Welcome to the first lecture in our fifth week of the course, Analysis of a Complex Kind. Today we'll learn about integration in the complex plane. But to begin with, let's review integration in R. Given a function f that's defined on some interval [a,b] and is real valued and continuous, we define the integral from a to b, of f(t)dt as the limit of these left hand sums. So we take f, an appoint tj, multiply it by the next tj+1- tj, And we add all these things up. What are these tj's? Well we take the interval from a to b. And we divide it up into these little chunks. So a is equal to t0, and the next little one is t1, then t2, and so forth, up to tn, and n is some large number. And we're going to let that go up to infinity later on. So right now, we're dividing up this interval into these n chunks. And we evaluate f at the left hand point of each of these intervals, so for example f(t0) is this value right here. We take that value and multiply it by t1-t0. What is t1-t10? It's this length right here. So this length is f(t0). t1- t0 is that length. So when I multiply those two I get the area of this little rectangle. Similarly I take f(t1), which is that value right here. So here's f(t1). And that's this length. I multiply that bit by t2 minus t1 and get the area of the next rectangle. And so by taking the sum of all these products I'm going to add up the values of all the areas of these rectangles. But that's an approximation for the area under the curve. So in the limit as n goes to infinity these rectangles become very narrow. They give a very good approximation for the area under the curve. In other words, if f happens to be above the x axis, just like I drew it right here, then this integral is actually simply the area under the curve. Otherwise, if f is sometimes above and sometimes below the x axis, we're going to take those portions that are above the x axis, and subtract from them the portions that are below the x axis. So that's the meaning of this integral. Here's the fundamental theorem of calculus. If f is a function as we discussed above we define a function F(x) as the interval from a of x, f(t) dt. So in other words, if f is above the x axis, the integral from a to x is simply the area under the curve between a and x. Now x is the variable, so x varies. So f(x) gives you this area depending on the value of x's. If x gets bigger in my picture, then F(x) gets bigger. If x is smaller, F(x) gets smaller because f happens to be above the x axis. But you can define a function uppercase F like this also if f is above and below the x axis. It turns out this function that you define is a differentiable function and its derivative is equal to little f(x). That's the Fundamental Theorem of Calculus. And why is this so useful? Uppercase F actually gives you an antiderivative of f. A function that, when you differentiate it, is little f. In general, we say that a function uppercase F that satisfies that its derivative is little f, such a function is called an antiderivative of f. But note the following. Suppose you found two different antiderivatives, F and G. So both of them have a property that F' is equal to little f, and also G'e is equal to little f. Suppose you found two different functions. What is (G-F)' Well, it's G'- F', but both of these derivatives are little f. So that's zero. So G- F is a function whose derivative is zero, and that makes it a constant function. So F and G, two different antiderivatives of the same function, f can only differ by a constant. The conclusion is that we can use any antiderivative to evaluate an integral, and you've probably seen this. Suppose little f is the function you want to integrate from a to b and you found an antiderivative,. Then the integral from a to b, f(t)dt, is the antiderivative evaluated at the upper bound b minus the antiderivative value at the lower bound, a. The reason for this is, if you wanted to prove such a fact, you would notice that G, this antiderivative that you found, is related to the antiderivative F from the previous page by simply a constant. Where F was this function, F(x) is the interval from a to x, f(t)dt. Of which we know by the fundamental theorem of calculus that is an antiderivative. So what is F(a) in that case? F(a) is the interval from a to a, so that's nothing, that's zero. And what's F(b)? Well that's the full interval that we're actually trying to find. So we know this is equal to, definitely, F(b)- F(a). Because F(b) is the integral, and F(a) is zero. But now F(b) is the same as G(b)-C. And F(a) is G(a)-C. And the constants cancel out. And so you get G(b)- G(a). So that's how easy that is to prove. So how will we generalize this to C? Instead of having a real valued function we have a complex valued function. And the function doesn't have the set of a real variable but a complex variable. So instead of integrating over an interval we're now in C, so what will we integrate over? The answer is we'll integrate over curves. Recall that a curve is a smooth, or piecewise smooth function gamma, defined on an interval, but into C. We could write this function as gamma(t)=x(t)+iy(t) for x is the real part, and y is the imaginary part. If f is a complex valued function that is defined on gamma, then we define the integral over gamma, f(z)dz to be the limit of these sums. We take f(zj)(zj+1- zj). And the zj's are the values of gamma at points tj when we divide the integral from a to b up into these little chunks again as we did before. So this is a direct generalization of the left hand sum. Let's look at the picture and understand what the sum really does. You can see the curve gamma. Gamma is actually the image of the interval from a to b. Under the function gamma. Again we divide this interval up into these little chunks. Where a is equal to t0, b is equal to tn, and this is t1, t2, and so forth. Then, we look at the image of these t1, t2, t3 and so forth. So, for example, the curve starts at z0 which is gamma of t0 or gamma of a. It ends at zn which is gamma of b or gamma of tn. And in between, we have the zj, which is gamma of tj, and the next point, zj plus 1 which is gamma of tj plus 1. So, we have all these points here. In order to form the sum, we evaluate f add zj, so what's the value of f right here and multiply that value by zj plus one minus zj. Then, we add up all these products and take the limit and goes to infinity. If that limit exist, we call it the integral. In fact, we call it the path integral of f over gamma. So again, it's the limit as n goes to infinity of the sum f(zj)(zj+1- zj) where the zjs are gamma(tj)s and the tjs are those little points that divide the integral from a to b up into n pieces. One can show the following. If gamma is a smooth curve and f is continuous on gamma, then this integral, f of z dz over gamma can be found by taking f evaluative of gamma of t. Multiply that by the derivative gamma prime of t and take the integral from a to b over that product. Why is that? Here's the idea of the proof. This is the sum that we use to define the integral. To get the actual integral you would put the limit as n goes to infinity in front of it. But let's look at the sum by itself. Zj is simply gamma of tj. Zj plus one is gamma of tj plus one. Zj is gamma of tj. That makes up this sum. But now, we just divide by tj plus one minus tj and then simply multiply by tj plus one minus tj so this won't affect the sum at all because we multiply and divide by the same number. But now, if you look at this term right here as you make tj plus one and tj be closer and closer to each other this approaches the derivative, gamma prime of tj. This is approximately the derivative of gamma prime of tj. So, the sum is over f of gamma tj times gamma prime of tj times tj plus one minus tj. We often call this delta tj. In the limit, this goes to the integral from a to b, of f of gamma of t, gamma prime of t, d t. Note, we haven't really defined how to integrate a complex valued function. So, let's quickly look at some examples. Suppose g is a function defined from a to b and maps into c, so we can right g(t) = u(t) +iv(t), then if we write something like interval from a to b g(t)dt, what that really means is the interval over u plus i times the interval over v. In other words, we integrate the real part, we integrate the imaginary part and then put them back together and that is how the integral over g is define. Let's look at an example. Suppose, I wanted to evaluate the integral from 0 to pi, e to the i dt. You then break up either the it as cosine t plus i sin t. And integrate the cosine function. And separately, integrate the sine function. An anti-derivative for cosine t is sine t because the derivative of sine's cosine. So, we need to take sine t and evaluate it at the upper bound. And subtract from it, its value at the lower bound. That's what this expresses right here. Next, we need an anti-derivative for sin t. An anti-derivative for sin t is minus cos t, because the derivative of minus cosine t is minus minus sin t, which is sin t. So, we need to take minus cos t, and evaluate it pi, and subtract from it the value it has at zero. Well, sine of pi is actually 0. Sine of 0 is also 0, so I put that all into this one 0 right here. Cosine of pi is negative 1. Cosine of 0 is 1, but we're subtracting that value, so minus another 1. And here's a minus i that's in front of it. All together this gives you 2i. So, the value of this integral is 2i. You actually don't have to break the integral up into real and imaginary part. One can actually alternatively simply find an anti-derivative right in the original form. So, if you can find an anti-derivative of e to the it, then you can just use the fundamental theorem with that anti-derivative. Well, it turns out an anti derivative either the it is minus i, e to the it, why is that true? Well, let's take the derivative of minus i, e to the it. Minus i is a constant that goes to the side and so I need to differentiate e to the it. Now, the derivative of e to the it is either the it times the derivative of the inside function it which is i. When I multiply through, i squared is minus 1. Together with this minus, it all cancels out, so it's e to the it. So, indeed, minus i, e to the it, is an anti-derivative of e to the it. So, I can take this anti-derivative and evaluate it at pi and subtract its value at zero. The value at pi is -i e to the i pi. The value at 0 is -i e to the 0, together with the fact that we're subtracting it we're getting this plus right here. Either the i pi is -1, together with this negative sign it's plus, so this whole Expression evaluates the i. If the zero is one, so this also evaluates the I so that i end up with 2i. So, these are two alternative methods defined the same interval. Lets look at another example. Interval from zero to one t plus idt. Again, you could break this up into the integral from zero to one of the function t dt. Quest i times the integral from zero to one of one dt. And then, find anti-derivatives separately and plug in the balance. Or, you could jut find an anti-derivative. For the whole function instead of finding each part separately. Anti-derivative for t is one half t squared, anti-derivative for i is i t. And so, one half t squared + i t is an anti-derivative for t + i. We need to evaluate that at 1, and subtract from that the value at 0. When I plug in t equals one, I get one half plus i. When I plug in t equals zero, this all vanishes. In other words, the value of the integral is one half plus i. Now, let's look at some examples of actual path integrals. Suppose, gamma of t is t + it, for t between 0 and 1. What does that look like? Well when t is equal to 0, gamma of 0 is 0. When t is equal to 1, gamma of 1 is 1+i. 1 and i is right here. How about in between? Well I'm graphing, x of t and y of t together. But x of t is equal to y of t. So we could write this as x(t)+iy(t). And I see that x sub t and y of t are always the same value. y is equal to x. Well, that's this line right here. So, the path, gamma, is really the line from the origin to 1+i. We can find the derivative as 1+i just take the derivative of t which is 1, the derivative of it is i. And supposed I wanted to integrate the function f(z)=z squared over that path. So I need to find the integral of f(z)dz. By our definition the integral of f(z) dz because f is a continuous function, the path is smooth I can take f(gamma(t))(gamma prime(t) and evaluate that integral from 0 to 1. What is f(gamma(t))? I need to plug in gamma (t) into my function, f. But the function f takes its input and squares it. In other words, this is gamma(t) squared. Gamma(t) squared is t+it, that's gamma(t), squared. So that's where this term comes from right here. And you need to multiply by the derivative of gamma. We found that as 1+i. And now I have an integral of a complex valued function and I know how to evaluate that. One way to evaluate that is to simply multiply through. So this first term here becomes t squared+2i times t times t- t squared. The minus t squared comes from the i squared right here in the second term. And I need to multiply that by 1+i. t squared and t squared cancel each other out so the first term is this 2it squared times 1+ i. If. I multiply through I have 2it squared plus this term right there + i2it squared. But i squared is negative one, so I end up with minus 2t squared. I could break this up into the real part and the imaginary part so I get -2 times the integral over t squared, that's from the second term here. +2i times the integral over t squared, that's the first term. An antiderivative for t squared is 1/3 t cubed. So I get -2/3 t cubed evaluated from 0 to 1 plus 2i again an antiderivative of t squared is 1/3 t cubed. You're going to need to evaluate from 0 to 1. When I plug in 1 for t I find -2/3. When I plug in 0, this term goes away. For the second term when I plug in t equals 1 I get 2i/3. Then again for t equals zero there is no term. All together I can combine these into 2/3(-1+i). We'll be using this fact later on again so I'll remind you of that when we need it. Here's another example. Suppose I wanted to define the integral of 1/z dz over the curve where the |z|=1. And this a notation that we haven't used so far. |z|=1 as a set denotes this circle of all the z's where the absolute value of z is 1. So it's the circle of radius one. When we write something like that, |z|= 1, we automatically assume that the circle's oriented this way. And most of the time the parameterization gamma(t) = e to the it is used for t runs from 0 to 2pi. So that's what's meant when I write something like this, integral over |z|=1. We'll see later than other parameterizations yield the same result with the integral. That's why this notation is actually your reasonable notation. So how do I find this integral? If gamma(t) is e to the it then gamma'(t), it's the derivative, is e to the it times the derivative of the inside function which is the exponent so times i. Now the integral over the curve gamma of 1/z dz is therefore the integral from 0 to 2pi of 1/gamma(t) times gamma'(t)dt. So wherever we see z, we plug in gamma(t). And dz is replaced with gamma'(t)dt. Next let's plug in what gamma(t) actually is. It's e to the it. And gamma'(t) is ie to the it, and we see that the e to the it term cancels out. All we're left with is i. Let's pull that outside of the integral, and we're left with the integral from 0 to 2pi of dt. You could write a 1 right here if you wanted to. An antiderivative of the function 1 is t, because the derivative of t is 1. Need to evaluate that from 0 to 2pi, so we get i times 2pi minus i times 0, which is 2pi i. So the integral of the function 1/z over the circle of radius 1 is 2pi i. Here's another example, same curve, z = 1, this time we're integrating over the function z. Instead of 1 over z we take z. So again, gamma(t) = e of it. Gamma'(t) = ie to it. So the integral over this curve gamma(z)dz is therefore, the integral from 0 to 2pi. We replace z gamma and dz with gamma'(t) dt. gamma(t) is e to the it gamma'(t) is ie to the it. And we can follow this i outside of the integral and get the integral from zero to 2pi and we can combine e to the it and e to the it into e to the 2it. So now what we need is an antiderivative of e to the 2it. And the claim is, an antiderivative of ie to the 2it is 1/2e to the 2it. We better check if that is true. How do we check it? Well, we take the derivative of 1/2 e to the 2it and check that it really is what we think it is. So what is the derivative of 1/2 e to the 2it? 1/2 is a constant that goes to the side. And the derivative of e to the 2it is e to the 2it times the derivative of the exponent, which is 2i. In other words, these 2s cancel out nicely. And we're left with ie to the 2it. So indeed an antiderivative of ie to the 2it is 1/2 e to the 2it. And we now take that antiderivative and evaluate it at 2pi. Subtract from it the value at 0. When I plug in 2pi for t, I get 1/2 times e to the 2 times 2pi i, so 4pi i. When I plug in zero, I get e to the zero. But e to the 4pi i Is equal to 1. e to the 0 is also equal to 1, and 1 minus 1 is equal to 0. So the integral evaluates to 0. Let's look at the same path gamma and integrate 1 over z squared dz. So, so far, we integrated 1 over z And then we got, that was 2 Pi i, then we integrated z, d of z, then we got that was 0, let's see what 1 over z squared gives us. Again, 1 over z squared means, I write one 1 over gamma squared or t over gamma of t squared times gamma prime of tdt. Gamma prime of t is less than the numerator and it's i e to the it, that's this part right here, and we divide by the square of gamma of t. What is gamma of t squared? That is e to the it squared, so e to the it times e to the it, and that is e to the 2it. So e to the 2it is here in the denominator and I can therefore cancel one of the e to the it's and left with one e to the it in the denominator or e to the- it in the numerator. What is an anti-derivative of ie to the -it? The claim is, it's -e to the -it. Again, let's check, how do I check? I take the derivative of -e to the -it, the derivative of that is -e to the -it times the derivative of the exponent which is -i. The two negative signs cancel out, I'm left with i times e to the -it, which is exactly what I wanted to get, so, -e to the -it is indeed an anti-derivative of the function I'm integrating and I need to evaluate the anti-derivative of 2 Pi and add 0. At 2 Pi, I get -e to the -2 Pi i, at zero I get e to the 0, I'm subtracting that but with this minus sign, I get this plus sign. The first term is -1, e to the -2 Pi i is 1 with a negative sign, it gives me a- 1, e to the 0 is 1, 1 -1 is 0. And again, the integral value is still 0. More generally, one can see the following. The integral over z to the m is equal to 2 Pi i when m = -1. When m is equal to -1, What I'm looking at is the integral over z = 1 of 1 over z dz, and we showed that that was 2 Pi i. For all other exponents, including n equals 1, in which case, I have the integral over z, m equals minus 2, in which case, gives us the integral 1 over z squared. All other cases, that integral is equal to 0. So all powers of z, the integral evaluates to 0, except for the function 1 over z. We we'll see more and why that is the case later. In the next lecture, we'll look at more examples and some first facts about complex path integrals.