### Video Transcript

A transformation which maps the
π§-plane to the π€-plane is defined by π taking π§ to one over π§, where π§ is
nonzero. Part one, find an equation for the
image of the modulus of π§ equals two under the transformation. Part two, find an equation for the
image of the argument of π§ equals three π by four. Part three, find a Cartesian
equation for the image of the imaginary part of π§ equals two. And part four, find a Cartesian
equation for the image of the modulus of π§ minus π equals one.

Our transformation takes π§ to its
reciprocal, one over π§. And so, π€ is one over π§. In the first part, we want to find
the image of the modulus of π§ equals two under this transformation. We can invert our transformation to
find π§ in terms of π€. Multiplying both sides by π§ and
then dividing through by π€, we find that π§ is one over π€. Substituting this then, we find
that our image has the equation the modulus of one over π€ equals two. But we can improve on this
equation. We use the fact that the modulus of
a quotient is the quotient of the moduli and that the modulus of one is simply
one. Rearranging then, we can rewrite
our equation as the modulus of π€ equals a half.

Now what does this look like on the
complex plane? Well, our original locus with
equation the modulus of π§ equals two is a circle with centre the origin and radius
two in the π§-plane. And its image under the reciprocal
transformation, weβve shown has equation the modulus of π€ equals a half, which we
recognise as a circle with centre the origin and radius a half in the π€-plane. This makes sense. We know that the image of a complex
number with modulus π under this transformation will have a modulus one over
π. And so the image of a complex
number with modulus two has modulus one over two which is a half. And so, when we map the circle of
all complex numbers with modulus two under this transformation, itβs natural that we
get the circle of all complex numbers with modulus a half.

Letβs now move on to find the image
of the argument of π§ equals three π by four. Again, we use the fact that π§ is
one over π€. And so the argument of one over π€
is three π over four. We use the fact the argument of a
quotient is the difference of the arguments. And also that the argument of one
is just zero. And so multiplying both sides by
negative one, we get the equation the argument of π€ equals negative three π by
four. Again, itβs helpful to look at the
diagram. We see that the half line of
complex numbers with argument three π by four is mapped to the half line complex
numbers with arguments negative three π by four. We know thereβs a complex number
with argument π will be mapped to a complex number with argument negative π. So this makes sense.

Now letβs find the image of the
imaginary part of π§ equals two. It may not be clear what to do with
the imaginary part of one over π€. What we do as we ask for a
Cartesian equation is to write π€ in terms of its real and imaginary parts, which we
call π’ and π£. How do we find the imaginary part
of one over π’ plus ππ£? We make the denominator real in the
normal way. And now with this number written in
algebraic form, we can just read off the imaginary parts. Itβs negative π£ over π’ squared
plus π£ squared. We might be tempted to stop
simplifying at this point. But if we divide by two and
complete the square in π£, we get something. This is recognisably the equation
of a circle. Letβs plot this on a diagram. We see that the line of complex
numbers with imaginary part two has been mapped to a circle in the π€-plane.

Finally, we find the image of the
modulus of π§ minus π equals one. We substitute one over π€ for π§
and then write what we have inside the modulus as a single fraction. This allows us to apply what we
know about the modulus of a quotient. So we can multiply both sides by
the modulus of π€. Now we can substitute π’ plus ππ£
for π€. We can simplify on the left-hand
side. And now, weβre ready to apply the
definition of the modulus. We square both sides. And negative π’ squared is the same
as π’ squared; so these cancel. Distributing on the left-hand side,
we see the π£-squared terms cancel as well. So we find that two π£ is negative
one. And hence, the equation of our
image is π£ equals negative a half. We see then that the circle with
centre π and radius one is mapped to the straight line of points in the π€-plane
with imaginary part negative a half.

Letβs recap then. In part one, the circle was mapped
to a circle. In part two, a line was mapped to a
line. Well, actually, a half line was
mapped to a half line. In part three, a line was mapped to
a circle. And in part four, a circle was
mapped to a line.