WEBVTT
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In this video, weβre going to learn about transformations of the complex plane.
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Really, weβre interested in complex functions.
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Thatβs functions which take a complex number as an input and return a complex number as an output.
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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.
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If we consider the real function π of π₯ equals π₯ squared, we can see key features of this function on its graph.
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For example, we see a root at π₯ equals zero.
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And we also have the minimum value of the function here.
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The minimum value the function outputs is zero.
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And we can read off the values of the function from the graph.
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For example, we see that π of negative one is one.
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Can we do the same for a complex function like π of π§ equals π§ squared?
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The only difference from our previous function is that the domain is now the complex numbers rather than the real numbers.
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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?
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Think about how you could read off the value of π of two plus π from this graph.
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Where is the point on this graph which corresponds to an input of two plus π?
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Can we find two plus π on our π§-axis?
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Maybe itβs there, for example.
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Well, unfortunately not.
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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.
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So to have any hope at all, weβre going to need two axes for our input.
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And the output of this function is also a complex number.
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In our case, π of two plus π is three plus four π.
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The output π€-axis needs to become a whole π€-plane.
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Only then can we represent the output.
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So instead of an π₯-axis for inputs, weβve got a whole π§-plane.
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And instead of a π¦-axis for outputs, weβve got a whole π€-plane.
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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.
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The graph of this function is therefore four-dimensional.
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And so itβs not particularly useful to help us visualise what a complex function does.
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We need a different idea.
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The idea is to think of the complex plane of inputs being transformed by the function.
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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.
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And then given an input like two plus π, we can represent it on the plane of inputs, which we call the π§-plane.
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And we transform the π§-plane to get the π€-plane of outputs.
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And we can represent the image of the input two plus π on this transformed π€-plane.
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The image is three plus four π, because two plus π squared is three plus four π.
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What is the image of the point negative one minus π?
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We use the formula π€ equals π§ squared.
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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.
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We use the fact that π squared is negative one to find that the image is two π.
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We mark this image point on the π€-plane.
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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?
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Given just these two points, you might think itβs a translation.
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But as the image of the complex number zero is zero, we see that something more complicated is going on.
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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.
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But it might take a lot of these points and a lot of work before we see whatβs going on.
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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.
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But doing things point by point is going to be very tiresome.
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We need a compromise.
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The idea is to consider the images of curves in the π§-plane.
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We draw a curve or perhaps a line on the π§-plane.
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And we see where the transformation map sits on the π€-plane.
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Letβs see how to do this using an example with a simpler function.
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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.
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We draw the π§-plane that weβre transforming and the π€-plane that weβre transforming at two.
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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.
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Now the transformation weβre dealing with takes π§ to π§ plus one plus π.
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And you might recognise this as a translation.
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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.
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So in other words, the point π₯π¦ on the Argand plane goes to π₯ plus one, π¦ plus one.
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And so, in fact, this is translation by the vector one, one.
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Now I can use this fact to draw the image of the modulus of π§ equals two.
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The centre moves from zero, the origin, to one plus π.
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And the circle is translated with it.
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And furthermore, the radius of the circle stays the same at two.
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I can use this drawing of the image to find its equation, which is after all what weβre looking for.
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But Iβd like to show you a more algebraic method.
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Imagine that we didnβt recognise this transformation as a translation and so we couldnβt draw the image.
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Instead, we used the formula of the transformation.
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For π§ on the π§-plane, its image is π€ equals π§ plus one plus π.
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And we can rearrange this to find the object π§ in terms of its image π€.
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π§ equals π€ minus one minus π.
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Why is this helpful?
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Well, we know that the modulus of π§ is two.
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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.
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This is the equation of the image that weβre looking for.
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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.
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The benefit of this algebraic method though is that it works for any transformation.
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We donβt have to rely on having a geometric interpretation of the transformation weβre given.
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We can use this algebraic method to gain geometric understanding of this transformation of the complex plane.
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Letβs apply this algebraic method to another example.
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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§.
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We solve this in four steps.
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We first write π€ equals π of π§.
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In our case, π of π§, the transformed value of π§, is a half π§.
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So π€ is a half π§.
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We then rearrange this equation to get π§ in terms of π€.
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Thatβs easy as π€ is a half π§.
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π§ is two times π€.
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We then substitute this for π§ in the equation for the π§-plane locus to get a π€-plane locus.
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The locus in the π§-plane is the modulus of π§ equals one, with π§ equal to two π€.
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We get the locus the modulus of two π€ equals one.
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And finally, we need to simplify this locus.
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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.
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To get the simplified locus, the modulus of π€ is a half.
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What weβre going to need to do to solve this problem, we can interpret the object that weβre transforming as the unit circle.
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Thatβs the circle centred at zero with radius one.
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And the image that it transforms to is the circle with centre zero again, but this time with radius a half.
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The circle in the π§-plane shrinks then.
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Its radius halves.
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And this shouldnβt be surprising if you recognise the transformation π as a dilation with scale factor a half.
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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.
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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.
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What Iβm going to do here is Iβm going to replace the half here by π.
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Now the transformation takes π§ to ππ§.
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Well, the steps are still the same.
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π€ equals ππ§.
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So π§ equals negative π times π€.
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And the π€-plane locus is then the modulus of negative π times π€ equals one.
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We again use the fact that the modulus function is multiplicative.
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And we know that the modulus of negative π is one.
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So the simplest form of the equation of the image is the modulus of π€ equals one.
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The image of the unit circle under this transformation is the unit circle.
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So itβs tempting to think that this transformation hasnβt done anything at all.
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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.
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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.
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This is because our transformation is a rotation by π by two radians or 90 degrees counterclockwise.
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Our unit circle has been rotated, therefore.
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While itβs hard to tell if a circle has been rotated, itβs straightforward when dealing with a line or half line.
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Letβs see an example.
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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§.
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We let π€ be the image of π§.
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So thatβs π to the negative ππ by three times π§.
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And by multiplying both sides by π to the ππ by three and swapping the sides, we find π§ in terms of π€.
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We can then substitute this expression for π§ in the equation of the object.
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We can then use the fact we know about the argument.
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The argument of a product is the sum of the arguments.
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And on the left-hand side, we get the argument of π to the ππ by three plus the argument of π€.
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And the argument of π to the ππ by three is just π by three.
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Subtracting this π by three and simplifying, we find an equation of the image.
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Itβs the argument of π€ equals negative π by 12.
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Both this equation and the equation we started with are the equations of half lines.
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Looking in the diagram, we can see that the image of the half line points in a different direction.
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Itβs been rotated.
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In fact, itβs been rotated by π by three radians clockwise.
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Well, we canβt tell whether a circle has been rotated.
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With a half line, itβs very straightforward to see.
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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.
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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.
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Looking at the diagram, itβs actually impossible to tell that the half line has undergone a dilation thatβs scale factor three.
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To tell what a transformation is doing, itβs a good idea to use both half lines and circles.
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The examples weβve seen were basic transformations of the complex plane being translations, dilations about the origin, or rotations about the origin.
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But we can compose such transformations to get more complicated ones.
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For example, we can compose a dilation scale factor π and a rotation by π radians counterclockwise.
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π§ is dilated to become ππ§.
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And this ππ§ is then rotated to become π to the ππ ππ§, or ππ to the ππ times π§.
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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.
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We can rewrite this statement using the modulus and argument of the complex number weβre multiplying by.
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You might like to pause the video to read this.
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A natural question to ask is, can we write the transformation which takes π§ to π§ squared as a composition of basic transformations?
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Can we therefore understand this transformation as some combination of translations, dilations, and rotations?
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Unfortunately, the answer is no.
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Letβs prove this.
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In fact, weβll prove a stronger statement.
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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.
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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.
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So letβs prove this.
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The first thing to note is that the three basic transformations all have this form.
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For a generic translation, π is one and π is π₯ plus π¦π.
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For a generic dilation, π is π and π is zero.
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And for a generic rotation, π is π to the ππ and π is zero.
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So this is indeed true.
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The second part of this proof is that the composition of two transformations of this form also has this form.
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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.
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Their composition takes π§ to π four of π five of π§.
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Well, π five of π§ is ππ§ plus π.
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And π four of something is π times that something plus π.
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Simplifying, we find that their composition is ππ times π§ plus ππ plus π.
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So the composition of transformations takes π§ to ππ times π§ plus ππ plus π.
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And so this composition also has the form π taking π§ to ππ§ plus π.
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And so weβve proved our statement.
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Any two basic transformations will have this form, and so their composition will also have the form.
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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.
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So their composition will also have this form, and so on.
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However many basic transformations you can pose with, you still end up with something of this form, π taking π§ to ππ§ plus π.
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The transformation taking π§ to π§ squared is not of this form and so canβt be thought of as a composition of basic transformations.
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Letβs finish by applying some of the techniques that weβve learnt to this transformation taking π§ to π§ squared.
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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.
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Find Cartesian equations for the images of the following loci under the transformation π taking π§ to π§ squared.
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Part a) The modulus of π§ equals two.
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Part b) The real part of π§ equals one.
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And part c) The imaginary part of π§ equals one.
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The first thing we do is draw our π§- and π€-planes.
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Now we can see that the locus, the modulus, of π§ equals two.
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This is a circle with centre at the origin and radius two on the π§-plane.
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But what is it on the π€-plane?
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Well, π€ is π§ squared.
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So π§ is the square root of π€.
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And so the locus becomes the modulus of the square root of π€ equals two.
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But weβd like the modulus of π€ on the left-hand side.
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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.
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So the effect of this transformation on this circle centred at the origin is two its square root radius.
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But remember, weβre looking for Cartesian equations.
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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.
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Now letβs consider the real part of π§ equals one.
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Such a π§ has the form one plus π¦π, where π¦ is a real number.
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Itβs here on the π§-plane.
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But what is its image on the π€-plane?
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Well, π€ is π§ squared, which is therefore one plus π¦π squared, which is one minus π¦ squared plus two π¦π.
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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£.
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And we can write down these values in terms of π¦.
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But we donβt want them in terms of π¦.
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We want an equation relating π’ and π£.
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So we eliminate π¦ by rearranging the second equation to get π¦ in terms of π£ and substituting this into the first equation.
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We find that π’ is one minus π£ over two squared, which we simplify to π’ equals one minus π£ squared over four.
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This is the equation of a parabola in the π€-plane.
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And so, for the first time, we see an example of when the image of a line is not itself a line.
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Itβs essentially the same procedure for part c).
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We find the equation to be π’ equals π£ squared over four minus one.
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This completes the question.
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And again this is a parabola in the π€-plane.
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Just looking at the image of the circle, we might think that the transformation is just a dilation, possibly followed by rotation.
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But looking at the images of the lines, we see something more interesting is going on.
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Here are the key points we covered in this video, the most important one being the first.
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That we can gain an understanding of transformations in the complex plane by considering their effect on lines and circles.