### Video Transcript

A complex number π§ satisfies the
modulus of π§ minus two plus four π equals the modulus of π§ plus two plus two
π. Describe the locus of π§ and give
its Cartesian equation. What is the minimum value of the
modulus of π§?

There are a number of ways of
describing the locus of a complex number π§ by using the modulus. And in fact, we can quote one of
these here. For two given constant complex
numbers, π§ one and π§ two, the locus of a point π§ which satisfies the modulus of
π§ minus π§ one equals the modulus of π§ minus π§ two is the perpendicular bisector
of the line segment that joins π§ one to π§ two. Weβre going to rewrite our equation
so it looks a little more like this. The modulus of π§ minus two plus
four π can be written as the modulus of π§ minus two minus four π. And we can rewrite π§ plus two plus
two π as π§ minus negative two minus two π.

So our equation is as shown. And so we compare our equation to
the equation in our rule. And we see that π§ one is equal to
two minus four π and π§ two, our second constant complex number, is negative two
minus two π. We can therefore say that the locus
of π§ is the perpendicular bisector of the line segment joining two minus four π
and negative two minus two π. The second part of the first part
of this question asks us to find the Cartesian equation of this. So letβs clear some space.

And what weβre going to do is
actually plot π§ one and π§ two on an Argand diagram. π§ one has a real part of two and
an imaginary part of negative four. So it has a coordinate of two,
negative four. π§ two has a real part of negative
two and an imaginary part of negative two. So its coordinates are negative
two, negative two. Weβre looking to find the Cartesian
equation of the perpendicular bisector of the line segment between these two
points. So weβll use the equation of a
straight line. Itβs π¦ minus π¦ one equals π
times π₯ minus π₯ one. Here, π₯ one π¦ one is a point that
the line passes through and π is its slope.

The equation for π is π¦ two minus
π¦ one over π₯ two minus π₯ one, change in π¦ divided by change in π₯. So weβre going to begin by finding
the slope of our line segment, the line segment that joins two, negative four to
negative two, negative two. Itβs negative four minus negative
two over two minus negative two. Thatβs negative two over four,
which is negative one-half.

Now, if we call π sub one the
slope of this line segment and π sub two the slope of its perpendicular bisector,
we know that the product of these two, π sub one and π sub two, is negative
one. This can alternatively be written
as π sub two is equal to negative one over π sub one. In other words, the slope of our
perpendicular bisector is the negative reciprocal of the slope of the original
line. And so the slope of our
perpendicular bisector is two.

But we need to find a point that it
passes through. Since itβs the perpendicular
bisector of that line segment, we know it must pass through the midpoint. And we find the midpoint by
essentially finding the average of our two coordinates. So thatβs negative two plus two
over two, negative two plus negative four over two. That gives us a midpoint of zero,
negative three. And so we substitute what we have
into our formula for the equation of a straight line. We get π¦ minus negative three
equals two times π₯ minus zero, which is π¦ plus three equals two π₯, and then we
rearrange by subtracting three from both sides. And we find that the Cartesian
equation for the locus of π§ is π¦ equals two π₯ minus three.

And so weβve answered the first
part of this question. The locus of π§ is the
perpendicular bisector of the line segment between two minus four π and negative
two minus two π. And its Cartesian equation is π¦
equals two π₯ minus three.

And so we go back to the second
part of this question. This says, what is the minimum
value of the modulus of π§? Well, remember, the modulus of π§
is the distance between a point π§ and the origin. So to find the minimum value of π§,
we need to find a point on the locus of π§ which is closest to the origin. Now, we know then that the line
segment that joins this point to the origin will be perpendicular to our line that
represents the locus of π§. And so the slope of this line will
be the same as the slope of the line segment joining our original two points,
joining negative two, negative two and two, negative four. And, of course, that was negative
one-half.

Letβs call this point π. And then the Cartesian equation for
the line that passes through the origin and π is π¦ equals negative one-half
π₯. π lies at the point of
intersection of our line with the line that represents the locus of π§. So we need to solve the equations
π¦ equals negative one-half π₯ and π¦ equals two π₯ minus three simultaneously. We can achieve this by equating
π¦. And we get negative a half π₯
equals two π₯ minus three. We add one-half π₯ and three to
both sides. And we find that three is equal to
five over two π₯. Then we divide through by five over
two, remembering that, to divide through by a fraction, we multiply by the
reciprocal of that fraction. And we get π₯ equals
six-fifths.

Weβre looking to find the
coordinates of π. So we substituted this value into
our earlier equation π¦ equals negative a half π₯. And we get π¦ equals negative a
half times six-fifths which gives us π¦ equals negative three-fifths. We can therefore say that point π
has Cartesian coordinates six-fifths, negative three-fifths. The minimum value of the modulus of
π§ is found by finding the distance between this point and the origin. So we use the Pythagorean
theorem. Itβs the square root of the sum of
the squares of six-fifths and negative three-fifths. Thatβs the square root of 36 over
25 plus nine over 25 which is the square root of 45 over 25.

Now, we know that the square root
of 25 is five, so we get a fifth times root 45. But the square root of 45 can be
simplified as three root five since itβs the square root of nine times the square
root of five, and the square root of nine is three. And we can therefore say that the
minimum value of the modulus of π§ is three root five over five.