### Video Transcript

In this video, weβre going to learn
about transformations of the complex plane. Really, weβre interested in complex
functions. Thatβs functions which take a
complex number as an input and return a complex number as an output. It would be great if you could draw
a diagram that helped us to understand such complex functions in the same way that
the graph of a real function helps us understand real functions.

If we consider the real function π
of π₯ equals π₯ squared, we can see key features of this function on its graph. For example, we see a root at π₯
equals zero. And we also have the minimum value
of the function here. The minimum value the function
outputs is zero. And we can read off the values of
the function from the graph. For example, we see that π of
negative one is one. Can we do the same for a complex
function like π of π§ equals π§ squared?

The only difference from our
previous function is that the domain is now the complex numbers rather than the real
numbers. Can we just draw the same graph
then call it the graph of π€ equals π of π§, where π§ can be a complex number? Think about how you could read off
the value of π of two plus π from this graph. Where is the point on this graph
which corresponds to an input of two plus π? Can we find two plus π on our
π§-axis? Maybe itβs there, for example.

Well, unfortunately not. Unlike real numbers, which we can
represent on a number line, we really need two dimensions or a whole complex plane
to represent complex numbers. So to have any hope at all, weβre
going to need two axes for our input. And the output of this function is
also a complex number. In our case, π of two plus π is
three plus four π. The output π€-axis needs to become
a whole π€-plane. Only then can we represent the
output.

So instead of an π₯-axis for
inputs, weβve got a whole π§-plane. And instead of a π¦-axis for
outputs, weβve got a whole π€-plane. Weβd like the point on our graph to
be something like two plus π, three plus four π, which with our four axes we have
to represent as a four-dimensional point two, one, three, four. The graph of this function is
therefore four-dimensional. And so itβs not particularly useful
to help us visualise what a complex function does. We need a different idea.

The idea is to think of the complex
plane of inputs being transformed by the function. So if again we think of the
function π of π§ equals π§ squared, we can think of it as the transformation π
that takes π§ to π§ squared. And then given an input like two
plus π, we can represent it on the plane of inputs, which we call the π§-plane. And we transform the π§-plane to
get the π€-plane of outputs. And we can represent the image of
the input two plus π on this transformed π€-plane. The image is three plus four π,
because two plus π squared is three plus four π.

What is the image of the point
negative one minus π? We use the formula π€ equals π§
squared. The image π€ when π§ is negative
one minus π is negative one minus π all squared, which distributing is one plus π
plus π plus π squared. We use the fact that π squared is
negative one to find that the image is two π. We mark this image point on the
π€-plane.

Looking at the π§-plane and the
π€-plane, can you see what kind of transformation that π which takes π§ to π§
squared is? Given just these two points, you
might think itβs a translation. But as the image of the complex
number zero is zero, we see that something more complicated is going on. We could continue adding points one
by one, looking at the before and after pictures to see if we can see how the
transformation affects the π§-plane. But it might take a lot of these
points and a lot of work before we see whatβs going on.

As we saw, by trying to consider
the graph of the complex function, we canβt think of the transformation of the
entire complex plane all at once. But doing things point by point is
going to be very tiresome. We need a compromise. The idea is to consider the images
of curves in the π§-plane. We draw a curve or perhaps a line
on the π§-plane. And we see where the transformation
map sits on the π€-plane. Letβs see how to do this using an
example with a simpler function.

Find an equation for the image of
the modulus of π§ equals two under the transformation of complex plane π takes π§
to π§ plus one plus π.

We draw the π§-plane that weβre
transforming and the π€-plane that weβre transforming at two. The object in the π§-plane that
weβre transforming is the locus modulus of π§ equals two, which is of course the
circle centred at the origin with radius two. Now the transformation weβre
dealing with takes π§ to π§ plus one plus π. And you might recognise this as a
translation.

A complex number π₯ plus π¦π goes
to π₯ plus π¦π plus one plus π, which is π₯ plus one plus π¦ plus one π. So in other words, the point π₯π¦
on the Argand plane goes to π₯ plus one, π¦ plus one. And so, in fact, this is
translation by the vector one, one. Now I can use this fact to draw the
image of the modulus of π§ equals two. The centre moves from zero, the
origin, to one plus π. And the circle is translated with
it. And furthermore, the radius of the
circle stays the same at two.

I can use this drawing of the image
to find its equation, which is after all what weβre looking for. But Iβd like to show you a more
algebraic method. Imagine that we didnβt recognise
this transformation as a translation and so we couldnβt draw the image. Instead, we used the formula of the
transformation. For π§ on the π§-plane, its image
is π€ equals π§ plus one plus π. And we can rearrange this to find
the object π§ in terms of its image π€. π§ equals π€ minus one minus
π.

Why is this helpful? Well, we know that the modulus of
π§ is two. And by writing π§ in terms of π€,
we get the modulus of π€ minus one minus π equals two, which is a locus in the
π€-plane. This is the equation of the image
that weβre looking for. We recognise this locus as a circle
with centre one plus π and radius two, which is exactly what we got by recognising
this transformation as a translation. The benefit of this algebraic
method though is that it works for any transformation. We donβt have to rely on having a
geometric interpretation of the transformation weβre given.

We can use this algebraic method to
gain geometric understanding of this transformation of the complex plane. Letβs apply this algebraic method
to another example.

Find an equation for the image of
the modulus of π§ equals one under the transformation of the complex plane π taking
π§ to a half π§.

We solve this in four steps. We first write π€ equals π of
π§. In our case, π of π§, the
transformed value of π§, is a half π§. So π€ is a half π§. We then rearrange this equation to
get π§ in terms of π€. Thatβs easy as π€ is a half π§. π§ is two times π€. We then substitute this for π§ in
the equation for the π§-plane locus to get a π€-plane locus. The locus in the π§-plane is the
modulus of π§ equals one, with π§ equal to two π€. We get the locus the modulus of two
π€ equals one.

And finally, we need to simplify
this locus. We use the fact that the modulus of
a product is the product of the moduli and that the modulus of two is just two. To get the simplified locus, the
modulus of π€ is a half. What weβre going to need to do to
solve this problem, we can interpret the object that weβre transforming as the unit
circle. Thatβs the circle centred at zero
with radius one. And the image that it transforms to
is the circle with centre zero again, but this time with radius a half. The circle in the π§-plane shrinks
then. Its radius halves. And this shouldnβt be surprising if
you recognise the transformation π as a dilation with scale factor a half.

Itβs important to note that even if
you have never come across the concept of dilation before, you could still get a
sense for what the transformation is doing by considering the images of various
objects in the complex plane. Letβs now see an example where
considering the image of a circle centred at the origin isnβt enough to understand
what the transformation is doing.

What Iβm going to do here is Iβm
going to replace the half here by π. Now the transformation takes π§ to
ππ§. Well, the steps are still the
same. π€ equals ππ§. So π§ equals negative π times
π€. And the π€-plane locus is then the
modulus of negative π times π€ equals one. We again use the fact that the
modulus function is multiplicative. And we know that the modulus of
negative π is one. So the simplest form of the
equation of the image is the modulus of π€ equals one.

The image of the unit circle under
this transformation is the unit circle. So itβs tempting to think that this
transformation hasnβt done anything at all. However, if you consider a point on
this circle, for example, the point π§ equals one, its image under the
transformation π§ goes to π times π§ is π. So while the image of the unit
circle is the unit circle itself, the image of a point on that unit circle isnβt
that point. This is because our transformation
is a rotation by π by two radians or 90 degrees counterclockwise. Our unit circle has been rotated,
therefore. While itβs hard to tell if a circle
has been rotated, itβs straightforward when dealing with a line or half line.

Letβs see an example.

Find an equation for the image of
the argument of π§ equals π by four under the transformation of the complex plane
π taking π§ to π to the negative ππ by three times π§.

We let π€ be the image of π§. So thatβs π to the negative ππ
by three times π§. And by multiplying both sides by π
to the ππ by three and swapping the sides, we find π§ in terms of π€. We can then substitute this
expression for π§ in the equation of the object. We can then use the fact we know
about the argument. The argument of a product is the
sum of the arguments. And on the left-hand side, we get
the argument of π to the ππ by three plus the argument of π€. And the argument of π to the ππ
by three is just π by three.

Subtracting this π by three and
simplifying, we find an equation of the image. Itβs the argument of π€ equals
negative π by 12. Both this equation and the equation
we started with are the equations of half lines. Looking in the diagram, we can see
that the image of the half line points in a different direction. Itβs been rotated. In fact, itβs been rotated by π by
three radians clockwise. Well, we canβt tell whether a
circle has been rotated. With a half line, itβs very
straightforward to see.

However, while itβs easy to tell
when a circle has been dilated because its radius has changed, itβs hard to tell
this for a half line. If we change the transformation so
it takes π§ to three π§ instead, as the argument of a third is zero, we find that
the argument of π€ is π by four. Looking at the diagram, itβs
actually impossible to tell that the half line has undergone a dilation thatβs scale
factor three.

To tell what a transformation is
doing, itβs a good idea to use both half lines and circles. The examples weβve seen were basic
transformations of the complex plane being translations, dilations about the origin,
or rotations about the origin. But we can compose such
transformations to get more complicated ones. For example, we can compose a
dilation scale factor π and a rotation by π radians counterclockwise. π§ is dilated to become ππ§. And this ππ§ is then rotated to
become π to the ππ ππ§, or ππ to the ππ times π§. So we see that the transformation
which corresponds to multiplying by a complex number in exponential form ππ to the
ππ is representing a dilation scale factor π about the origin, followed by a
rotation by π radians counterclockwise.

We can rewrite this statement using
the modulus and argument of the complex number weβre multiplying by. You might like to pause the video
to read this. A natural question to ask is, can
we write the transformation which takes π§ to π§ squared as a composition of basic
transformations? Can we therefore understand this
transformation as some combination of translations, dilations, and rotations?

Unfortunately, the answer is
no. Letβs prove this. In fact, weβll prove a stronger
statement. The only transformations that we
can write as a composition of any number of the basic transformations above β thatβs
translation, dilation, and rotation β are those of the former π takes π§ to ππ§
plus π, where π and π are complex numbers. As a result of this theorem, the
transformation taking π§ to π§ squared cannot be written as a composition of basic
transformations, as it does not have this form. So letβs prove this.

The first thing to note is that the
three basic transformations all have this form. For a generic translation, π is
one and π is π₯ plus π¦π. For a generic dilation, π is π
and π is zero. And for a generic rotation, π is
π to the ππ and π is zero. So this is indeed true.

The second part of this proof is
that the composition of two transformations of this form also has this form. We prove this by taking two
arbitrary transformations of this form, π four which takes π§ to ππ§ plus π and
π five which takes π§ to ππ§ plus π, where, remember, weβre allowing these
coefficients to be complex numbers. Their composition takes π§ to π
four of π five of π§. Well, π five of π§ is ππ§ plus
π. And π four of something is π
times that something plus π.

Simplifying, we find that their
composition is ππ times π§ plus ππ plus π. So the composition of
transformations takes π§ to ππ times π§ plus ππ plus π. And so this composition also has
the form π taking π§ to ππ§ plus π. And so weβve proved our
statement. Any two basic transformations will
have this form, and so their composition will also have the form. And if you go on to compose that
composition with another basic transformation, well, both of those transformations
have the form π taking π§ to ππ§ plus π. So their composition will also have
this form, and so on. However many basic transformations
you can pose with, you still end up with something of this form, π taking π§ to
ππ§ plus π. The transformation taking π§ to π§
squared is not of this form and so canβt be thought of as a composition of basic
transformations.

Letβs finish by applying some of
the techniques that weβve learnt to this transformation taking π§ to π§ squared. Weβll see if we can gain any
insight into what the transformation does to the complex plane and hence what the
complex function does to complex numbers.

Find Cartesian equations for the
images of the following loci under the transformation π taking π§ to π§
squared. Part a) The modulus of π§ equals
two. Part b) The real part of π§ equals
one. And part c) The imaginary part of
π§ equals one.

The first thing we do is draw our
π§- and π€-planes. Now we can see that the locus, the
modulus, of π§ equals two. This is a circle with centre at the
origin and radius two on the π§-plane. But what is it on the π€-plane? Well, π€ is π§ squared. So π§ is the square root of π€. And so the locus becomes the
modulus of the square root of π€ equals two. But weβd like the modulus of π€ on
the left-hand side. So we square both sides using the
fact that the product of moduli is the modulus of the product, to find that the
modulus of π€ is two squared, which is four.

So the effect of this
transformation on this circle centred at the origin is two its square root
radius. But remember, weβre looking for
Cartesian equations. If we call the real part of π€ π’
and the imaginary part of π€ π£, then the Cartesian equation is π’ squared plus π£
squared equals the radius four squared, which is 16.

Now letβs consider the real part of
π§ equals one. Such a π§ has the form one plus
π¦π, where π¦ is a real number. Itβs here on the π§-plane. But what is its image on the
π€-plane? Well, π€ is π§ squared, which is
therefore one plus π¦π squared, which is one minus π¦ squared plus two π¦π. But remember, we want a Cartesian
equation in terms of the real part of π€, which is π’, and the imaginary part of π€,
which is π£. And we can write down these values
in terms of π¦. But we donβt want them in terms of
π¦. We want an equation relating π’ and
π£. So we eliminate π¦ by rearranging
the second equation to get π¦ in terms of π£ and substituting this into the first
equation.

We find that π’ is one minus π£
over two squared, which we simplify to π’ equals one minus π£ squared over four. This is the equation of a parabola
in the π€-plane. And so, for the first time, we see
an example of when the image of a line is not itself a line. Itβs essentially the same procedure
for part c). We find the equation to be π’
equals π£ squared over four minus one. This completes the question. And again this is a parabola in the
π€-plane. Just looking at the image of the
circle, we might think that the transformation is just a dilation, possibly followed
by rotation. But looking at the images of the
lines, we see something more interesting is going on.

Here are the key points we covered
in this video, the most important one being the first. That we can gain an understanding
of transformations in the complex plane by considering their effect on lines and
circles.