### Video Transcript

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.