Welcome to our third week in our course, Analysis of a Complex Kind. Today, we'll learn about the complex derivative. Before we go there, let's quickly review the derivative from calculus. Given a real value function f of a real variable so there's nothing complex in here. And a point x0 in the integral in which f is defined we say the function is differentiable at x0 if the limit of the difference quotient exists. And the difference quotient is f (x)- f (x ) divided by x- x 0. If that limit as x goes to x 0 exists, we call the limit the derivative of f at x 0. And denote it by f prime of x 0. Let's quickly review what this means geometrically. Here's a graph of the function in this blue graph here is the graph of f in this example. Now we pick a point x0 where f is defined and we look at f ( x0), that's this point right here. So this point that I'm circling has coordinates x0 and f (x0). And then we take a second point x. And we look at f ( x) and so this point that I'm now circling is the point whose coordinates are x and f ( x). Next we look at the numerator of our difference quotient f (x)- f ( x0). F(x) is this value right here. F(x0) is this value. The difference between the two of them is this length. So this is f(x)- f(x0). And we divide that by x- x0. X- x0 you can see in this length right here, so this thing right here. F(x)- f(x0) divided by x- x0 is rise over run for this red secant line that I drew. In other words it gives us the slope of the red secant line. It's the change in y divided by the change in x. So this difference quotient f(x)- f(x0) divided by x- x0. It's the slope of the secant line through the points (x0, f(x0)) and (x,f(x)). Next we need to let x approach x0. And by doing that, the slope of the secant line will change. As this point, x moves toward x0, the slope of the red line that's the secant line will change, here's an example. X moved closer to x0 and the line changed quite a bit. As x moved closer and closer and closer to x0, the secant line becomes a tangent line in the limit. In the limit x and x0 have kind of overlapped. And the slope of the secant line becomes the slope of the tangent line. In other words, if this limit exists, it is the slope of the tangent line to the graph of f at x0. The derivative is thus the slope of the tangent line to the graph at x0. The slope can be negative, it can be zero, it can be positive. Here you see an example where the slope and that's the derivative is negative, the tangent line has a negative slope, it's sloping downward. In the middle example you see an example where the tangent line is horizontal. It has a zero slope and therefore the derivative is zero at this point x0 where the minus is tangent to the craft. In this example here, the slope of the tangent line is positive and therefore the derivative is positive. And it becomes more positive as you move along this graph because the line would get steeper and steeper. The derivative does not have to always do this however. Here's an example where the graph has some kind of corner. So if this is my point x0 f(x0) and that word to approach this point x0 with x values from the left. And these secant lines all look a little bit like this and definitely have a positive slope even the limit of slope is positive. However, if I approach this point from the right. My secant lines have a negative slope and approach a limit that's negative in slope. And I can not find a one tangent line to the graph of f at this point x0. There is no tangent line and so there's no unique slope which means this function does not have a derivative at x0. So in this example f prime of x0 does not exist. Let's return to complex numbers now. The derivative is defined in a completely analogous way to the real value case. We say a complex valued function f of complex value is complex differentiable. And typically we don't really say complex differentiable. We say differentiable if it is clear from the context that we mean complex differentiable. So a function is differentiable at a point z0 which must be on the domain of f, obviously. If the limit of the difference quotient exists we look at f(z)- f(z0) divided by z- z0 and let z approach z0. Now notice, though if you pick a point z0 in the complex plane. There are lots of ways of approaching that point. It's not just along the real axis like we have to look at in the real value case. You could also come in the complex direction from the imaginary axis. You could come at a 45 degree angle. Or you could even approach in spirals. There's lots of different ways of having z approach z0. And so for this limit to exist it's a little harder to check than in the real case. If this limit exists it's denoted f prime of z0. Sometimes we also use the notation df dz as a derivative of f with respect to z at z0, or just d dz of f(z) at z = z0 Let's look at some examples. Suppose f (z) is a constant function, so no matter what z you plug in, you get the same constant value out of it. If you prefer, you can just think about a specific constant like ten right here. Take a z0 in c and look at this difference quotient, what is f(z)- f(z0) divided by z- z0? F(z) is this constant c, but f(z0) is also equal c because f is a constant function. Divided by z- z 0. The numerator evaluates to 0, the denominator's non zero. So the quotient's equal to 0. And as z approaches z0 it doesn't matter how z approaches z0, this quotient is always zero and clearly has the limit zero. Therefore the derivative of f is equal to zero for all points In the complex plan. Sometimes instead of writing the difference quotient as f(z)- f(z0) divided by z- z0. We write z as z0 + h. Where h is a complex number. And the difference quotient, then would become instead of writing z here, we're going to write z0 + h, so f(z0 + h)- f(z0). And z- z0 then becomes, if we write z as z0 + h, and subtract from that z0, what's left over is h. So this is the exact same quotient, just written in a different way. In terms of the difference h, rather than in terms of c. So the difference quotient is f(z0 + h)- f(z0) divided by h, or often we just write f(z + h)- f(z) over h when we get tired of writing z0 all the time. And we'll take the limit as h approaches 0, because when z and z0 get closer and closer to each other, it means that the difference z- z0, which is h, goes to 0. So here's another example, let's look at the function f(z) = z. So the function that assigns to each point just the same point. So what is f(z0 + h)- f(z0) divided by h? Well f(z0 + h) is just z 0 + h, which you see right here. F(z0) is simply z0. F just returns exactly the argument that you passed into it. We divide by h, and you see in the numerator the z 0's cancel out with each other, and you're left with h over h. H over h is 1. No matter what the value of h is. And as h approaches zero, it remains one. No matter how H approaches zero. Which means this limit exists. And that means the derivative of f is equal to one for all of Z in the complex claim. So this is differential everywhere with derivative one. >> Let's next look at the function f of z equals z squared. The difference quotient then becomes f(z0) plus h, and the function f just takes its input, in this case, z0 plus h, and squares it. So that becomes (z0 + h) quantity squared- f(z0), which is z0 squared, divided by h. This term, (z0 + h) quantity squared multiplies through to z0 squared + 2 times z0 times h + h squared. And then we subtract z 0 squared from this expression. So we have a z 0 squared here and we subtract one here. Which means the numerator simplifies to 2 z 0 h + h squared. Next we can cancel an h out of the numerator, an h out of the denominator. So, we are left with 2z0 plus h. And as h goes to zero, this plus h term becomes less and less important so that in the limit we are left with two times zo. So the limit exists and therefore f prime of z is two z for all z in the complex plain. Let's look at another example f(z) = z to the n. So, the function takes an argument and raises it to the nth power. Therefore, the difference quotient becomes f of z0 plus h which z 0 plus h to the nth power minus f(z0), which is just z0 to the n divided by h. Now how do you multiply add z0 + h to the nth power? We can do that using the binomial theorem. But, we're actually only entered in the first few terms. The binomial theorem says that this is equal to the sum of n choose k times z 0 to the k times h to the n minus k. And this runs from k to zero to n. You don't really need to know this binomial theory. All you need to know is that when you multiply through, you have different powers of z zeroes, and different powers of h which always add up to n. And the binomial coefficients tell you how many of those terms you have. So for example you'll have a z0 to the nth power and no h term. Then you have a Z0 to the n minus 1st power and an h term. Then a Z0 to the n minus second power and then h squared terms, that would be is Z0 to the n minus third in h cube and so forth, all the way to Z0 to the no power times h to n. And the number of times these terms occur are given by the binomial coefficients. Z0 to the n occurs once, h times z0 to the n- 1 occurs n times, which is n choose 1. This coefficient here is n choose 2, and so forth. None of these really matter for the limit. In the limit the only terms that are really important are these first two terms. And you notice we have, again, a Z0 to the n term that we're going to subtract, so that term goes away. We're left with terms that all have an h or higher power of h in them. The only one that has a power one of h is this first term n times h times z0 to the n minus one. So if we cancel out an h of the numerator and denominator, this term has no h's left in it. It becomes n times z0 to the n minus one. All the other terms just lose one power of their h. But all have an h left in them. You could factor out an h. And have all these remaining terms left in here. So as h approaches 0, we have some number here that gets multiplied by h, which becomes smaller and smaller and smaller, and in the end becomes negligible. In other words the limit of this expression is n times z0 to the n- 1. Therefore the function z to the n is differentiable in the complex sense with derivative n times z to the n minus one for all z and c. Now that we have taken a few derivatives it would be great to understand how we can combine functions whose derivatives we already know. So let f and g be differentiable functions and differentiable at z, and suppose h is another function that's differentiable at f(z). We want to look at the composition that's why we want to have h differentiable at f(z). And suppose c is a complex constant. And the derivative of a constant times f Is the same thing if the constant times the derivative of f. So constants can go out to the side when taking derivatives. Next, the derivative of a sum of two functions is the sum of the derivatives. In other words you can individually differentiate these two functions, and then just add the results together. The product is a little bit more complicated, it's not sufficient to just simply differentiate each function, but there's a product rule. You differentiate f and multiply it with g. Next, you don't differentiate f and multiply that by the derivative of g. There's also a rule for the derivative of the quotient f over g. Obviously you can't do that if g(z) is 0. So you want to require here that g(z) is not 0. But if it is not 0 the quotient rule is the same as it is in calculus. You find the derivative of the quotient by squaring the denominator, writing down the denominator again times the derivative of the numerator, minus this time the numerator times the derivative of the denominator. So you almost see the product rule in the numerator except for that minus sign. And so you need to remember which term comes first, which term is the one with the minus sign and which term is the one without the minus sign. The way I remember that is I square the denominator and write it down right away again and then times the derivative of the numerator, therefore I know which one is the term that comes first. Finally, the derivative of the composition h of f of z can be found by taking the derivative h prime at f(z), and multiplying that by f prime of z. Rather than showing you all these proofs which are the same as the proofs in calculus, I'm going to show you a number of examples. These are important rules to practice. If you are not so familiar with these rules from calculus anymore, I highly recommend Let you do some practice. You could, for example, practice with all the problems I'm about to present but by trying to find the derivatives by yourself first before looking at my solutions Let's start with a function f(z) = 5z cubed + 2z squared- z + 7. So we have a sum of a bunch of functions. Difference is the same as the sum, you can therefore differentiate these four functions individually. So what are all these derivatives? The derivative of 5 times z cubed is a constant times z cubed. Constants can go to the side, 5, times the derivative of z cubed. We just found the derivative of z to the n. That was n times z to the n- 1. In other words, the derivative of z cubed is 3 times z squared. So, the derivative of 5 z cubed is 5 times 3 z squared. Similarly we find the derivative of 2z square, the 2 goes to the side times the derivative of z squared which is 2z. The derivative of z which is either the one that we found earlier, which was 1, or it would also apply this as the derivative of z to the first power which is 1 times z to the zeroth power, which is also 1. And 7 is simply a constant whose derivative is 0. Therefore, I didn't write down the plus 0 part. Now if you wanted to simplify this a bit, this is 15z squared + 4z- 1. Next, let's look at 1 over z. We can use the quotient rule for that one. Even though there's also easier ways of finding this derivative, but that's just fine by using the quotient rule. So we square the denominators, z squared, we write down the denominator again, that's z, times the derivative of the numerator, which is 0. The numerators are constant. Its derivative is 0. Minus the numerator, 1, times the derivative of the denominator. The derivative of z is 1. The numerator there for simplifies to 1 with a negative sign in front of it. The denominator is z squared. So the derivative of 1 over z is -1 over z squared. A different way of finding this derivative would have been to look at this function as the function f of z = z to the power -1. And realizing that this rule for the derivative of z to the n also holds for negative exponents. And so the derivative is therefore -1 times z to the n -1. So -1 -1, which is -z to the -2, and if you rewrite that, that is -1 divided by z squared. Let's look at f(z) = (z squared- 1) to the nth power. So here we have an inside function and an another function on the outside to the power n. We can use the Chine rule. The rule that said that the derivative of h(f(z)). Is h prime of f of z times f prime of z. So we need to find the derivative of the outside function, which is the function that raises its argument to the nth power. Well, the derivative of that is n times the argument to the n-1 power. And then multiply that by the derivative of the inside. So z squared- 1, well the derivative of z squared -1 is 2z. Therefore the derivative in this case is n times z squared minus 1 to the n minus 1st power times 2z. Let's look at another equation. f(z) = (z squared- 1) (3z + 4), which is the product of two functions. Of course we could multiply through and then differentiate term by term. But I want to show you a use of the product rule here. The product rule said that we need to take the derivative of the first function which in this case is 2Z, the derivative of z squared- 1. And multiplying with the second function, 3z + 4 plus the first function, z squared- 1 times the derivative of the second function, but the derivative of 3z + 4 is simply 3 because 4 is a constant and its derivative is 0. Finally, let's look at z divided by z squared + 1, we can. Let's use the quotient rule which means we have to square the denominator z squared + 1, 1 is z squared. Write down the denominator again and multiply with a derivative of the numerator. But the numerator is z, its derivative is 1. So I didn't even write down the times one right here. Subtract from that the numerator z times the derivative of the denominator which is 2z. And if we wanted to simplify that, we have a (z squared + 1)- 2 z squared, so the numerator becomes 1- z squared, and the denominator is the same as it was before. I reordered as (1 + z squared) quantity squared just to make it look nice. Here's another example of function that is not conflicts differential. f(z) = Re(z). In other words, if you write z as x + iy then f of x+iy is simply the real part of what we plug in. So just the x. Let's use the f (z + h)- f(z) over h form of the difference quotient. And let's write h as hx, the real part of h, + i h y, the imaginary part of h. Then what is f of z + h? z + h would be x + h x + i y + h y. And if we only look at the real part of that, that real part is x + hx. We subtract from that f(z), f(z), this is x. We see that there is an x here and we subtract x here, so that cancels out. The numerator becomes simply hx. The denominator is h, hx was the real part of h. So the difference quotient is simply the real part of h divided by h. What happens as h goes to 0? As we noted earlier, there are lots of different ways of approaching a point, h could go to 0 along the x axis. Or it could go to 0 along the imaginary axis, or along all kinds of other ways. But let's check out for starters what happens when h approaches along the real axis. So h in other words is real elements of the form h x+ IX0 does not have an imaginary part. In other words the real part of h which is hx is the same as h itself because there is no imaginary part. In other words in this quotient right here the numerator is equal to the denominator which means the whole fraction evaluates to 1 and as h goes to 0 just along the real axis the limit is 1. So then it exists as h approaches 0 along the real axis. Next let's have h approach 0 along the imaginary access. Then h has not real part and only has an imaginary part, so H is of the form = + i times hy. In other words the real part of h is 0. So again in this fractional real part of h over h the numerator is 0. So the whole fraction evaluates to 0. And in the limit as h approaches 0 along that imaginary axis, the limit is still 0. That's a different limit than we obtained coming along the real axis. Let's next look at a sequence hn that approaches 0 in a spiral type way. hn is i to the n over n. We looked at the sequence before. We know the sequence approaches 0. What is the real part of hn divided by hn? The dividing by n happens in both numerator and denominator, and so therefore, it cancels out. The numerator is real part of i to the n over n and the denominator is i to the n over n, which simplifies to real part of i to the n over i to the n. i to the n is 1 or -1 or i or -i. And you look at the quotient for even n's, provided that n is real, and so the quotient is 1. For odd n's, i to the n is purely imaginary, it's i or -i, and the quotient is therefore 0, because the real part of i to the n is 0. The sequence therefore bounces back and forth between 1 and 0 and does not have a limit as n goes through infinity, because it does not approach one of these numbers, it bounces back and forth. Altogether we see that this function f is not differentiable anywhere because the limit of the difference quotient does not exist. It is different depending on how we approach the origin. Sometimes the limit doesn't even exist at all. So this function does not have a derivative anywhere in the complex plane. Here's another non-example, f(z) equals the complex conjugate of z. So what is f(z + h)-f(z) / h? Well, (z + h) gets mapped under f to z + h conjugate. And we've learned that we can pull that inside, so we have the conjugate of z plus the conjugate of h. And that's what you see right here. We subtract from that f(z), which is the conjugate of z, and again, we see, there's the conjugate of z, we subtract the conjugate of z, those two canceled out, and the numerator is just the conjugate of h, the denominator is h. Does that have a limit as h goes to 0? I don't think so. Again, if h is a real value, then the conjugate of h is simply h. And the quotient is equal to 1 approaches 1 as h goes to 0. If h is purely imaginary, then the conjugate of h is the opposite of h. And so, the quotient's value is to -1, and has a limit of -1. And if h approaches the origin in any other way, it gets even more complicated. Certainly, the limits are not the same, and that already proves to us that this limit of the difference quotient does not exist as h approaches 0. And therefore, f is not differentiable anywhere in the complex plane. Here's a fact. It's the same as is true in real value calculus. If f is differentiable at z0, then it must be continuous at z0, not the other way around. A function can be continuous without being differentiable. But if it happens to be differentiable, that's a stronger concept than being continuous. It applies continuity. Here is how you would prove this fact. Instead of showing that the limit of f(z)=f(z0), which is what we need to show for the function to be continuous at z0, we'll bring this f(z0) to the other side, and we look at f(z)-f(z0), which then needs to be equal to 0 in the limit. So we look at f of z-f of z0 as z approaches z0. And here comes the trick. We divide by z-z0, well, you can't just do that. But if you also multiply by z-z0, then we really just multiply f(z)-f(z0) by 1. So, that's allowable. But interpreting this as the difference quotient times z-z0, if the function f has a derivative, then the difference quotient has a limit as z approaches z0, that limit is f'(z0). z-z0 also has a limit as z approaches z0, namely, 0. So the limit is f'(z0) times 0, and that is 0. Since f(z)-f(z0) approaches 0 as z goes to z0, the function f is therefore continuous at z0. Again, a function can be continuous without being differentiable. But it can't be differentiable without being continuous. We'll see an example about this shortly. Here's another definition. A function f is analytic in an open set in the complex plane if it is (complex) differentiable at each point in U. We need differentiability in a whole open set. We're not interested in functions that are simply differentiable just at a point. We want them to be differentiable in a whole open set. If that open set happens to be all of the complex plane, so if the function is differentiable at each point in the complex plane, we've seen some examples, then we also say this is an entire function. Here are some examples. Polynomials are analytic in C because we showed that anything of the form z to the n is differentiable everywhere in the complex plane. We can even put a constant in front of it. That's still differentiable. And then we can add to that another term d times z to the m with a different power, and more and more of these. So these form polynomials, and therefore, polynomials are analytic in the complex plane, and therefore entire functions. Rational functions, which are polynomial divided by another polynomial, are by the quotient rule analytic wherever we're not dividing by 0. So whenever the denominator is nonzero, a rational function is also analytic. We've seen that the function f(z) equals the complex conjugate of z is not differentiable anywhere, and so it's also not analytic anywhere. Similarly, the real part of z wasn't differentiable anywhere, so it's also not analytic anywhere. Let's look at another example, f(z) is the absolute value of z, squared. Remember how to find the absolute value squared of a complex number? We had various ways that was defined. One way was to take x squared + y squared if z was of the form x + iy. But once we learned about the complex conjugate, we also learned that this is equal to z times the conjugate of z. Because z times the conjugate of z was x+iy times x-iy. And if you multiply through, this becomes x squared-i squared y squared. i squared is -1, so we're back to x squared + y squared. Let's apply that in our case here. We need to look at the difference quotient, which is f(z+h)-f(z)/h. f(z + h) is the absolute value of z+h squared. f(z) is the absolute value of z squared. And now we use the fact that the absolute value squared can be obtained by taking the complex number and multiplying it with its conjugate. The conjugate of z plus h is the conjugate of z plus the conjugate of h. And you subtract from that's still z squared. When we multiply through in z + h times z conjugate + h conjugate, we get a z times z conjugate, which is just the absolute value of z squared plus z times h conjugate which you see right here plus h times z conjugate, this term, plus h times h conjugate, and there's still the minus z squared here. Again, you see the z squared terms is added and then subtracted so they cancel out. What we're left with are these three terms in the numerator divided by h. The second and the third term have an h in them, so we can cancel out that h. The first term however, does not have an h. And so we cannot cancel anything out there. What we're left with therefore is a z bar from the second term where the h cancels out, and an h bar from the third term where the h cancels out, but a z times h bar over h, where the h didn't cancel out, and that is the troublesome term. As h goes to 0, z bar is not affected. That's just regular z bar. h bar just goes to 0 as h goes to 0. But h bar over h, as we saw earlier, does not have a limit. So, unless this term isn't there, miraculously, it will cause trouble and will make the limit not exist, except when this z is equal to 0, therefore making this term nonexistent. So if z is nonzero, then this limit of the difference quotient simply does not exist because of the h bar over h issue. If z is equal to 0, that issue's gone and the limit is simply 0, because if z is 0, z bar is also 0. So add 0, the limit of the difference quotients exists, so f is differentiable at 0 with derivative 0. It is not differentiable anywhere else. One point is not enough to call the function analytic. We need the function to be differentiable in th whole open set in order for it to be called analytic. So f is not analytic anywhere. Note, however, that this function is a perfectly continuous function, it's continuous everywhere in C. It's only differentiable at the origin, it's not analytic anywhere. Next stop are the Cauchy Riemann equations. Those are equations that relate the derivatives of u and v, if we write f as u+iv, it relates the derivatives in the x and the y direction of u and v with each other. I'll explain that in the next lecture.
