[BLANK AUDIO] Welcome to the third lecture in the fourth week of our course, Analysis of a Complex Kind. Today we'll start learning about Mobius transformations, which are special conformal mappings. This is a two lecture part series. So we will also learn about Mobius transformations in part four. Let's start with a definition. A Mobius transformation is also sometimes called a fractional linear transformation, and it is simply a function of the form f of z equals az plus b over cz plus d, with these numbers a, b, c, and d are complex numbers such that ad minus bc is not equal to zero. So for example, you could say something like f(z) = 3z- 4 divided by z + 15i. In this case, ad which is 3 times 15i minus bc which is 4 is not equal to zero, and so this is a Mobius transformation. In other words, we have a linear function at the top and a linear function at the bottom. What happens to s if you let z go toward infinity. What the term ad becomes more and more and important than term b because more and more important than the same thing happens in the denominator. So in the limit f of z is something like az over cc because b and d have lost their importance given that az and cd are so big. And that is equal to a over c. Actually you need c to be non-zero for this be true. If c was equal to zero, then the denominator is just d. And az + b over d goes to infinity, c goes to infinity. So we say f of z goes to a over c as long as c is non-zero and it goes to infinity if c=0. We therefore allow z to be equal to infinity and define f of infinity to be a over c if c is non-zero, and f over infinity to be infinity if c=0. We can do a similar thing for the point that gets mapped to infinity, namely the point z. For which you're going to be dividing by zero. When do you divide by zero in this function? Well, when cz + d = 0. When is cz + d = 0? Well, that is the case when cz = -dz or when z =- d over c. So we say f of -d over c is = to infinity, as long is c is non zero. If c = 0, you can't make the denominator = 0. We therefore regard these Mobius transformations f, not only as mapping from c, but this mapping is defined on C hat, the extended complex plane and mapping the extended complex plane C hat to the extended complex plane C hat. What's the derivative of a Mobius transformation? Well we can just simply use the quotient rule. We square the denominator, which is (cz + d) squared. We write down the denominator again, (cz + d) multiplied with the derivative of the numerator, which is a minus the numerator times the derivative of the denominator, which is c. When you multiply through here on the top, you have a cza term + da- azc- bc. And you notice that cza and azc are the same thing, they cancel each other out. So all that's left is ab- bc in the numerator and (cz+d) squared in the denominator. Therefore this condition ad-bc to be non 0. That condition simply guarantees that the derivative is non 0. Because if the derivative was 0, it would be to 0 everywhere and the function would be constant. So this condition ad- bc not equal to 0 guarantees that f is not a constant function and automatically we know that f prime of z is then never equal to 0, for all z. Now notice if you multiply each of these parameters, a, b, c, d by a constant k, you get the exact same mapping. You would get the mapping kaz + kb over kcz + kd and you'd notice that you can factor out a k in top and bottom. And therefore the top then is az + b. And the bottom is you can factor in the k times cz plus d. You can cancel these k's. And you're left again with az + b over cz + d. So you get the exact same function again. So when multiplying all these parameters a,b,c,d by a constant you get the same mapping and therefore for a given mapping, the constants a, b,c,d are not uniquely determined. You can multiply all by a non-zero number and get the same mapping. A Mobius transformation is actually one-to-one and onto from C hat to C hat. If you wanted to prove this you would do something like the following. Pick a w in C hat and ask yourself when if f(z) = w? f(z) = w when az + b over cz + d is equal to w. You multiply both sides of that equation by the denominator cz + d, that becomes az + b needs to be w times (cz + d). If we solve that equation for z you can find az times a over here and az time wc over here. If we bring that to the other side we find z times a minus wc. Similarly, if we bring all the non z terms to the other side we have a wd- b on the right hand side. Dividing by a- wc you can solve for z, and find that z needs to be wd- b over -wc + a. In other words, no matter what w you pick, you can find exactly 1z. That gets mapped under f to w. That makes the function onto but also one-to-one because there's exactly 1z. There's not a second z, my formula gave me this is what z has to be. There's not another z that is also mapped onto w. That doesn't exist and therefore, the function is also one-to-one. So for each w, there is one and only one z such that f(z)=w and you can include infinity in this discussion by just looking carefully enough and understanding carefully enough what the images and pre-images of infinity are. Mobius transformations are therefore conformal mappings from C hat to C hat. In fact, Mobius transformations are the only conformal mappings from C hat to C hat. If you are asking for a conformal mapping, that's defined on C hat. And maps to 2 C hat. The only thing you will find are the Mobius transformations. Let's look at some examples. Suppose c is equal to 0, so this number is equal to 0, and d is equal to 1. The denominator is therefor just 1 and the function becomes f(z) equals az + b. These are also called affine transformations, because they simply multiply z by the number a, which corresponds to a rotation dilation. And after that add b, which means add a translation. These mappings map infinity to infinity, and therefore also map C to C. No point in C is mapped to infinity and therefore they map C to C. They're also therefore conformal mappings from C to C in addition to being conformal mappings from C hat to C hat. And again, in fact, these are the only conformal mappings from C to C. In particular, if you in addition let b equal to 0, then the mapping is of the form f(z) is equal to az. Therefore, you're multiplying z by a complex number which corresponds to a rotation and a dilation, if you write a as its absolute value times e to the i theta. You're rotating z by the angle theta and you're stretching or shrinking it by a factor of absolute value of a. On the other hand if a is equal to 1 and b is arbitrary, then f(z) becomes z+b, and that's simply a translation. You take your point z and move it to z + b. Let's look at another example. Suppose a is 0, b is 1, c is 1, and d is 0. So the function becomes f(z) is equal to 0 x z, az, + b, which is 1 / 1 x z, that's my z, + 0. But that is simply 1 / z. This is called an inversion. If you write z as re to the i theta, what is 1 / z? What's 1 / r times 1 / e to the i theta, when 1 / e to the i theta we proved is the same as e to the -i theta. So 1 / z is 1 / r times e to the -i theta, which means f(z) is obtained from z by taking the inverse of the length and taking the negative of the angle. In other words, if you're given the point z, The certain angle, theta, then you take the negative of that angle and you take 1 over the length of z. So if z was bigger than 1 in absolute value, then 1 / z is going to be smaller than 1, so maybe this is 1 / z. Now we notice if z was of absolute value equal to 1, Then f(z) is still of absolute value equal to 1. So the circle as a whole set of points is preserved under f, even though point wise points move around, but they stay on the circle. So f maps the circle to the circle, how about points that are outside of the circle? Points outside of the circle have a radius bigger than 1. After being mapped with f, they're going to have a radius less than 1. So, they're going to be mapped into the circle. On the other hand, points inside the circle start having a radius less than 1, putting 1 over that, they're going to be outside of the circle. So f interchanges the outside and the inside of the unit circle while keeping the unit circle fixed as a set. Not point wise, but as a whole set the circle is matched to this one. Furthermore a circle centered at 0, where is that mapped to? Let me again draw the circle of radius 1. And now let's draw an additional circle of radius, let's say, one-half. What is its image? Every point on that circle has radius one-half. After mapping it with f, the images are going to have radius 1 over one-half, which is 2. And so the image is going to be another circle, but of radius 2. And the same is true for any other circle. You start with an even smaller circle, its image is going to have an even larger radius, but they're all going to be circles. So a circle centered at the origin is mapped under this one mapping, 1/ z, to another circle centered at the origin. How about other circles? Let's look at an example. What are the images of circles? Let's look at the circle, a set of all z, such that z -3 is equal to 1. So what is that? That is the circle that is centered at 3 and has radius 1. So this is K. The question is, what is the image of K under f. Let's say we have a point w that is in f(K). How did w get to be in f(K)? Well 1 / w needs to be in the original circle. When is 1 / w in K? When 1 / w satisfies this condition, so 1 / w- 3 must be of absolute value 1. Let's multiply both sides of this equation by absolute value of w. So we find that 1- 3w needs to be equal to w in absolute value, and then we can square both sides of the equation as well. So 1- 3w squared needs to be w squared. How do we find 1- 3w absolute value squared? Remember, the absolute value squared of a complex number can be found by taking that complex number and multiplying it with its conjugate. But what is the conjugate of (1- 3w)? You can take the conjugate individually, it's the conjugate of 1, that's just 1 because 1 is a real number. Conjugate of 3 is 3, so all we have to do is the conjugate of w. Now we can multiply through. This is 1- 3 times the conjugate of w- 3w, minus, minus, plus, 3 times 3 is 9, and w times its conjugate is the norm of w squared. That's this next line that I wrote out right here. So 1- 3w- 3 times the conjugate of w + 9w squared is equal to w squared. We'll bring this 1 w squared over to the left hand side and find that 8w squared- 3w- 3 times the conjugate of w is equal to -1, we'll bring this 1 over to the other side instead. And now, we divide both sides of the equation by 8. And what we end up with after dividing by 8, Is |w| squared- 3/8w- 3/8 times the conjugative w = -1/8. We notice that we can factor this. We can factor this left hand side into (w- 3/8) (w conjugate- 3/8). Because when we multiply this through, we get w times conjugate of w, which is this w squared right here. We get w times -3/8, which you have right here. You have -3/8 times the conjugative w, and you get this one extra term. You get an extra term plus nine sixty fourths. That wasn't there before so in order to make this a true equation, we need to also add 9/64 on the right hand side of the equation, and that's what we did right here. So we see that (w- 3/8)(conjugate of w- 3/8) needs to be equal to 9/64- 1/8. Now we see on the left hand side, that is simply |w- 3/8| quantity squared. And on the right hand side, we see 9/64- 8/64, which is 1/64 which is 1/8 squared. And then, if you take a square root of both sides of the equation, you end up with |w- 3/8| = 1/8. So, what have we just seen? We have seen that if w belongs to the image of the circle K. When w-3/8=1/8. But that's another circle. That is the circle, centered at 3/8. Here's 1, there's 1/2, 3/8 is here, and the radius 1/8. So it's this little circle right there. So w belongs to the image of the original circle K if, and only if, it's on this circle. Other words the original circle K gets mapped to this little tiny circle. Here's the picture again. The circle K is right here, the circle of radius 1 centered at 3, here's the unit circle in blue. Under f it is mapped to this little tiny circle inside the unit circle, but it's a circle. So again a circle was mapped to a circle. Let's check another example. Suppose were looking at the circle z. Such that|z-1|=1. So it's the circle of radius 1, centered at 1. Here's the unit circle. And we're looking at the circle of radius 1, centered at 1. So here's the circle we're looking at. Oops, it's supposed to go through its origin. What is its image under f? Again, we do the same thing as before. w belongs to f(k). Well how do they get there? 1/w has to belong to K. When this 1/w in this circle K. Well when the distance of of being 1/w and 1 is equal to 1. We multiple by w square both sides of the equation we get |1-w| squared = |w| squared. We do the same thing, we break 1-w quantity squared is 1-w times 1-w conjugate. We multiply through we find that 1-w- the conjugate of w is less than |w| squared is equal to |w| squared. But now something funny happens. The w squared term goes away. They cancel each other out and we're only left with w + the conjugate of w. We brought this from the other side is = 1. What is w + the conjugative w? That is two times the real part of w. If we divide that equation by 2, we find that the real part of w must be equal to 1/2. So w belongs to f(K) if it's real part is equal to 1/2. Which ones have real part equal to 1/2? Any point on this vertical line through the point 1/2 has a real part equal to 1/2. In other words, the orange circle gets mapped to that green line. Here's the picture. Here's the circle of radius one centered at one and I drew the unit circle as well. Well, its image under f is the vertical line through the point one half. Well, because f has the property that f(f(z)) is f(1/z) = z. We also find that f maps the line who's real part is 1/2 back to the circle of radius 1, centered at 1. And similarly the previous example that little circle of radius 1/8 centered at 3/8 is now under f also to the circle of radius 1, centered at 3. So it appears that circle somehow can get back to circles or lines. Let see what can happen to a line. Let l be the line, through the origin given by the set of all z such as z is of the form t + it. In other words, it's the set of all points where real and imaginary parts agree. What does that look like? Which points have the same real and imaginary parts, what's that line through the origin? If z is of the form t + it, what's f(z)? What's 1/z? Another way of writing 1/z is multiplying top and bottom by the conjugate of z, which turns the denominator into the |z| squared and numerator into conjugate of z. The conjugate of z is t- it. And what is z squared? Well z squared is |{t + it}| squared. And you get the norm of a complex number by squaring its real part and its imaginary part and then taking the square root. But since we want the norm squared it's simply real part squared plus imaginary part squared. That's 2t squared. So f(z) is then t- it / 2t squared. We can pull that apart into the real part and the imaginary part. The real part is there, 1/2t. The imaginary part is -1/2t. If we call this number here s then we see this is also s. So our number f(z) of the form s- is. Where is s-is? Those are complex numbers with the imaginary part as the opposite of the real part. This have these numbers of the form. So, the orange line a piece that be map on the green line. Here's the image again. L seems to be mapped onto f(L). Now notice, this function f(z) = 1 / z maps 0 to infinity, and it maps infinity to 0. So if we include infinity in the line, then the line L gets knocked to the line f(L), which is the line through the origin just at the opposite angle. So images of lines and circles seem to be lines and circles. And indeed. That is true. Every Möbius transformation maps circles and lines to circles or line. So the statement is to be read as follows. The image of a circle is a circle or a line. The image of a line is a circle or a line. You could actually view a line simply as a circle through infinity and you could then say Möbius transformations map circles to circles. We will see how to prove this in our next lecture. Let me finish up by showing you one more fact about Mobius transformations. Given three distinct points z1, z2, and z3 in the extended complex plane, you can find a unique Möbius transformation that maps z1 to zero, z2 to one, and z3 to infinity. So these three points determine the Möbius transformation uniquely, but you can also find one if that's exactly what you wanted to do. You can actually prove this fact by simply writing out this Möbius transformation, and here it is. Let's check that this Möbius transformation actually does what I promised it would do. What is F (Z1) for example. F(Z1) = Z1-Z1, that's 0. Divided by Z1- Z3. That's not zero. Because Z1, Z2 and Z3 are supposed to be distinct to each other times Z2-Z3 divided by Z2-Z1. Because this one term up here is equal to 0 and everything else is non zero that is equal to 0. So, indeed f map Z1 to 0 and it is a Möbius transformation. We have this extra factor here that makes it look a little bit different from a Möbius transformation but it's off the form k times Z-Z1 over Z-Z3, and you could simply make that into K Z- K Z1 divided by Z-Z3. And now you recognize it is indeed a Möbius transformation, we just wrote it in a somehow different form. Let's next check what is F of Z2. When I plug in Z2, I have a Z2 and a Z1 right here divided by Z2-Z3 and then I multiply a Z2-Z3 over Z2-Z1. And I see these terms cancel each other out. And this simplifies to 1. So Z2 indeed does map to 1. Finally what's the image of Z3? Z3-Z1 is a non-zero number, because Z1 and Z3 are not the same. But in the denominator we get Z3-Z3 which is 0. All the other numbers are non-zero, so this image is defined to be infinity. So indeed Z3 is mapped to infinity. Next up are constructions of more examples. How can you actually find Möbius transformations that you are interested in. And this fact that we just proved in this particular Möbius transformation we just wrote down will be extremely helpful in creating your own Möbius transformations.