### Video Transcript

In this video, weβll learn how to
draw and interpret loci in the complex plane, expressed in terms of the
argument. Just as we can use the modulus to
define loci in the complex plane, by considering the geometry of this plane, we can
also use the arguments of a complex number to interpret the loci of points which
satisfy certain criteria. Weβll consider the loci of half
planes, major arcs, semicircles, and minor arcs, and the Cartesian equations which
correspond to these in this video.

Remember, for a complex number
represented on an Argand diagram which is joined by a line segment or a half line to
the origin, the argument is the angle this line segment makes with the positive real
axis. And itβs always measured in a
counterclockwise direction. To calculate the argument, we first
consider which quadrant the point representing the complex number will lie in. For a complex number of the form π§
equals π plus ππ, its argument is given as the arctan of π divided by π, for
complex numbers which would be represented in the first and fourth quadrant. For complex numbers which will be
represented in the second quadrant, the argument is the arctan of π divided by π
plus π. And if weβre looking at a complex
number thatβs plotted in the third quadrant, its argument is arctan of π divided by
π minus π.

Now, rather than asking what the
argument of a complex number is, we could alternatively ask what is the locus of a
point of a fixed argument, say the argument of π§ is equal to π by three
radians? This represents the set of complex
numbers that lie on the ray or the half line which makes an angle of π by three
with the π₯-axis in a counterclockwise direction. The locus of π§ is therefore this
half line. Remember, though, the argument is
not defined when π§ is equal to zero. So the locus cannot include the
origin.

We can generalize this for any half
line in the complex plane, by considering a transformation by subtracting a fixed
complex number, π§ one. We can say that the locus of the
point π§ such that the argument of π§ minus π§ one is equal to π is a half line
from but not including the point at π§ one. This half line makes an angle of π
with the horizontal half line extending from π§ one in the positive
π₯-direction. And itβs measured in a
counterclockwise direction. Letβs have a look at an example of
this.

Sketch the locus of π§ when the
argument of π§ plus two plus π is equal to π by four.

Remember, the locus of the point π§
when the argument of π§ minus π§ one is equal to π is a half line from but not
including π§ one. This half line makes an angle of π
to the horizontal half line in the positive π₯-direction. And itβs measured in a
counterclockwise direction. Weβll begin then by writing the
argument of π§ plus two plus π in the form the argument of π§ minus π§ one, to
ensure that we can correctly identify our value of π§ one. We factor negative one, and we get
the argument of π§ minus negative two minus π. And this means we can rewrite our
equation as the argument of π§ minus negative two minus π equals π by four.

We can now see that π§ one is equal
to negative two minus π. This is the endpoint of the ray or
half line. And we do need to recall that the
half line doesnβt actually include this point. On an Argand diagram, π§ one can be
plotted by the point whose Cartesian coordinates are negative two, negative one as
shown. And weβve added this empty circle
to show that we donβt want to include this point in our locus.

Next, we use the argument. The argument of π§ plus two plus π
is π by four radians. This means the locus of π§ is the
set of points that make an angle of π by four radians in a counterclockwise
direction from the horizontal. π by four radians is equivalent to
45 degrees. So we add a line, as shown, at this
angle. And this means that the locus of π§
when the argument of π§ plus two plus π is equal to π by four is as shown. We can also reverse this process to
form the equation given a diagram of the locus of π§.

In our next example, weβll look at
how the locus can be expressed as a Cartesian equation.

Find the Cartesian equation of the
locus of π€ such that the argument of π€ plus three plus π is equal to π by
three.

Remember, the locus in this form is
a half line. Weβre looking to find the Cartesian
equation of this half line. So a sensible starting point is to
find the gradient of this line. We can find the gradient of this
line by considering the argument, which is π by three radians. Now, the formula for gradient is
rise over run. Thatβs the same as opposite over
adjacent. And, of course, thatβs equal to the
tangent function. We can say then that the gradient
of our line is equal to tan of π by three which is the square root of three.

Our next job is to find the point
which this line must pass through. We use the definition of the locus
to rewrite our equation. We factor negative one. And we can see that this is the
same as the argument of π€ minus negative three minus π equals π by three. And we can see then that our line
begins at the point representing the complex number negative three minus π. This will have Cartesian
coordinates negative three, negative one. Letβs substitute these values into
the formula for a straight line, π¦ minus π¦ one equals π multiplied by π₯ minus π₯
one.

When we do, we see that π¦ minus
negative one equals root three times π₯ minus negative three. We distribute the parentheses and
simplify as far as possible. And we can see that the Cartesian
equation of the line is π¦ equals root three π₯ plus three root three minus one. Remember, though, this is a half
line. And it doesnβt actually include the
point at negative three, negative one. So weβre going to need to add a
restriction on π₯ or π¦. We can say that π₯ must be greater
than negative three. And we found the Cartesian equation
of the locus of π€, given that the argument of π€ plus three plus π is π by
three.

The next locus we are interested in
is that of a circle. We can say that the locus of the
point π§ such that the argument of π§ minus π§ one over π§ minus π§ two is equal to
π is the arc of a circle which subtends an angle of π between the points
represented by π§ one and π§ two, as shown in the diagram. If π is less than π by two
radians, the locus is a major arc. If π is equal to π by two, the
locus is a semicircle. And if π is greater than π by two
radians, the locus is a minor arc. Now, remember, the endpoints are
not part of the locus. So we include open dots
representing these points, as shown. Letβs look at an example that uses
this idea.

The figure shows a locus of a point
π§ in the complex plane. Write an equation for the locus in
the form the argument of π§ minus π over π§ minus π equals π, where π and π,
which are complex numbers, and π, which is greater than zero and less than or equal
to π, are constants to be found.

Remember, the locus of a point π§
in this form is the arc of a circle which subtends an angle of π between the points
represented by π§ one and π§ two. We have three conditions on π. If itβs less than π by two, the
locus is a major arc. If itβs equal to π by two, itβs a
semicircle. And if itβs greater than π by two,
the locus is a minor arc. And, remember, the endpoints are
not part of this locus. We can see by looking at the
diagram that the locus of our π§ is the major arc of a circle. And this makes sense because π is
equal to π by five radians.

The endpoints of our locus lie at
π΄ and π΅ whose Cartesian coordinates are four, negative three and negative three,
one, respectively. These represent the complex numbers
four minus three π and negative three plus π. And, remember, this locus is traced
in a counterclockwise direction. Since the starting point is that
represented in the complex number four minus three π, we can say that the equation
of our locus is the argument of π§ minus four minus three π over π§ minus negative
three plus π equals π by five. We were actually told to find the
value of the constants π, π, and π. π is four minus three π, π is
negative three plus π, and π is π by five.

In our next example, weβll practice
sketching a locus in this form.

The point π§ satisfies the equation
the argument of π§ minus six over π§ minus six π equals π by four. Plot the locus of π§ on an Argand
diagram.

The locus of π§ is the arc of a
circle which subtends an angle of π by four radians between the points represented
by six and six π. Remember, these are plotted in a
counterclockwise direction from six to six π. But they donβt actually include
these points themselves. These are the points on the Argand
plane whose Cartesian coordinates are six, zero and zero, six, respectively. And since π by four is less than
π by two, we know that we have a major arc. So we begin by adding the points
six, zero and zero, six on our Argand diagram.

And then we come across a
problem. How do we know where the major arc
sits? Sure, itβs the major arc of a
circle. But without knowing the centre of
the circle, we canβt use that information to find the arc. It could actually be either of
these two arcs shown. Here, we recall the fact that the
locus is drawn in a counterclockwise direction. We need the arc that begins at the
point six, zero and ends up at the point zero, six to be a major arc, when drawn in
this direction.

This means we have to choose this
arc on the right. And, therefore, the locus is as
shown. Itβs not actually necessary to add
the cords shown. But by doing so, we can see that we
will get an angle thatβs less than π by two radians. It is also possible to find the
Cartesian equation of loci in this form. Occasionally, you can take a
geometric approach. But, in general, an algebraic
approach is sensible.

In our final example, weβll
consider one such algebraic approach.

The locus of π§ satisfies the
equation the argument of π§ minus three π over π§ minus five π equals two π by
three. Plot this locus on an Argand
diagram and find its Cartesian equation.

Remember, the locus of the point π§
such that the argument of π§ minus three π over π§ minus five π equals two π by
three is the arc of a circle which subtends an angle of two π by three radians
between the points represented by three π and five π. This time, it will begin at three
π and travel in a counterclockwise direction. Two π by three is greater than π
by two. So we know that this forms a minor
arc. And as always, the endpoints are
not part of the locus. The points represented by three π
and five π have Cartesian coordinates zero, three and zero, five, respectively.

So once weβve plotted these points
on an Argand diagram, how do we decide where the minor arc sits? Once again, we donβt know the
centre of the circle. So we canβt use that information to
find the location of the arc. We do, however, know that the locus
is drawn in a counterclockwise direction. We need the arc from zero, three to
zero, five to be a minor arc, when drawn in that counterclockwise direction. This means we travel along this arc
as show.

Next, we need to find the Cartesian
equation of this locus. In some scenarios, we can find this
equation by finding the centre and radius of the circle. Here, thatβs not so easy. So weβre going to need to
substitute π§ equals π₯ plus π¦π into our equation. When we do, we get that the
argument of π₯ plus π¦π minus three π over π₯ plus π¦π minus five π is two π by
three. Letβs begin by evaluating π₯ plus
π¦π minus three π over π₯ plus π¦π minus five π. To evaluate this problem, we need
to multiply both the numerator and the denominator of our fraction by the conjugate
of the denominator.

To find the conjugate of a complex
number, we change the sign for the imaginary part. So the conjugate of π₯ plus π¦
minus five π is π₯ minus π¦ minus five π. Weβre going to multiply both the
numerator and the denominator of this fraction by this number. On the numerator, we end up with π₯
squared minus π₯ π¦ minus five π plus π₯ π¦ minus three π minus π¦ minus three
times π¦ minus five times π squared. And on the denominator, we have π₯
squared minus π₯ times π¦ minus five times π plus π₯ times π¦ minus five times π
minus π¦ minus five squared π squared. And we can see that negative π₯
times π¦ minus five times π plus π₯ times π¦ minus five times π is zero.

Then using the fact that π squared
is equal to negative one and distributing our parentheses, we have the expression
shown. Next, we collect the real and
imaginary parts. And now we can find the argument of
π₯ plus π¦π minus three π over π₯ plus π¦π minus five π. If we take π to be the real part
of our complex number and π to be the imaginary part, that is two π₯ over π₯
squared plus π¦ squared minus 10π₯ plus 25, we can say that π divided by π, the
imaginary part divided by the real part, must be equal to tan of two π by
three. Now, normally, we would be worrying
about which quadrant the complex number lies in. But since time is periodic with a
period of π radians, adding or subtracting multiples of π to our value of π has
no effect on the value of tan π. Letβs clear some space for the next
step.

We can say that two π₯ divided by
π₯ squared plus π¦ squared minus eight π¦ plus 15 is equal to tan of two π by
three, which is equal to negative root three. We multiply both sides of this
equation by π₯ squared plus π¦ squared minus eight π¦ plus 15. And then we can simplify a little
by multiplying through by negative root three. And then, we add two root three π₯
to both sides of this equation. We now need to complete the square
for both π₯ and π¦.

Remember, weβre trying to find the
equation of the circle. Thatβs π₯ plus root three all
squared minus three plus π¦ minus four all squared minus 16 plus that 15. Negative three minus 16 plus 15 is
negative four. So we add four to both sides of
this equation. And we get π₯ plus root three all
squared plus π¦ minus four all squared equals four. And we can see that the centre of
our circle now lies at the point with Cartesian coordinates negative root three,
four. We do, of course, need to add a
restriction on π₯ and π¦ to ensure that the points three π and five π donβt
actually lie on the locus itself. This restriction is that π₯ is
greater than zero. So the Cartesian equation of our
locus is four equals π₯ plus root three all squared plus π¦ minus four all squared,
only when π₯ is greater than zero.

In this video, weβve seen the we
can use the arguments in the same way that we can use the modulus to define loci in
the complex plane. Weβve seen that the locus of a
point π§ which satisfies the argument of π§ minus π§ one equals π is a half line
from but not including π§ one. And it makes an angle of π to the
horizontal half line extending from π§ one in the positive π₯-direction.

Weβve also seen that π must be
measured in a counterclockwise direction. We saw how the locus of a point π§
which satisfies the equation the argument of π§ minus π§ one over π§ minus π§ two
equals π is an arc. When π is less than π by two,
itβs a major arc. When itβs equal to π by two, itβs
a semicircle. And when itβs greater than π by
two, the locus is a minor arc. And we saw that the endpoints
cannot be part of this locus. And the locus is measured in a
counterclockwise direction.