Welcome to Lecture 5 in the 4th week of our course analysis of a complex kind. We're finally ready to study the Riemann mapping theorem, one of the biggest theorems in complex analysis. So far we've studied conformal mappings and we've seen the following. The conformal mappings from C hat, which was simply the complex plane extended by the point at infinity, to C hat are all of the form az+b/cz+d. Those are the Mobius transformations we've studied during the last two lectures. We've also seen that the conformal mappings from C to C are affine mappings of the form az+b. These are special Mobius transformations for which this number c is equal to 0. More is true, in fact. There is no conformal mapping from C to a domain D where D is a subset of C and not equal to C. Furthermore, there's also no conformal mapping form C hat to anything that is contained in C. Therefore, we can ask ourselves the question, what other conformal mappings are there? So far we've only see the Mobius transformations. For example, what are conformal mappings that map the unit disk, that I denoted with a double-bar D here, to some other set D that is contained in the complex plane? And here is the famous Riemann mapping theorem. Suppose D is a simply connected domain. Now domain meant open and connected. Simply connected means the domain has no holes. So you're not missing part of the domain in the middle of it. Such a thing is not allowed. So suppose we have a simply connected domain. It's an open connected domain with no holes in the complex plane, but not the entire complex plane. Then there exists a conformal mapping. And those are again analytic, one-to-one and onto mappings from D onto the unit disk D. You could in addition require that f(z0) = 0, but you don't have to require that. There are many choices, and in order to make this mapping unique, you're going to have to find a point that you want mapped under f to the origin. And you need even a little bit more to make the mapping unique. We'll look at that in a second. We say that this domain D is conformally equivalent to the unit disk. This theorem thus says that any simply connected domain that is not the entire complex plain is conformally equivalent to the unit disk. In order to make this mapping unique, as we mentioned before, we're going to have to specify 3 real parameters. For example, you can specify, as we said, a point z0 that is to be mapped under f to 0. That gives us 2 real parameters, the x-coordinate of z0 and the y-coordinate of z0, so we still need a third real parameter. You could, for example, specify the argument of the derivative at z0, which gives you another real parameter, for example our requiring that the derivative be positive, therefore the argument is equal to 0 at z0. The proof of the Riemann mapping theorem is beyond the scope of this course. Instead, we'll look at some examples and applications. For our first example let's look at the upper half plane. Let's say D is the upper half plane, so the set of all D's where the imaginary part of z is positive, and we're looking for a mapping that maps D conformally onto the unit disk. In fact, the restriction of a Mobius transformation can do this trick. Suppose f is the Mobius transformation that maps 0, 1, and infinity to 1, i, and -1. So 0 to 1, 1 to i, and infinity to -1. A Mobius transformation maps circles and lines to circles and lines. So in particular, the line through 0, 1, and infinity, for that is the real axis over here, has to be mapped to the line or circle through i, 1, and -1. Well, that is the unit circle. So this mapping has to map the real axis to the unit circle. And then what else is it supposed to do? It has to map the upper half plane either to the outside or the inside of the circle, and the lower half plane to the other part. But it turns out, if you look from 0 to 1 to infinity, we're kind of giving the real axis an orientation. And with that orientation, the domain D is to the left of the real axis. And their images of 0, 1, and infinity are 1, i, and -1, and that gives the unit circle an orientation. And again, the image is going to be to the left of this line that I just drew. Mobius transformations have that property. Therefore, we know that the upper half plane is mapped to the unit disk. And by the way, the lower half plane is mapped to the outside of the unit circle. So therefore, the restriction of the Mobius transformation f maps the upper half plane D to the unit disk. Can we find a formula for f? We sure can, we practiced that last class. f is to map 0 to 1, 1 to i, and infinity to -1. f is a Mobius transformation that has to be of the form az+b/cz+d. Since f does not map infinity to infinity, we know that this number c is not equal to 0, and therefore we can assume it is equal to 1. So f has the form az+b/z+d. Since f does map infinity to -1, we need a to be -1. Therefore, f is of the form -z+b/z+d. Furthermore, we know that 0 is to be mapped to 1. So when you plug in 0 you're supposed to get 1. So that means that b/d has to be equal to 1 or b has to equal d, which means f(z) is of the form -z+b/z+b. I just replaced d with b. Finally, we have that f(1) = i. So when you plug in 1 into this expression right here, you need to get i. So -1+b/1+b needs to be equal to i. If we solve this for b, we find b=i. How did we do that? Multiply both sides of the equation by 1+b. Then you find -1+b = i(1+b). We can collect b's on one side of the equation. There's this is one b from here, and then there's an i times b on the right-hand side. And then collect all the other terms on the right-hand side. So here we have an i, and we're going to bring this -1 to the other side as a +1. In other words, b = i+1/1-i. And I claim that is equal to i. Why is that the case? Let's multiply top and bottom by the conjugate of the denominator. So (i+1)(1+i), which is the conjugate of the denominator. And in the denominator I have (1-i)(1+i). If I multiply through on the top, I have an i squared which is -1. I have 1 times 1 which is 1. And I have 2 i's. On the bottom I find a 1. I have a -i squared which is a +1. And I have a -i and a +i and they cancel each other out. I'm thus left with 2i/2, and that is i. Thus, b is equal to i and therefore f maps the upper half plane D conformally onto the unit disk D. By the way, because we'll need this in a second, what is f(i)? i is in the upper half plane and f(i) = 0. So i is the point that is mapped to the origin under this mapping that maps the upper half plane conformally onto the unit disk. So this point i right here is mapped to the origin. Let's look at another example. Let Q be the first quadrant, so the domain in the complex plane that is bound by the positive real axis and the positive imaginary axis. Looking at the map from the previous example, it maps 0 to 1, it maps i to 0, and it maps infinity to -1. So it must map the line through i, 0, and infinity, which is the imaginary axis, to the line through 1, 0, and -1. That's the real axis. So the mapping also maps the imaginary axis to the real axis in addition to mapping the real axis to the unit circle. So what's the image of this quadrant Q which is stuck between the imaginary axis and the real axis? It must map it to something stuck between the unit circle and the real axis. And it turns out that it's the upper half of the unit disk. So the restriction of the Mobius transformation f constructed before to the upper quadrant here maps that first quadrant to the upper half of the unit disk, and I'll denote that D with a + right next to it. Another example, let's look at the map z squared. We know the map z squared is injective and analytic in that first quadrant. And it maps it conformally onto the upper half plane, which we named D in our first example. So the first quadrant is mapped conformally under the mapped z squared to the upper half plane. Now let's put all this together. Let's start with upper half of the unit disk. We know that the map f, the restriction of f to the first quadrant, maps the first quadrant conformally onto the upper half of the unit disk. Also its inverse, f inverse, maps the upper half of the unit disk to the first quadrant. The map z squared maps the first quadrant conformally onto the upper half plane, D. And the map f itself, restricted now to the upper half plane, maps the upper half plane conformally to the unit disk. If I look at the composition of all these mappings, so first f inverse, then g, then f, then this mapping maps the upper half of the unit disk to the entire unit disk. We have therefore constructed the Riemann mapping from the upper half of the unit disk to the unit disk. Now why is this important? Why is the Riemann mapping theorem so powerful? Many problems are easier to solve in a standard region, like the unit disk or the upper half plane or some other nice standard region, than in the region they're actually formulated in. A solution to a problem can then be found in the standard region and afterwards transported back to the original region via a Riemann map. Here's an example. Fluid flow can be modeled quite nicely in the upper half plane. For example, a flow in x-direction is depicted right here, or a flow from a source at the origin in this middle picture, or a combination of those two, like a flow together with the source at the origin as depicted right here. Understanding a similar flow in the different region would be way more complicated. So suppose we had a river that looked like this, and we knew water was flowing into the river from here but there was also source of pollution maybe coming in here and you wanted to track where things are going. That would be pretty difficult to model. However, you could map this river using a Riemann map onto the upper half plane, model the flow right there, and then transport the flow back to the river using a Riemann map. It's hard to find the Riemann map, but there are approximations that can be found, using a computer, of the Riemann map. There are many other examples, such as electrostatics, heat conduction, aerodynamics, how does air move around an airplane, and many more. Next week we'll begin integration. We'll study integration of analytic functions along curves and get to some very powerful theorems, amongst them the Cauchy integral theorem.