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 the plane, we can
also use the argument of a complex number to interpret the locus of points which
satisfy certain criteria. Weβll consider the loci of half
lines, major arcs, semicircles, and minor arcs and the Cartesian equations which
correspond to these.
Remember, for a complex number
plotted on an Argand diagram joined by a line segment or half line to the origin,
the argument is the angle this line segment makes with the positive real axis
measured in a counterclockwise direction. Now, rather than asking what the
argument of a complex number is, we could alternatively ask, what is the locus of a
point given a fixed argument, say, the argument of π§ is π by three?
This represents all the complex
numbers that lie on the ray or half line which makes an angle of π by three radians
with the π₯-axis in a counterclockwise direction. The locus of π§ then is this half
line. Now, remember, the argument isnβt
defined when π§ is equal to zero, so the locus cannot include the point at the
origin. Letβs generalize this idea for any
half line in the complex plane by considering a transformation by subtracting a
fixed number π§ one. We say that the locus of a point π§
given that the argument of π§ minus π§ one is equal to π is a half line from but
not including π§ one, which makes an angle of π to the horizontal half line that
extends from π§ one in the positive π₯-direction. And of course this is measured in a
counterclockwise direction. Letβs now look at an example of
this.
Sketch the locus of π§ when the
argument of π§ plus two plus π is equal to π over four.
Remember, the locus of a point π§
given that the argument of π§ minus π§ one equals π is a half line from but not
including π§ one. This half line makes an angle of π
with the horizontal half line that extends out from π§ one in the positive
π₯-direction. And itβs important we remember this
is measured in a counterclockwise direction. So to begin, weβre going to write
the equation of our locus in the form the argument of π§ minus π§ one equals π. We can factor by negative one, and
we get that π§ plus two plus π is the same as π§ minus negative two minus π. So the argument of π§ minus
negative two minus π equals π by four. Comparing our equation to the
general form, we see we let π§ sub one be equal to negative two minus π and π is
equal to four.
Remember, π§ one is the endpoint of
our ray or our half line, and the half line doesnβt actually include this point. On an Argand diagram, π§ one is the
point whose Cartesian coordinates are negative two, negative one. And we add an empty circle here to
show that we donβt want to include this point in our locus. We add a horizontal half line from
π§ one in the positive π₯-direction. Then weβre going to measure π by
four radians from this line in the counterclockwise direction. When we do, we find the locus of π§
is shown. We can also reverse this process to
form an equation given a diagram of the locus of π§.
Next, weβre going to 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, a locus in this form is a
half line. Given the argument of π§ minus π§
one equals π, the half line extends from but doesnβt include π§ one. It makes an angle of π with the
positive horizontal half line, and thatβs measured in a counterclockwise
direction. Now, letβs rewrite our definition
as the argument of π€ minus π€ one. And weβre going to need to write
the equation the argument of π€ plus three plus π equals π over three in this
form. To achieve this, we factor by
negative one, and we get the argument of π€ minus negative three minus π is equal
to π by three. Then, we can let π€ sub one be
equal to negative three minus π and π is π by three.
We know that our half line extends
from the point π€ one but doesnβt include it. We can plot π€ one using the
Cartesian coordinates negative three, negative one. Then we measure an angle of π by
three radians from the positive horizontal half line extending from π€ one. And so the locus of π€ is as
shown. Essentially, itβs a straight
line. So to find its Cartesian equation,
weβre going to use the formula π¦ minus π¦ one equals π times π₯ minus π₯ one. We know π₯ one, π¦ one is a point
that this line passes through. And so we can say that π₯ one, π¦
one here is the point negative three, negative one. π, however, is the slope of the
line. Now, if we add a right-angled
triangle onto our loci, we know that π is rise divided by run.
With our included angle of π by
three, thatβs the same as opposite over adjacent. But of course, thatβs the tan
ratio. And so here our slope π must be
equal to tan of π by three, which is root three. We substitute everything we know
into our formula for the equation of a straight line. And we get π¦ minus negative one
equals the square root of three times π₯ minus negative three. And then we distribute our
parentheses such that π¦ plus one equals the square root of three π₯ plus three root
three.
If we subtract one from both sides,
we see that we have a Cartesian equation of the straight line. But we want the Cartesian equation
of the locus. We know that our half line doesnβt
include the point at negative three, negative one. So we impose this condition on π₯
such that π₯ is greater than negative three. And so the Cartesian equation is π¦
equals the square root of three π₯ plus three root three minus one, where π₯ is
greater than negative three.
The next locus weβre interested in
is that of an arc of a circle. The locus of a point π§ given the
argument of π§ minus π§ one over π§ minus π§ two equals π is the arc of a circle
which subtends an angle of π between the points represented by π§ one and π§
two. This is traced in a
counterclockwise direction from π§ one to π§ two and doesnβt include the endpoints
as part of its locus. Now, if π is less than π by two,
as in this diagram, the locus is a major arc. If itβs equal to π by two, our
locus is a semicircle. And if itβs greater than π by two,
we have a minor arc.
Letβs have a look at an
example.
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 elements of the set of complex numbers, and π, which is greater than zero
and less than or equal to π, are constants to be found.
We recall that the locus of a point
π§ such that the argument of π§ minus π over π§ minus π equals π is an arc of a
circle which subtends an angle of π between the points represented by π and
π. Itβs important that we realize that
this locus travels in a counterclockwise direction from π to π. So for our locus, thatβs this
direction. Now, if π is less than π by two,
the locus is a major arc. Well, we do indeed have a major
arc, but we can see that our value of π is equal to π by five. So that makes a lot of sense. The endpoints lie at π and π
whose Cartesian coordinates are four, negative three and negative three, one,
respectively.
The first of these points
represents the complex number four minus three π. We know that in order to travel in
a counterclockwise direction, the locus must begin here. So we let π be equal to four minus
three π. Similarly, it ends at the point
negative three, one, which represents the complex number negative three plus π. We can therefore say that the
equation of the locus is the argument of π§ minus four minus three π over π§ minus
negative three plus π is π by five.
In our next example, weβll practice
how to sketch a locus in this form.
The point π§ satisfies the argument
of π§ minus six over π§ minus six π equals π by four. Sketch the locus of π§ on an Argand
diagram.
We notice that the equation in this
question looks a lot like the argument of π§ minus π§ one over π§ minus π§ two
equals π. Now, the locus of π§ in this case
is an arc of a circle which subtends an angle of π between the points represented
by π§ one and π§ two. And of course this is sketched in a
counterclockwise direction from π§ one to π§ two. Comparing the general form to the
equation in our question, and we find that π§ one must be equal to six and π§ two
must be equal to six π. On an Argand diagram, these are
points represented by the Cartesian coordinates six, zero and zero, six,
respectively. We also see that π is equal to π
by four. And we know if π is less than π
by two, then our locus represents a major arc.
We also know that the endpoints are
not included in our locus, but here we do have a bit of a problem. We know our locus is the arc of a
circle. But without knowing the center of
the circle, we canβt use this information to find the correct arc. In fact, it could be this major arc
or this major arc. But we do know that the locus is
drawn in a counterclockwise direction from π§ one, well thatβs six, zero, to π§ two,
which is zero, six. For that to be the case, we have to
use this arc on the right, and we see that the locus is shown. We can also choose to add the angle
of π by four radians here.
Itβs also possible to find the
Cartesian equation of loci in this form. Occasionally, weβll be able to take
a geometric approach, but in general, an algebraic approach is sensible.
In our final example, weβll
consider one such approach.
The locus of π§ satisfies the
argument of π§ minus three π over π§ minus five π is equal to two π by three. Sketch this locus on an Argand
diagram and find its Cartesian equation.
Remember, the locus of a point π§
such that the argument of π§ minus π§ one over π§ minus π§ two is π is the arc of a
circle which subtends an angle of π between the points represented by π§ one and π§
two. Itβs traced in a counterclockwise
direction from π§ one and π§ two, but the endpoints arenβt part of the locus. Now, we know that if π is greater
than π by two radians, the locus is a minor arc. Similarly, if itβs greater than π
by two, itβs a major arc. And if itβs equal to π by two,
itβs a semicircle.
Here, we can say that π§ one must
be three π, π§ two must be five π, and π is two π by three radians. So that is greater than π by two,
and itβs a minor arc. Now, to plot π§ one and π§ two on
the Argand plane, we plot the points with Cartesian coordinates zero, three and
zero, five, respectively. But here we have a bit of a
problem. How do we know where the minor arc
sits? We know itβs the arc of a
circle. But without knowing the center of
the circle, we canβt use this information to find the arc. We do, however, know that the locus
is drawn in a counterclockwise direction from π§ one to π§ two. And for the arc from zero, three to
zero, five to be a minor arc when drawn in this direction, we need to draw the locus
shown. And we can add two π by three
radians if necessary.
The next part of this question asks
us to find the Cartesian equation of the locus. Now, in some scenarios, we can find
the Cartesian equation by finding the center and radius of the circle. In this case though, weβre going to
use an algebraic approach. Weβre going to substitute π§ equals
π₯ plus π¦π into the complex equation. So we get the argument of π₯ plus
π¦π minus three π over π₯ plus π¦π minus five π equals two π over three. Letβs rewrite this a little
bit. And then weβre going to perform the
complex division π₯ plus π times π¦ minus three divided by π₯ plus π times π¦
minus five.
To do this, we multiply the
numerator and denominator of our fraction by the conjugate of the denominator. Remember, to find the conjugate of
our denominator, we change the sign of the imaginary part. So weβre going to multiply the
numerator and denominator of our fraction by π₯ minus π times π¦ minus five. Distributing the parentheses on our
numerator and remembering of course that π squared is negative one, we get π₯
squared minus π₯π times π¦ minus five plus π₯π times π¦ minus three plus π¦ minus
three times π¦ minus five. And on our denominator, we simply
get π₯ squared plus π¦ minus five squared.
Weβre now going to split this into
the real and imaginary parts. And we see that the real part is π₯
squared plus π¦ minus three times π¦ minus five over π₯ squared plus π¦ minus five
squared. And the imaginary part is negative
π₯ times π¦ minus five plus π₯ times π¦ minus three, which Iβve written as π₯ times
π¦ minus three minus π₯ times π¦ minus five all over π₯ squared plus π¦ minus five
squared.
Now, in fact, distributing the
parentheses, and we get π₯ squared plus π¦ squared minus eight π¦ plus 15 as the
numerator of our first part and two π₯ as the numerator of our imaginary part. Now, we know that the argument of
this complex number is equal to two π by three. And for a complex number of the
form π plus ππ, we know that tan of π, where π is the argument, is equal to π
over π. This means that tan of two π over
three will be equal to π, which is the imaginary part, divided by π, which is the
real part of our complex number. And when we divide the imaginary
part by the real part, the denominators cancel. So we find that tan of two π over
three equals two π₯ over π₯ squared plus π¦ squared minus eight π¦ plus 15. tan of two π over three, though,
is negative root three.
Letβs clear some space and
rearrange our equation. To do so, weβre going to multiply
both sides by the denominator of our fraction and then divide through by negative
root three. And when we do, we find that π₯
squared plus π¦ squared minus eight π¦ plus 15 is equal to two π₯ divided by
negative root three, which we can write as negative two root three over three
π₯. We add two root three over three π₯
to both sides, and now weβre going to complete the square. Completing the square for the
expression π₯ squared plus two root three over three π₯ gives us π₯ plus root three
over three squared minus a third.
Similarly, completing the square
for π¦ squared minus eight π¦, and we get π¦ minus four squared minus 16. And of course, we add 15 and set
this equal to zero. Negative one-third minus 16 plus 15
is negative four-thirds. So we add four-thirds to both
sides. And we get the equation π₯ plus
root three over three squared plus π¦ minus four all squared equals four-thirds.
We can now see that this is the
equation of a circle as we were expecting. Now, we do need to be a little bit
careful. We need to state a restriction on
π₯ and π¦ to ensure that the points lie on the locus. And of course, thatβs the minor arc
we sketched earlier. And so weβre only going to consider
the points where π₯ is greater than zero. And so weβve sketched the locus on
an Argand diagram and found its Cartesian equation to be π₯ plus root three over
three squared plus π¦ minus four squared equals four-third for values of π₯ greater
than zero.
In this video, weβve seen we can
use the argument of a complex number in the same way that we might use the modulus
to define loci in the complex plane. We can use our understanding of the
geometry of the complex plane to interpret the loci of points which satisfy certain
equations and investigate their properties. We saw 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, which makes an angle of π to the horizontal half line in
a counterclockwise direction.
Similarly, we saw that the locus of
a point π§, which satisfies the argument of π§ minus π§ one over π§ minus π§ two is
equal to π, is an arc of a circle which subtends an angle of π between the points
represented by π§ one and π§ two. Itβs traced in a counterclockwise
direction, and the endpoints are not part of the locus. And finally, we learned that, in
this case, if π is less than π by two radians, the locus is a major arc. If itβs equal to π by two, the
locus is a semicircle. And if itβs greater than π by two,
the locus is a minor arc.