Welcome back to our third week in our course, Analysis of a Complex Kind. This is the second lecture this week and we'll discuss the Cauchy-Riemann Equations. These are equations that result from a function being complex differential As a reminder, remember that a function, f, that is a complex function, can be split up into its real part view and its imaginary part v. When we write z as x + iy, then f(z) becomes u(x,y) + iv(x,y), where now u and v are functions that spit out real numbers and take two real parameters, x and y. Let's look at an example just to remember this setting. Suppose f(z) is the function z squared, with z being x + iy, that becomes x + iy quantity squared. And if I multiply through, this is x squared plus 2xiy. I wrote that term back here. Plus, i squared, y squared, but i squared, y squared, is minus y squared, and so it's purely a real, and I wrote it here in the front. If I collect all the real terms, I get x squared minus y squared, so that's my u(x,y) and if I collect all of the purely imaginary terms, I get 2xy, which is v(x,y). And so my function u(x,y) would be x squared minus y squared and v(x,y) would be 2xy. For example, if z is the number 2 + i, I could just calculate f(z) by taking 2 + i and squaring it, which is 4 + 4i + i squared. Since i squared is -1, this simplifies to 3 + 4i. Alternatively, I could use my functions, u and v, to evaluate the function, f. So, writing z as x + iy, 2 + i means that x is 2 and y is 1, the number that i gets multiplied with here. And so, u(x,y) and I plug in x = 2 and y = 1. So, it's u(2,1). And since the function, u, is x squared minus y squared, it takes the first parameter, that's everything squares it, and subtracts from now the second parameter squared, so 2 squared minus 1 squared, that's 4- 1, that's 3. And similarly, v(x,y) would be v(2,1). And since v takes those two parameters, multiplies them and multiplies the result by 2, I would get 2 times 2 times 1, which is 4. So f(2 + i) can also be evaluated by taking u(2,1) + iv(2,1), u(2,1 )was 3, v(2,1) was 4, so 3 + 4i, and that is indeed the same result we obtained by just plugging in z after the function, f. So these are two alternative ways of looking at the same function, and they'll become helpful when we want to look at the function, u, and the function, v, and their differentiability properties. What do we already know about the function z squared? Well we know this function is differentiable everywhere in C, in fact, we found its derivative to be 2z for all z and C. Let's just start by looking at the function u(x, y) more carefully. Again, u(x, y) is simply x squared minus y squared. So, for example, if we fix the variable, y, at a certain value, for example, y = 3, then the function, u, only depends on x, because u(x,y) is then u(x,3) if we always plug in 3 for y, and the function u(x,3) becomes x squared minus 9. This function only depends on x, because y is fixed. And therefore we can differentiate this function with respect to x. There's no y in there. I can find the derivative as I do in calculus of the function x squared minus 9. The derivative of that function is 2x. We write du/dx(x,3) is equal to 2x, that derivative we just found. This is a notation for that derivative. We write du/dx, it's called the partial derivative of u with respect to x, because there also would have been a choice of finding a derivative with respect to y, and to indicate that I had a choice, I use this funny, curly d. So it's a partial u with respect to partial x, that's what that's called. But all that means is, we fix y at a certain value and find the derivative with respect to x. Another notation is to omit this partial u, partial x, and put a little x next to u. That also indicates I'm going to take the derivative with respect to x. So that was 2x. More generally, if I plug in a different value for y, for example 4, then u(x,4) would be x squared minus 16, the derivative is still 2x. And we notice, no matter what value of y I plug in, as long as I fix it, the derivative of u with respect to x is going to be 2x. And so we write more generally, du/dx(x,y), or ux(x,y) is 2x. And again, this is called the partial derivative of u with respect to x. But we can do the same thing for the function, v, v(x,y) is the function 2xy. So for example, again, if we fix y = 3, v(x,3), we plug in 3 for y, it's 2 times x times 3, that's 6x. The derivative of this function with respect to x is 6. And so we would write dv/dx(x,3) or vx(x,3) equals 6. More generally, if I have a different value for y, for example, y = 4, then v(x,4) would be 2 times x times 4. So that would be 8x. And so, the derivative, vx(x,4) would be 8. If I have yet another value, try v(x,5), v(x,5) is 10x. So that derivative would be 10. So the derivative seems to depend on y, even though we're fixing y, the x derivative depends on y. If we look at this function very carefully, v(x,y) = 2xy. If you think about this abstractly and say to yourself, two was a constant, and I'm going to keep y constant as well, I'm not going to change it, I'm going to plug in a value, I just don't know which value that is, but if I fix these two numbers, than this is really equal to these two constants, 2y, and those are both constant, times x. And the only variable that really varies is x. The derivative of such a function, with respect to x, how do you find the derivative while constants go to the side? So I keep the 2y, and then differentiate x. But what is the derivative of x? It's 1. In other words, the derivative of this function is simply 2Y. The X derivative of V is two times Y and that agrees with these results what we got eight when Y was equal to four, ten when Y was equal to five, six when Y was equal to three. So, generally we have the X derivative of V is equal to two times Y. This is called the partial derivative of V with respect to X. Obviously, we can do the same thing by fixing X and differentiating with respect to Y. For example, when X is equal to two, then U of XY is U of 2Y which is four minus Y squared and the derivative of four minus Y squared with respect to Y is minus 2Y. This is the regular derivative from calculus. More generally, if I look at this function U of XY, as a function of Y and X is something fixed, even though typically X is the variable that varies. If I now think in my head, this is a constant and I find the derivative of this function. The derivative of a constant sitting there by itself is zero. The derivative of minus Y squared if Y is the variable, that's minus 2Y. So more generally, the partial derivative of U with respect to Y at X and Y is minus 2Y. Look at the notation. It's very similar to the notation we looked at earlier, except the partial X in the denominator here was replaced by a partial Y. Similarly, we put a little Y next to the U to indicate we're using the partial derivative of U with respect to Y. You can use either of those notations, almost to use the U sub Y notation. I can do the same thing for a V. Again, let's fix an X value, two in this example, and look at V of X and Y. V of X and Y is then two times two times Y, so four Y. And the derivative of the function four Y, with respects to Y, is equal to four. So we would write Div V, Div Y of two and Y or V Sub Y of two and Y equals four. More generally, if I look at this function two times X times Y and X is fixed and Y varies, then the function V of X and Y could be written as two X, these are my two constants, times Y. So that if I want to find the Y derivative of this function, it's a constant times Y, the derivative of a constant times Y is simply the constant, 2X. So more generally the partial derivative of V with respect to Y is two times X. Let's summarize what we have found. We started with a function, F of Z equals Z squared, which we broke up into a real part U, and its imaginary part V. U was X squared minus Y squared. V is 2XY, and then we found all kinds of derivative. The derivative in the complex sense of F was 2Z. And then we had these somewhat real types of derivatives, the partial derivatives. UX is 2X, UY is minus 2Y, VX was 2Y, and VY was 2X. We notice certain things. UX is 2X. So is VY, so UX is equal to VY. We also notice VY is minus 2Y, and VX is 2Y. They seem to be opposites of each other. UY is minus VX and finally, we notice when you look at the derivative of F, F prime. That's two times Z. We could write that as 2X plus I times 2Y. 2X, that's my UX derivative. 2Y, that's my VX derivative. So I could write this derivative F prime, I could put that together from UX plus IVX. We call that the FX derivative. Alternatively, I could also use UY and VY. Since UX is equal to VY. I could write this as VY plus I times and now VX is equal to minus UY. And if I've factor out an I. This becomes minus UY minus IVY. Because I times I is minus one together with this minus one that gives me just this VY. If I indeed factor a minus I as I did right here, I find that F prime is the same as minus I times UY + IVY. That's the FY derivative. Taking all of F and taking the derivative of F simply with respect to Y. So F prime is the same thing as taking FX or minus I times FY. Was that coincidence? Was it just true for the function Z squared? Let's look at another example. Let's look at the function F of Z equals 2Z cubed minus 4Z plus one. Again we write Z as X plus IY and when we plug that in, we find two times X plus IY quantity cubed minus four times quantity X plus IY plus one. Now we need to multiply this through unfortunately, so it gets a little bit ugly. X plus IY quantity cubed is X cubed plus 3X squared IY plus 3X times IY squared, so I squared Y squared plus IY cubed, which is I cubed, Y cubed. Then we would subtract from that four times X plus IY, so 4X and 4IY, and then add the one in the end. Let's collect turns. You want to find the function U and the function V. U is all the real parts and V consists of all the purely imaginary parts. So, what's real? X cubed is real. Well, that two in front of it that gives me 2X cubed. X squared IY, that's not real. 3XI squared Y squared, I squared is negative one, so this is another real term. With a two in front of it, that becomes 6XY squared and with a negative sign because of the I squared. I cubed is minus I, so that's imaginary. Minus 4X is right here. And then there's finally the one that's right here. So we've now collected all the real terms. The rest is imaginary terms. 3X squared IY goes under B with a two in front of it. That becomes 6X squared Y, a factor of the I. Then I have I cubed Y cubed. I cubed is negative I. So this is negative Y cube with a two in front of it, so negative 2U cubed. And finally minus 4IY shows up as minus four in my function V. In other words, u of x and y is 2x cubed- 6xy squared- 4x + 1 and v of x and y 6x squared y- 2y cubed- 4y. Let's find again, the x and the y derivatives of these functions. Remember to find the x derivative of u, we pretend y is a constant such as 4 or 16, just the constant. And we just take the derivative with respect to x as we're doing calculus. So, the x derivative of u becomes 6x squared- 6y squared. Y squared is a constant, so that goes to the side, derivative of x is simply 1- 4. The derivative of this constant, 1 which has no constant attached to it is 0. So, that gives us the x derivative of u. If you wanted to find the y derivative of u, you would have to keep x constant. Now, 2x cubed is a constant with no y attached to it. The derivative of such a constant is 0. The derivative of -6xy squared, with respect to y, is found by realizing -6x are constants that go to the side. y squared is the function that varies, depending on y. Its derivative is 2y, times the constant, I find -12xy. - 4 x doesn't depend on y, + 1 doesn't depend on y. The derivatives of these charts are just 0. Similarly, I find the x derivative of v. X derivative of v, y is constant now, is 12xy. 2y cubed and 4y are constants that don't depend on x, their derivatives are 0. The y derivative of v, 6x squared is a constant multiplied by y. So, the derivative of that is 6x squared. The derivative of 2y cubed is 6y, the derivative of 4y is 4. So, I find all of these derivates. Let's see what happened. Let's look at ux and vy and we realize, again, ux is equal to vy, let's look at uy and compare that to the vx. Again, they're opposites of each other. So again, ux is vy and uy is -vx but this fraction was a whole lot more complicated than the previous example. So now, I'm starting to think maybe this was not coincidence. Indeed, it was not. Let's look at how this was all related to the derivative of f. Again, I wrote down ux, uy, vx and vy for us to remember. Let's find f prime, f prime in the complex sense. You find the complex derivative of 2z cubed- 4z + 1. The derivative of 2z cubed is 6z squared, the derivative of -4z is -4. The derivative of 1 is just the constant of 0. So, the derivative of f is 6z squared- 4. If we plug in z being x + iy, this becomes 6(x + iy) quantity squared- 4. You can multiply through. x+iy quantity squared isx squared- y squared + 2ixy, 2ixy times 6 is where this 12ixy comes from. And here, I wrote down the real part, which consists of 6x squared i multiply x squared by 6. - 6y squared, again, the -y squared times six. And -4 coming from this 4 right here. But when I look at this carefully, I recognize that this real term here, 6x squared- 6y squared- 4, that's ux. And this term, 12ixy, that's i times vx, in other words, f prime again is ux + i times vx. Or again I could also write as -i times uy + ipy. So again, f' = fx(z) or -ify(z). So, it's no longer a coincidence, here's a theorem. Suppose, f is broken up into u + iv. And suppose that it is differentiable at a point z0. Then, these partial derivatives all exist at z0 and satisfy these equations. Ux is a vy and uy is -vx. This is a consequence of being differentiable. These equations are called Cauchy-Riemann Equations. The very famous equations named after Cauchy Riemann. Also, f prime, if f is differentiable in a complex sense can then be written as ux + ivx which is the x derivative of f at z0 or -i times uy + ivy which is -i times the y derivative of f at z0. Again, what this theorem is saying that if a function is differentiable in the complete sense, then, it must satisfy the Cauchy-Riemann Equations. It doesn't say that the Cauchy-Riemann Equations are satisfied for all functions, only for those that are differentiable in the complex sense. If you wanted to prove this theorem, which we're not going to do in this lecture, it would actually not be that hard. You'd look at the difference quotient. f(z0 + h0 -f(z0)/h. And the limit of this different quotient must exist if f is differentiable at z0 and that's your assumption. The limit of the difference quotient existing as h goes to 0 means we can let h go to 0 whichever way we want to. In particular, we could look at h go on to zero along the real axis and we could also look at h approaching 0, along the imaginary axis. We could also look at h goes to 0 all kinds of other ways. But it's actually sufficient in this proof to look at these two directions. So, you'd let h approach 0 along the real axis. And then, along the imaginary axis line. And both times, this limit must exist and both times you must get the same limit because the limit must exist and be the same no matter how h approaches the origin. But by looking at these two limits and realizing that they must be the same, the Cauchy-Riemann Equations follow once you recognize the partial derivatives in the expressions. Let's look at another example. Let's look at the function f(z) which is the conjugate of z. So, x- iy(z) is x + y and the conjugate of that is x- iy. In other words, my function u which is the real part is the x and my function v is -y. It's very simple to find the x in the y derivatives. The x derivative of the function x is 1, the y derivative for this function doesn't even depend on y. So, this is a completely constant function for all ys concerned. And so, the y derivative is simply 0. For the function v, it doesn't depend on x, so the x derivative is 0. And the y derivative is simply -1. Now, when you compare ux and vy, they're not the same. So, the Cauchy-Riemann equations are not satisfied. uy and vx indeed Are opposites of each other as 0 is the same as minus 0. But ux is not equal to vy while uy could be regarded as minus vx. And therefore, the function cannot be differentiable anywhere. We don't have to check if f of z plus h minus f of z over h exists in the limit. We know, since the Cauchy-Riemann equations are not satisfied, this function cannot be differentiable. because if it were differentiable, then we would know from the previous theorem that the Cauchy-Riemann equations would have to be satisfied. Again, remember, if f is differentiable at a point z0, then the Cauchy-Riemann equations must hold at that point. Now, is the reverse statement also true? Is it true that if at a point the Cauchy-Riemann equations are satisfied, then that implies the function is differentiable at that point? The answer is almost. Here's the theorem. Suppose f is defined on a domain D in the complex plane. Again, a domain is simply an open connected set. Then f is analytic in that domain if and only if u and v have continuous first partial derivatives on D that satisfy the Cauchy-Riemann equations. So this theorem says two things. It says, if f is analytic, then u and v satisfy the Cauchy-Riemann equations and have continuous first partial derivatives. It's kind of what we already knew. But it also says, if u and v have continuous first partial derivatives, then f is analytic if these derivatives satisfy the Cauchy-Riemann equations. Continuous is the important part here. The first partial derivatives, it's not enough for you to be able to just find ux, uy, vx, vy, these are the first partial derivatives, they need to be continuous. If they are continuous and satisfy the Cauchy-Riemann equations, then f is analytic. It's a very subtle point, and this would mostly not be important for us. But one can actually find counter examples, one can find a function that has these first partial derivatives and they satisfy the Cauchy-Riemann equations, yet the function is not differentiable. And the reason is that there is a point where these functions are not continuous. But for now let's just look at an example. Let's look at the function f(z) = e to the x cosine y + i e to the x sine y. This is our regular exponential function from real value calculus and cosine, our regular cosine function, and sine is our regular sine function. In other words, the function u is the function e to the x times cosine y, u of x and y is e to the x cosine y. And v is the function e to the x sine y. I can again find the x derivative of u, the y derivative of u, the x derivative of v, and the y derivative of v. Let's start with the x derivative of u. In that case, cosine y is simply a constant that I multiply e to the x with. The constant goes to this side and all I need to find is the derivative of e to the x. But e to the x has this amazing property that its derivative is the function itself. In other words, the x derivative of u is the same as u's, e to the x cosine y. If I wanted to find the y derivative of u? Now e to the x is my constant. And I need to find the derivative of cosine y. The derivative of cosine y is minus sine y, so the derivative becomes minus e to the x sine y. Similarly, the x derivative of v is e to the x sine y, and the y derivative of v is e to the x cosine y. If I look at these functions, I realize ux is equal to vy, and uy is equal to negative vx, so the Cauchy-Riemann equations are satisfied, and these are all products of continuous functions, so ux, uy, vx and vy are continuous in C. Therefore, by the theorem, this function f is an analytic function in C. In fact, that makes an entire function because it's analytic in the entire complex plane. This function f( z) = e to the x times cosine y + i times e to the x sine y, is going to be our complex exponential function. We'll study it in the next lecture.