know to be outside of the circle. Letβs choose the point whose
Cartesian coordinates are one, zero. This is the complex number one. Weβre going to substitute this into
the first inequality and see if the statement makes sense.

This statement is the modulus of
one is greater than or equal to two times the modulus of one plus 12 minus nine
π. Or the modulus of one is greater
than or equal to two times the modulus of 13 minus nine π. Well, the modulus of one is
one. And the modulus of 13 minus nine π
is root 250. Well, itβs not true that one is
greater than two root 250. So this statement is false. And that tells us weβre going to be
interested in the inside of the circle. Thatβs the region that satisfies
that first condition. Weβll fully shade a region when
weβve considered the other two situations.

Now for two, we know that the
equation, the modulus of π§ minus two π equals the modulus of π§ plus six plus four
π, represents the perpendicular bisector of the line segment joining the point
which represents two π and negative six minus four π. Thatβs the line segment between
zero, two and negative six, negative four. We could find the exact equation of
the perpendicular bisector of this line segment by considering the gradient and
midpoint of the line segment it bisects. Alternatively, in this example, we
can do this by inspection. And we can see that the line passes
through the point zero, negative four and negative four, zero. And in fact, it also passes through
the centre of our circle.

Once again, weβll substitute π§
equals one into the inequality and see if the statement makes sense. The modulus of one minus two π is
the square root of five. And the modulus of one plus six
plus four π or the modulus of seven plus four π is the square root of 65. Once again, we can see that itβs
actually not true that the square root of five is greater than or equal to the
square root of 65. And we can see that weβre
interested in the other side of the line. And of course, remember, we drew a
solid line because our inequality is a weak inequality.

Letβs now consider the third
region. Weβre told that the imaginary part
of π§ must be less than 12. The boundary of this region is the
horizontal line passing through 12 on the imaginary axis. And because itβs a strict
inequality, we draw a dash line. Weβre interested in the region
below this dash line. To satisfy all three regions in
this question, we need the intersection of the regions. So we shade the overlap between the
three regions. Itβs the sector of the circle
shown. And we are done. We have represented the region on
an Argand diagram.

In this video, weβve seen that we
can use our understanding of loci to represent regions in the complex plane. We used dashed lines to represent
boundary points that are not included and solid lines to represent regions that
include their boundary points. And we can use set operations such
as unions, intersections, and complements to define compound regions in the complex
plane.