### Video Transcript

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.