Video: Möbius Transformations

In this video, we will define Möbius transformations and learn how to interpret them in the complex plane.

17:36

Video Transcript

In this video, we’re going to learn about Möbius transformations, a class of transformations of the complex plane named after the German mathematician August Ferdinand Möbius, who also gives his name to the Möbius strip. This class of transformations has some very nice properties, some of which we’ll see in the video, which lead to applications to mathematics and beyond. Let’s jump straight in with the definition.

A Möbius transformation is a transformation of the complex plane of the form 𝑇 taking 𝑧 to 𝑎𝑧 plus 𝑏 all over 𝑐𝑧 plus 𝑑, where 𝑎, 𝑏, 𝑐, and 𝑑 are complex numbers and 𝑎𝑑 minus 𝑏𝑐 is nonzero. So essentially, the image of 𝑧 is given by the quotient of linear polynomials in 𝑧, with this added condition involving the coefficients of those polynomials. Let’s first see why we need this condition. Well, suppose 𝑎𝑑 minus 𝑏𝑐 was zero, then 𝑎𝑑 would be 𝑏𝑐. And hence, we could write 𝑎 in terms of the other coefficients. What would this tell us about the image of 𝑧? We use the formula we have and substitute our expression for 𝑎. And now we simplify. And here is a trick.

In the numerator, we see there is a common factor of 𝑏. And in the denominator, we see there is a common factor of 𝑑. Factoring 𝑏 and 𝑑 out, we see the other factors cancel. And the image of any complex number 𝑧 is just the constant quotient of coefficients, 𝑏 over 𝑑. When 𝑎𝑑 minus 𝑏𝑐 is zero, the whole complex plane is mapped to a single point, the points corresponding to the complex number 𝑏 over 𝑑. We don’t want to consider such transformations. So we require that 𝑎𝑑 minus 𝑏𝑐 is nonzero to stop this from happening.

Now that we understand this, we can focus on the form of 𝑇. A Möbius transformation is a transformation that can be written in this form. And 𝑎, 𝑏, 𝑐, and 𝑑 can be any complex numbers you like, as long as 𝑎𝑑 minus 𝑏𝑐 is nonzero. To get an example of the Möbius transformation, we just have to pick complex values for the coefficients 𝑎, 𝑏, 𝑐, and 𝑑. Here, I’ve made 𝑎 two plus 𝑖, 𝑏 negative two, 𝑐 one and 𝑑 negative three minus four 𝑖. Of course, to make sure that this is really an example of a Möbius transformation, we have to check that 𝑎𝑑 minus 𝑏𝑐 is nonzero. Well, with the choices we’ve made, we get a value of negative 11𝑖 for 𝑎𝑑 minus 𝑏𝑐, which is nonzero. And hence, this really is an example of a Möbius transformation.

This example is pretty complicated though. If we want to understand Möbius transformations, we should start by considering the simplest Möbius transformations we can. What we can do is choose 𝑐 to be zero and 𝑑 to be one. Then the denominator, zero 𝑧 plus one, is just one. And so the image of 𝑧 simplifies nicely to just 𝑎𝑧 plus 𝑏. Now, what does our condition become? 𝑏 times zero is zero, and 𝑎 times one is just 𝑎. So the condition for this to be a Möbius transformation is just 𝑎 is nonzero. We see then that any transformation of the form 𝑇 one taking 𝑧 to 𝑎𝑧 plus 𝑏, where 𝑎 is nonzero, is a Möbius transformation. And all the so-called basic transformations that we’ve seen before have this form.

For example, we can narrow our focus further and set 𝑎 equal to one. We see then that any transformation of the form 𝑇 two taking 𝑧 to 𝑧 plus 𝑏, where 𝑏 is a complex number, is a Möbius transformation. There’s no extra condition required on the coefficients as 𝑎 is one and not zero. And you recognise 𝑇 two. Any translation in the complex plane will have this form. And hence, all translations of the complex plane are Möbius transformations. The class of Möbius transformations contains the class of translations. Now instead of setting 𝑎 equal to one in 𝑇 one, we can set 𝑏 equal to zero. Doing this leaves us with transformations of the form 𝑇 three taking 𝑧 to 𝑎𝑧, where 𝑎 is a complex number.

Here, we do need to be careful about the condition that 𝑎 is nonzero. If 𝑎 were zero, then 𝑇 three would just map every complex number 𝑧 to zero. The entire complex plane would be mapped to the origin. We can include this condition by making 𝑎𝑏 a nonzero complex number. Again, we recognise the form 𝑇 three. All dilations of the complex plane with centre the origin have this form, as they’re all rotations about the origin. In general, this is the form of a composition of such a dilation and rotation. And so, we can see that all dilations with centre the origin, rotations about the origin, and compositions thereof are Möbius transformations.

And, in fact, although we won’t prove this, the form 𝑇 one includes all dilations in the complex plane and no matter what the centre is and all rotations of the complex plane no matter which point in the complex plane we’re rotating about. So Möbius transformations include all translations, all dilations, and all rotations of the complex plane. However, these are the simplest cases of Möbius transformations. And so we might not be getting a good idea of the general Möbius transformation just by looking at translations, dilations, and rotations. For a general Möbius transformation, 𝑐 is not zero and 𝑑 is not one. And so, we have to do some division as well. Let’s look at the simplest Möbius transformation which involves some division.

We set 𝑎 to be zero, 𝑏 to be one, 𝑐 to be one, and 𝑑 to be zero. And so, simplifying, we see that the image of 𝑧 is its reciprocal, one over 𝑧. The condition on the coefficients is satisfied. And so this really is a Möbius transformation. I claim that with our understanding of translations, dilations, and rotations, in order to understand what a general Möbius transformation does, it’s enough to understand this reciprocal transformation. I’ll back this claim up later. But, first, let’s look at the reciprocal transformation in detail.

It’s easiest to see what the reciprocal transformation does by considering its effects on the modulus and the argument of the complex number. So we write our complex number in exponential form. Its image is then one over 𝑟𝑒 to the 𝑖𝜃, which by DeMoivre’s theorem is 𝑟 to the negative one times 𝑒 to the 𝑖 negative 𝜃. So the modulus 𝑟 goes to its reciprocal, one over 𝑟. And the argument 𝜃 goes to its opposite, negative 𝜃. Given the point in the complex plane, we can find its image on the reciprocal transformation in two steps. First, we take the modulus to its reciprocal leaving the argument unchanged. And then, we take the argument to its opposite, leaving the modulus unchanged.

So if we started with 𝑧 one, then this is the image of 𝑧 one, which is one over 𝑧 one. The first step, where we take the modulus to its reciprocal, is slightly mysterious. It’s important to realise that if we start with a complex number whose modulus is less than one, then upon taking the reciprocal of the modulus, we’ll end up with a greater modulus. And so, the image of 𝑧 two has a greater modulus than 𝑧 two itself. However, we might recognise the second step where we take the argument 𝜃 to its opposite. This step on its own takes a number to its complex conjugate. Geometrically, this represents reflection in the real axis.

You might like to spend some time exploring where this transformation maps points in different parts of the complex plane. You will see, for example, that for this transformation to make sense, 𝑧 must be nonzero. In particular, it’s worth exploring where the reciprocal transformation maps points inside, on, and outside the unit circle. So pause the video and have a think about that.

We continue by finding the images of various loci in the complex plane under this reciprocal transformation. A transformation which maps the 𝑧-plane to the 𝑤-plane is defined by 𝑇 taking 𝑧 to one over 𝑧, where 𝑧 is nonzero. Part one, find an equation for the image of the modulus of 𝑧 equals two under the transformation. Part two, find an equation for the image of the argument of 𝑧 equals three 𝜋 by four. Part three, find a Cartesian equation for the image of the imaginary part of 𝑧 equals two. And part four, find a Cartesian equation for the image of the modulus of 𝑧 minus 𝑖 equals one.

Our transformation takes 𝑧 to its reciprocal, one over 𝑧. And so, 𝑤 is one over 𝑧. In the first part, we want to find the image of the modulus of 𝑧 equals two under this transformation. We can invert our transformation to find 𝑧 in terms of 𝑤. Multiplying both sides by 𝑧 and then dividing through by 𝑤, we find that 𝑧 is one over 𝑤. Substituting this then, we find that our image has the equation the modulus of one over 𝑤 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 𝑤 equals a half.

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

Let’s now move on to find the image of the argument of 𝑧 equals three 𝜋 by four. Again, we use the fact that 𝑧 is one over 𝑤. And so the argument of one over 𝑤 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 𝑤 equals negative three 𝜋 by four. Again, it’s 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 there’s a complex number with argument 𝜃 will be mapped to a complex number with argument negative 𝜃. So this makes sense.

Now let’s find the image of the imaginary part of 𝑧 equals two. It may not be clear what to do with the imaginary part of one over 𝑤. What we do as we ask for a Cartesian equation is to write 𝑤 in terms of its real and imaginary parts, which we call 𝑢 and 𝑣. How do we find the imaginary part of one over 𝑢 plus 𝑖𝑣? 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. It’s negative 𝑣 over 𝑢 squared plus 𝑣 squared. We might be tempted to stop simplifying at this point. But if we divide by two and complete the square in 𝑣, we get something. This is recognisably the equation of a circle. Let’s 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 𝑤-plane.

Finally, we find the image of the modulus of 𝑧 minus 𝑖 equals one. We substitute one over 𝑤 for 𝑧 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 𝑤. Now we can substitute 𝑢 plus 𝑖𝑣 for 𝑤. We can simplify on the left-hand side. And now, we’re ready to apply the definition of the modulus. We square both sides. And negative 𝑢 squared is the same as 𝑢 squared; so these cancel. Distributing on the left-hand side, we see the 𝑣-squared terms cancel as well. So we find that two 𝑣 is negative one. And hence, the equation of our image is 𝑣 equals negative a half. We see then that the circle with centre 𝑖 and radius one is mapped to the straight line of points in the 𝑤-plane with imaginary part negative a half.

Let’s 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. These examples were carefully chosen to illustrate all the possibilities. The reciprocal transformation maps any circle to either a circle or a straight line, and any straight line to either a straight line or a circle. We can define a circline to be a curve that is either a circle or a straight line. Then it is a theorem that we won’t prove in this video that the reciprocal transformation maps circlines to circlines.

The image of a circline under the reciprocal transformation is a circline. The reciprocal transformation is representative of all Möbius transformations in this way. Any Möbius transformation maps circlines to circlines. The image of a circle or line under a Möbius transformation will always be a circle or a line. There are no other possibilities. This is one of the nice properties of Möbius transformations which makes them so useful. It’s worth spending some time convincing yourself that this is clearly true for translations, rotations, and dilations, all of which are special cases of Möbius transformations.

Now that we hopefully understand the reciprocal transformation a bit better, let me justify my claim that this helps us to understand any Möbius transformation. We take an arbitrary Möbius transformation. Then if 𝑐 is nonzero, we define some auxiliary transformations. Notice they are all Möbius transformations. In particular, we have two translations, a dilation with centre the origin, rotation about the origin or combination thereof and the reciprocal transformation we’ve just looked at. Hopefully, all of these transformations are now familiar. Then it turns out that our arbitrary Möbius transformation is a composition of these auxiliary transformations.

I’m not going to go through the algebra here. But you can check this if you’d like. If we understand the transformations 𝑇 one through 𝑇 four individually, we can understand the original transformation as a composition. If 𝑐 is equal to zero, then it’s even easier. We can prove many of the properties of Möbius transformations by decomposing them in this way. For example, we know that 𝑇 one as a translation takes circlines to circlines. And we have good reason to expect that 𝑇 two does too. 𝑇 three is a dilation or rotation about the origin or maybe a combination of those. And so, it certainly takes circlines to circlines. And 𝑇 four is another translation which takes circlines to circlines.

So what does 𝑇 do to circlines? Well, 𝑇 one takes the circline to a circline which is mapped by 𝑇 two to a circline which is mapped by 𝑇 three to a circline which is finally mapped by 𝑇 four to a circline. So the arbitrary Möbius transformation 𝑇 maps circlines to circlines. Or, at least we can reduce the problem of checking that an arbitrary Möbius transformation maps circlines to circlines to checking that the reciprocal transformation maps circlines to circlines. Möbius transformations behave nicely under composition. Let’s prove that the composition of two Möbius transformations is itself a Möbius transformation.

Prove that the composition of two Möbius transformations is a Möbius transformation.

We let 𝑇 one and 𝑇 two be two arbitrary Möbius transformations with coefficients 𝑎, 𝑏, 𝑐, 𝑑 and 𝛼, 𝛽, 𝛾, 𝛿, respectively. Our task is then to show their composition is a Möbius transformation. By the definition of composition, we see that the transformation takes 𝑧 to 𝑇 one of 𝑇 two of 𝑧. And we know what 𝑇 two of 𝑧 is; so we substitute it. It’s 𝛼𝑧 plus 𝛽 over 𝛾𝑧 plus 𝛿. Now we just have to apply 𝑇 one to this. We use the formula for 𝑇 one of 𝑧 substituting 𝛼𝑧 plus 𝛽 over 𝛾𝑧 plus 𝛿 for 𝑧. Now, we simplify and rearrange the terms to write it in the expected form.

So it looks like we have the form of a Möbius transformation. But, we’re not quite done yet. We still need to check the constraint on the coefficients that the product of these coefficients minus the product of these coefficients is nonzero. We apply the distributive property and notice that some terms cancel, that two of the remaining terms have a common factor of 𝑎𝑑. And the other two terms have a common factor of 𝑏𝑐. Now we can see the common factor of 𝛼𝛿 minus 𝛽𝛾, which allows us to factor completely. Now, how does this help us show that this is nonzero?

Well, we know that 𝑎𝑑 minus 𝑏𝑐 is nonzero and also that αδ minus 𝛽𝛾 is nonzero. And hence, this quantity as a product of nonzero numbers is itself nonzero. Hence, as required, the composition of 𝑇 one and 𝑇 two is a Möbius transformation. And as 𝑇 one and 𝑇 two were arbitrary Möbius transformations, we’ve proved the composition of two Möbius transformations is a Möbius transformation as required.

The key points covered in this video are as follows. A Möbius transformation is a transformation of the form 𝑇 taking 𝑧 to 𝑎𝑧 plus 𝑏 over 𝑐𝑧 plus 𝑑 where 𝑎, 𝑏, 𝑐, and 𝑑 are complex numbers and 𝑎𝑑 minus 𝑏𝑐 is nonzero. Möbius transformations map lines and circles to lines and circles. And the transformation you get by composing two Möbius transformations is always itself, a Möbius transformation.

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