Lesson Video: Transformations of the Complex Plane | Nagwa Lesson Video: Transformations of the Complex Plane | Nagwa

# Lesson Video: Transformations of the Complex Plane Mathematics

In this video, we will learn how to translate and rotate a complex number in the complex plane.

17:56

### Video Transcript

In this video, weβre going to learn about transformations of the complex plane. Really, weβre interested in complex functions. Thatβs 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 itβs 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, weβre 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, weβve got a whole π§-plane. And instead of a π¦-axis for outputs, weβve got a whole π€-plane. Weβd 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 itβs 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 itβs 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 whatβs going on.

As we saw, by trying to consider the graph of the complex function, we canβt 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. Letβs 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 weβre transforming and the π€-plane that weβre transforming at two. The object in the π§-plane that weβre 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 weβre 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 weβre looking for. But Iβd like to show you a more algebraic method. Imagine that we didnβt recognise this transformation as a translation and so we couldnβt 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 weβre 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 donβt have to rely on having a geometric interpretation of the transformation weβre given.

We can use this algebraic method to gain geometric understanding of this transformation of the complex plane. Letβs 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€. Thatβs 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 weβre going to need to do to solve this problem, we can interpret the object that weβre transforming as the unit circle. Thatβs 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 shouldnβt be surprising if you recognise the transformation π as a dilation with scale factor a half.

Itβs 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. Letβs now see an example where considering the image of a circle centred at the origin isnβt enough to understand what the transformation is doing.

What Iβm going to do here is Iβm going to replace the half here by π. Now the transformation takes π§ to ππ§. Well, the steps are still the same. π€ equals ππ§. 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 itβs tempting to think that this transformation hasnβt 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 isnβt 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 itβs hard to tell if a circle has been rotated, itβs straightforward when dealing with a line or half line.

Letβs 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 ππ by three times π§.

We let π€ be the image of π§. So thatβs π to the negative ππ by three times π§. And by multiplying both sides by π to the ππ 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 ππ by three plus the argument of π€. And the argument of π to the ππ by three is just π by three.

Subtracting this π by three and simplifying, we find an equation of the image. Itβs 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. Itβs been rotated. In fact, itβs been rotated by π by three radians clockwise. Well, we canβt tell whether a circle has been rotated. With a half line, itβs very straightforward to see.

However, while itβs easy to tell when a circle has been dilated because its radius has changed, itβs 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, itβs actually impossible to tell that the half line has undergone a dilation thatβs scale factor three.

To tell what a transformation is doing, itβs a good idea to use both half lines and circles. The examples weβve 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 ππ§. And this ππ§ is then rotated to become π to the ππ ππ§, or ππ to the ππ times π§. So we see that the transformation which corresponds to multiplying by a complex number in exponential form ππ to the ππ 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 weβre 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. Letβs prove this. In fact, weβll prove a stronger statement. The only transformations that we can write as a composition of any number of the basic transformations above β thatβs translation, dilation, and rotation β are those of the former π takes π§ to ππ§ 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 letβs 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 ππ 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 ππ§ plus π and π five which takes π§ to ππ§ plus π, where, remember, weβre 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 ππ§ plus π. And π four of something is π times that something plus π.

Simplifying, we find that their composition is ππ times π§ plus ππ plus π. So the composition of transformations takes π§ to ππ times π§ plus ππ plus π. And so this composition also has the form π taking π§ to ππ§ plus π. And so weβve 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 ππ§ 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 ππ§ plus π. The transformation taking π§ to π§ squared is not of this form and so canβt be thought of as a composition of basic transformations.

Letβs finish by applying some of the techniques that weβve learnt to this transformation taking π§ to π§ squared. Weβll 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 weβd 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, weβre 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 letβs consider the real part of π§ equals one. Such a π§ has the form one plus π¦π, where π¦ is a real number. Itβs 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 donβt 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. Itβs 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.

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