Welcome to Week 4 in the course Analysis of a Complex Kind. Today we'll learn about Inverse Functions of Analytic Functions, we'll learn that these are themselves analytic under reasonable assumptions. As a motivation, let's start with the Logarithm Function. The logarithm function is going to be an inverse of the exponential function, so to remember things, let's remember what the exponential function does. Remember if w is of the form u + iv, then e to the w was defined to be e to u and that's our regular exponential function from u analysis times e to the iv. Where e to the iv simply stands for cosine v plus i sine v and again these are the real value of cosine in sign functions. So with that in mind, suppose we have the following problem, given a certain z in the complex plane not the origin. Find a w in the complex plane, such that e to the w is equal to z. In other words, we want to start with a z over here in the z plane, which is not the origin, and we want to find a w in the w plane, such that the exponential function maps e to the w to z. So this point w is to be mapped to z under the exponential function. So what we're looking for is an inverse to the exponential function. The exponential function maps w to z, we're looking for the function that starts at z and finds the w but on to the exponential function goes back to z. Now we know the exponential function is periodic with period 2pi i, so once we have found one w, we know that w+2pi i also gets mapped to the exact same z value. And so this function starting with a z finding a w is not going to be unique, we're going to have to pick which one of these possible many values we want to choose. So, how do we do this? Let's write z in polar form, as the norm of z times e to the i-th theta where theta is an argument of z and let's write w in cartesian form as u + iv. Then e to the w is equal to z becomes e to the w is equal to absolute value of z e to the i theta because we wrote z in polar form. And since we wrote w as u + iv, this then become e to the u, e to the iv, which is just e to the w = norm of z e to the i theta. Here we see two complex numbers in polar form, when are two complex numbers in polar form the same? Well, their distance from the origin needs to be the same, so e to the u needs to be the same thing as the opposite value of z, and their angles need to agree, module 2pi or 4pi or multiple of 2 pi. So, v and theta need to agree up to a multiple of 2pi, which means v needs to be an argument of z. We therefore can define the following. We define the logarithm of z with an uppercase L to be the natural log of the absolute value of z + i times uppercase argument of z and that's defined to be the principal branch of the logarithm. Remember I told you we have many choices for the argument of z. If we choose the principal argument of z, then we get the principal branch of logarithm, and it's defined as the natural logarithm of the absolute value of z, plus i times uppercase argument of z. For the more lower case logarithm of z is ln absolute value of z plus i times argument of z with lowercase argument, it's a multi-valued function it allows all possible values. For the argument down here and it can be obtained from uppercase argument of z which is just adding the multiple of 2pi i. Both of these functions have the property that e to the log of z is equal to z. Let's look at some examples. So remember, the logarithm of z, the principle branch is defined as the natural logarithm of the absolute value of z plus i times uppercase Arg z. What is the logarithm of 1 for example? Well, we have to take the natural logarithm of the absolute value of 1 + i uppercase argument of 1. The absolute value of 1 is simply 1. The natural logarithm of 1 is 0. Remember the natural logarithm function looks like this, and it actually crosses the x-axis here at 1. So the natural logarithm of 1 is equal to 0, plus i times the argument of 1, but the argument of 1 is the angle that the line segment from the origin to 1, forms with a positive real axis and that is 0. Remember, the argument uppercase argument is between negative pi and pi. Let's look at another example. What's the logarithm of i? I is this number right here. Again, we need to find its distance from the origin which again is 1 plus i times its argument. Argument here is pi over 2, so the logarithm of i is natural log of 1, which is 0, plus i pi over 2, so just i pi over 2. How about the logarithm of negative 1? Again, the distance from the origin of negative 1 is 1. Natural logarithm of 1 is 0. The argument is pi. In other words, the logarithm of -1 is i pi, how about the logarithm of 1 + i? 1 + i is this number right here, its distance from the origin is square root of 2. So we get that the logarithm of 1+i is the natural log of square root of two, plus i times the argument, and the argument here is pi/4. So the logarithm of (1+i) is the natural logarithm of root two, plus i times pi/4. What continuity properties does the logarithm function have? Again, here's the definition, log z is the natural logarithm of the absolute value of z plus i times argument of z. Let's look at the parts that this function is made up of. First of all, let's look at the function z maps to the absolute value of z. Well that's a continuous function. It's varies continuously at C, varies in the complexity/h distance from the origin, varies continuously, it never jumps. The absolute value of z is the Number that is non negative, it could be a 0 when Z is equal to 0, otherwise it is a positive number, so we can take it's natural Logarithm. The natural Logarithm function is a continuous function, and so, the composition of these two is a continuous function as long as you stay away from 0, because the natural logarithm of 0, is undefined. This whole function here is a continuous function. Now, let's look at this second part, the argument of z. Well, we studied that already. We noticed, it's a continuous function as long as you stay away from the negative real axis, on the negative real axis the argument jumps. Because we said the argument of a number that's close to the negative real axis is almost pi, but if you're close to the real axis from below, then the argument is close to negative pi. And so, as you're crossing the negative real axis, the argument jumps from pi to -pi thereby not being continuous. Everywhere else in the complex plane, the argument is a continuous function. Angle that appoint forms with a positive real axis changes continuously except the jumps over the negative real axis. So, the argument function is continuous on the complex plane minus the negative axis therefore, the sum of these two continuous functions is continuous where the both are continuous. In other words, in the complex plane minus the negative real axis. However, the negative real axis is a real issue as z approaches an x value on the negative real axis. So, this is my negative x value if you approach that point from above, the logarithm approached the natural logarithm of the absolute value of minus x which is x plus i times pie coming from above. Whereas, if you are approaching this point from below, the logarithm approaches the logarithm of x minus and those are clearly not the same which means the Logarithm Function just like the argument function is not continuous on the negative real axis. And it's also not defined at the origin of course. In other words, the Logarithm Function is a continuous function on C- the negative real axis and nowhere else. More is true in fact. In fact, it is true that the principal branch of the logarithm function is analytic in C- the negative real axis. We haven't proved this yet. We won't really prove it, but we'll look at that in more general fact. But let's first find the derivative of this function. How do you find the derivative? Well, we do know that e to the Log z = z, that's how we define the logarithm function, so it undoes what the exponential function does, so e to the Log z = z. And we now differentiate both sides of this equation wIth respect to z. Differentiating the right hand side is pretty simple. The derivative of z is 1. What's the derivative of e to the Log z? That's a composition of two functions and we need to apply the chain rule. The derivative is does, the derivative of the exponential function, so e to the Log z again, times the derivative of the insight, just the derivative of Log z that's we don't know what that is. So that's kind of still our question mark right here. But we can now solve this equation for that derivative by dividing both sides of the equation by e to the Log z. But what is e to the Log z? E to the Log z is simply z, so we divide both sides of the equation by z and find that the derivative of Log z is 1 over z. Just like it was in real analysis. The derivative of l and x, Was nothing but one over x. So now, the derivative of the principle branch of the logaritm is 1 over z. Here's the more General Theorem. Suppose that f is an analytic function defined on some set of U. So here, I drew U down here and f is this function. And it maps you into the complex plane. And suppose there exists a continuous function g, it's this one down here, that's defined on some domain D, this is my D. And maps D into U, it doesn't have to go onto U but somehow into U. So here, I drew g(D). Such that, if I start with the point z over here, follow it with g to g of z and then, apply the f that was defined on all of U so in particular at this point end up where I started, f of g of z. So, f is sort of an inverse of g, f of g of z, they undo each other f of g of z is equal to z. If that holds for all z and d, where the function g is continuous, then g in fact is analytic and has derivative one over f prime of g of z. That's our inverse function there. Its enough to find an inverse function that is continuous if the original f was analytic, that guarantees that g is analytic and we even have its derivative. And this derivative formula is found just like we found the derivative formula for the logarithm function. Let's look at some examples, suppose f of c is the function z squared. That's clearly an analytic function, we know it's derivative of prime of z is 2z and it's defined in the whole complex plane, so U is equal to the whole complex plane here in this case. Let's now look at the principal branch of the square root function. Remember, we couldn't really define the principal branch of the square root function on the entire complex plain we had the same issue we had to cut open the complex plain. Take away the negative real access, for example and that made the principle branch of the square root function the continuous function. Remember how that was defined? The principle branch of the square root function was defined as taking the square root of the absolute value of z times e to the i. R z over 2. That's the principle branch of the square root function and again because the argument functions continuous in the complex plane except for the negative real axis. This function is going to be a continuous function in the complex plane minus the negative real axis. And f of g of z is equal to z for all z in the complex plane minus the negative real axis because if I take the square root and then square a number, I'm at the number that I started with. So, if I start with a point over here, I take the square root which is going to put the number somewhere over here. And then, I subsequently square it again I end up again where I started, namely over here. G is a continuous function and by the theorem it's therefore an analytic function in D. And its derivative is one over f prime of g of z but f prime is 2z. So that's one over two times g(z), and g(z) was the function square root of z. So the derivative of the square root function is 1/2 times the square root function. That's exactly the same that we had in calculus with the derivative of the square root of x. But this time it's this principal branch of the square root function, which is defined in the entire complex plane we have to exclude the negative real axis because the function was not continuous across the negative real axis. Let's look at another example. f is the same function as before. f(z) is z squared. Again, the derivative is 2s. Again, we define an inverse of this function, but on a separate domain. We're going to define a function that is continuous, but this time we want it to be continuous on the domain D tilde, which is the complex plane minus the positive real axis. If we want to define a square root function here, let's figure out what the square root of z looks like. All square roots of z have the form square root of the f that value of z times e to the i argument of z over 2 plus optionally i pi. Since e to the i pi is 1, we get two possibilities for the square root function. It's + or- the square root of the absolute value of Z x e^1 argument of Z/2. In other words, that's + or- the principle square root of z. Those are my two choices with square roots for a point. So now I want to define a square root function on the complex plane minus the positive real axis. If I choose the regular square root function on the upper half plane, then as my z's approach a point -x on the negative real axis, the square root of those points wants to be on the positive imaginary axis. So if I wanted to continuously continue this to the lower half plane. I can now choose the principle branch of the square root function which maps points close to the negative real axis. So the negative imaginary axis. Instead I need to choose the negative square root function in the lower half plane and the positive square root function, on the upper half plane. And that's exactly what we're going to do. When the imaginary part of z is positive, we chose the square root of z. When it's negative, we chose the negative square root of z. This function has the property that it is continuous on the negative real axis but it's going to have a jump somewhere and that jump is going to happen on the positive real axis. That's why we need to exclude the positive real axis because as z approaches a positive value on the real axis, the square root of z approaches a real value that's in the positive real axis, but minus square root of z is coming from below approach a value on the negative real axis. So this function h(z) is not continuous across the positive real axis but continuous everywhere else in the complex plane. Again If I say square either of those values. Square root of the z or minus square root of z I get z. So f of h of z is z for all in the plane minus the positive praxis. H is continuous in this new domain and therefore by the theorem this function h is also analytic in D tilde. And this plane minus the positive real axis and its derivative is 1/f prime (h(z)), which is 1/2h(z), so this function h(z) right here. So that is equal to 1/2 square root of z. If the imaginary part of Z is positive and 1 over -2, square root of z, if the imaginary part of z is negative. Let's finish up by just remembering some terminology. Given any function f from set U to set V, and this doesn't have to be in the complex plane, this is general terminology. We say such a function f is injective, or 1 to 1, provided that whenever you start with two separate points A and B, and you said U, they get matched to separate points f(a) and f(b) in your set V. They cannot be matched to the same point. So that is the case when we say the function is injective. If A and B get mapped to the same point of functions no longer injective. But if we're all possible combinations of A's and B's they're not the same, f(a) and f(b) are also not the same then f is injective. f is surjective, also called onto, provided that no matter which v you start with, you can always find some value in u that is mapped under f to that v value. In other words, every value in v is an important number, it is hit by some x value. Could be hit by more than one x value, but there's at least one x value that is mapped onto this y value. F is a bijection, we also say it's 1-1 and onto if it is both injective and surjective. Let's look at some examples. The function z^2 is a bijection if I choose my domains appropriately. So here I chose for my u, the set of all z's whose real part is positive. So my u is the right half plane. And my v that I chose up here, is the whole plane minus the negative real axis, I'm taking away the negative real axis. And the claim is that the function z squared is now a bijection from U to V. Whenever I choose two distinct values in U, they're going to be mapped to distinct values in V because I restrict my function to the right half plane. We know that the function z squared is typically not a bijection because for every point I can just choose the negative of that point, when I square these two they get mapped to the same value, somewhere over here. However, I'm not allowing any point n is negative to be contained in my set U at the same time. Therefore, that makes my function z squared injective. Moreover, every point in V has a preimage. I just divide this argument in half and take the square root of the distance from the origin by dividing the argument in half. Everything lands over in the right-half plane, and there's in fact for every point in V, I can find a point in U that it came from. And therefore, that makes this function, z squared, a bijection from U on to b. If on the other hand, I allow U to be the entire complex plane and V to be the entire complex plane, and look at the functions z squared between those two. Well, it's not injective, as we saw before, because for every one z, there's another one z. It gets mapped to the same point over here. But it is surjective because for every point in the complex claim they can find the square root that is coming from somewhere over here. So it is surjective but no longer injective. Finally if I do restrict my u to C minus the negative real axis and I allow v to be the entire complex plane and I look at the function square root of z. I know that function is injective, no two points get mapped to the same value over on the right hand side under square root of z. But it is not surjective, because all of this under the principal branch of the square root function gets mapped into the right half plane. And so any point over on the left does not have a pre image under the square root function where it would've come from. So no point in the plane minus the negative real axis gets knocked to the left half plane under the principle branch of the square root function. So that's enough for today. Next up are conformal mappings, Moebius transformations, and looming in the distance the Riemann mapping theorem.
