Welcome to Lecture 2, in the second week of our course Analysis of a Complex Kind. In this lecture, we'll study sequences and limits of complex numbers. Both of these are topics necessary for us to understand Julia sets and the Mandelbrot set. So let's start with sequences of complex numbers. Consider for example the following sequence of complex numbers. This is for listing 1, 1 half, 1 3rd, 1 4th, 1 5th, and so forth. It's a sequence of complex numbers. What happens as these numbers get further and further out to the right? If you were to graph these, we could graph the complex plain and say, say 1's here, 1 half is here, 1 3rd here, 1 4th, 1 5th so these points seem to accumulate near the origin. Let's look at another example i, i over 2, i over 3, i over 4, and so forth. Where are those numbers? Well i is over here, i over 2 is here, i over 3, i over 4, i over 5. So, these points are going to go on the imaginary axis, but again they seem to be getting awfully close to the origin as we go further and further to the right along the sequence. Let's look at another sequence, i minus 1 half. Minus i over 3 and so forth, so i again is this point right here. Minus 1 half, that seems to be here. Minus i over 3 over here. 1 4th, i over 5, minus 1 6th, minus i over 7. So these points kind of spiral around the origin but again. They're getting closer and closer to the origin. We say all these three sequences converge to zero. Informally, we say a sequence {sn} converges to a limit s if the sequence eventually lies in any (ever so small) disk, centered at s. This means that no matter how small a disk I draw around the origin in this example, for example eventually, far enough out to the right, every element of my sequence is going to be somewhere in this little disk. And if I were to choose a smaller disk, I just have to wait a little bit longer. And eventually, the sequence will be found inside that disc. How do you make this mathematically precise? We don't really need to have a mathematic precise notion, but I want to show you how to make this mathematically precise, in case you're interested. So here's the actual definition. A sequence {sn} of complex numbers converges to s, another complex number, if for every epsilon, this is going to describe the radius of that little disc we were talking about. There exists an index N, a point in the sequence from which on xn minus s is less than epsilon. So what is sn minus s? That is how far sn is from s. So if I were to draw a picture, and suppose s was not the origin necessarily, but suppose this is my limit, and I draw my circle of radius epsilon around that point, then s is the center of the circle. And if I want sn minus s to be less than epsilon, that means sn has to be inside this circle. So this definition simply makes mathematically precise what we want, and we want sn to be inside that disc of radius epsilon, far enough out along the c's. In this case, we also use the notation, the limit, lim, limit, of sn is equal to s. And we put n goes to infinity underneath that limit sign, to indicate what's the n that we want to go to infinity. Here's an example, so example from the previous page, the limit as n goes to infinity of 1 over n is equal to 0. How would we show that? Well we would have to show that the limit is 0, which means no matter how small a disk of radius epsilon that shows around the origin, eventually my sequence is going to be in it. So I'm going to have to show that eventually 1 over n minus 0. You don't really have to write that down but I'm just going to write it down for completeness. Is less than epsilon for a given epsilon so we would pick some number epsilon. And by the way, this epsilon is again one of those Greek symbols and the mathematicians often use the Greek symbol epsilon to indicate a really small number. If you really want to confuse a mathematician, make epsilon a large number. So what does it mean for 1 over n to be less than epsilon? This actually simply simplifies to 1 over n, less than epsilon, and it can solve this equation. For which values of n is this true, that is true when n is greater than 1 over epsilon, so for my given epsilon. I calculate one over epsilon and then I just pick an index and that is bigger than that one over epsilon and I'm guaranteed from that index on, my entire sequence will be located inside this disc. Another example is the sequence 1 over n to some power p, p can be anything, it can 2s, 3, 4, it can be one half. So, for example, when p is equal to 2 the sequence we'd be looking at would be 1 over 1 squared, 1 over 2 squared, 1 over 3 squared and so forth, so 1, 1 4th, 1 9th. That sequence converges to 0 but also, when p is equal to 1 half. What's it mean to raise a number to the power of one half, it means taking a square root of that number. So in that case, the sequence would be one over square root of one, one over square root of two, one over square root of three and so forth and that sequence would also go to 0 and we can show that very similarly. To how we show the sequence one over n converges to 0. And in general, for any power P, you fix that power and the sequence will go to 8. It'll take it longer to go to 8, it'll be slower when P is a really small number, and it'll be faster when P is a large number. Furthermore, I could put a constant in the numerator, I could multiply the entire sequence by a complex number and it still would go to 0. Because anything that goes to 0 I can multiply it by 2 but it just takes twice as long to go to 0, it still goes to 0. Another example is the sequence of numbers q to the power n, where q is a number between 0 and 1. So, for example, if q is equal to let's say 1 3rd, then the sequence we would be looking at is, 1 3rd to the power 1, 1 3rd squared which is 1 9th. 1 3rd cubed, just 1 over 27. 1 3rd to the 4th, which is 1 over 81, and so forth. So as long as q is a number less than one, whenever you raise that to a higher and higher power, the numbers will get smaller and smaller and smaller and again, they approach 0. If q was the number 1, itself, you would be raising 1 to the power n, that would be constantly equal to 1, that would not converge to 0. And if q was bigger than 1, these numbers would blow up, definitely not converge to 0. More generally, if instead of a real number, q between 0 and 1, we plug in a complex number here, whose length is less than 1. That will still converge to 0 so, let's look at a picture of why that is happening. So, suppose here is the circle of radius 1 and, my z is some number here. If I square this number z, what happens is, we square the distance to the origin, which makes that number smaller, because the distance from the origin is less than 1 to begin with and we double the argument, so maybe z squared will be over here. Z cubed, we're tripling the argument and even closer to the origin. Z to the fourth and so forth. So, these numbers will sort of spiral around the origin, getting closer and closer to the origin because the distance from the origin is just going to 0 another example. Suppose we're looking at the nth root of 10. So that sequence starts out with being the first root of 10. What's the first root of 10? Well, that's 10. The second root of 10 is just the square root of 10. So that's 3 point something then the cubed root of 10. So that's the number whose cube is equal to 10. So that's between 2 and 3 and the fourth root of 10, so that's a number, so when that raises the power of 4, it's equal to 10. You can see these numbers get smaller and smaller and smaller and they actually approach the number 1, for large enough value of N. And finally, the nth root of n that's an interesting sequence, and here we're not so sure where it really happens. Let's look at what happens for n is equal to 1, that's the first root of 1, so it's just 1. For n is equal to 2, we get the square root of 2, we know that's 1.4 something, then we get the cubed root of 3. The 4th root of 4, the 5th root of 5 and it's pretty unclear because the number with which we are taking are getting larger and larger, but we're also taking larger and larger roots of these numbers and this requires some more thought. What one can prove that, that sequence converges to 1. Here are some rules and facts about limits, one fact is that convergent sequences are bounded. Remember what it means to be bounded? That means the sequence is contained in some large disk around the origin. So why is that true, well here is our coordinate axes and suppose we have a sequence that converges to some point s. That means, for, any disc we choose around this point s, eventually, the whole remainder of the sequence will lie inside the disc of radius s. So, only a few elements of that sequence can be outside of the disc of radius s and that's only final the menu, so that's one largest one. And we can just pick a disc that's large enough, to contain all of these final [INAUDIBLE] elements, plus the disc around the point s, and that disc then shows us that the sequence is bounded. Another fact is that if you have two convergent sequences, one called sn which converges to s and one called tn, suppose its limits is t, then I can add these two sequences to each other, and the sum converges to s plus t. Let's look at an example, we know that if sn is equal to 1/n. Then that sequence converges to 0. We also knew that if tn is equal to say 1 3rd to the n. Then tn also converges to 0. So now, according to this fact, I can add these two things up. I can look at s and plus tm which is 1 over m plus 1 3rd to the power n and together, this will still converge to 0. Another fact is that sn times tn will converge to s times t, so I could have multiplied these two sequences or any other two convergent sequences. And their product will converge to the product of the limits, in particular, I can multiply a convergent sequence by some number. And that new sequence will converge to the old limit times that number. I can also form the quotient of two sequences and it will converge to the quotient of the limits. Obviously I can't divide by 0 so I can't do that for the example we've been looking at above. 1 3rd to the n is no good, it converges to 0 and so 0 in the denominator is not something that we would want. And so here we would need another example, something that converges to 1 for example. So let's apply these facts that we just learned and find some limits. Let's look at the sequence n over n plus 1 well, n itself is not a convergence sequence X just gets bigger, 1, 2, 3, 4, 5. The sequence n plus 1 also gets bigger and bigger, so looking at it like this is not useful but, we can pull the n out of both numerator and denominator. And if we do that the numerator becomes 1 and the denominator becomes 1 plus 1 over n. And all of a sudden we find ourselves with a constant sequence in the numerator, which obviously converges to 1, because it's constant. And the denominator has the sequence 1 over n, which we know converges to 0. We add to that the number 1, which is a constant sequence which is going to break up into 1. So, the denominator converges to 1. 1 divided by 1 is 1. So, the whole sequence converges to 1. So, here this denominator, we know converges to 1. the numerator is constant so it converges to 1. So, the limit of the quotient is 1. Here's another example again if we look at it in it's original form 3 n squared plus five divided by i n squared plus 2 i n minus 1, it's hard to see what the limit is. Because both numerator and denominator seem to go off to infinity. They're not bounded, they're not convergent sequences, convergent sequences are unbounded. But if I pull the n squared out of numerator and denominator, I need to do that correctly. I need to pull that out of each term, so the numerator becomes 3 plus 5 over n squared and the denominator becomes i plus 2 i divided by n. There's only one n here, so we need to pull the second n out right here, minus 1 over n squared. Now I know 1 over n squared converges to 0, so 5 over n squared converges to 0, so the whole numerator converges to 3. In the denominator, I have i over n squared going to 0, I have 2 i over n going to 0 and so the denominator goes to i. So the whole limit is 3 over i which is not how we write complex numbers, we know by now. Imagine the only number is in the denominator, so this is equal to minus 3i so the limit of the sequence is minus 3i. Here's another example. n squared over n plus 1. Again, both numerator and denominator go to infinity, so we try to pull out an n. If I pull out one n, I have another n left in the numerator. The denominator is 1 plus 1 over n. So the denominator behaves really well. It converges to 1. So the denominator can be treated as almost equal to 1, a margin of n. However, the numerator gets bigger and bigger and bigger and bigger. And if I divide a really big number by a number that's almost 1. The really big number doesn't change a whole lot. And so this sequence is not bounded, it just gets bigger and bigger. The numerator is simply stronger than the denominator. The sequence is not bounded, and so it does not converge. By r fact, convergence sequence are bounded. Since this one is not bounded it cannot converge and, finally, let's look at this example 3n plus 5 divided by in squared plus 2yn minus 1. Again, let's pull up n squared out of both numerator and denominator, the numerator becomes then, 3 over n plus 5 over n squared. The denominator becomes i, plus 2 i over n minus one over n squared. In the numerator, 3 over n goes to zero, 5 over n squared goes to 0. In the denominator, one over n squared goes to zero, 2 i over n goes to 0 I, ios constant, equal to i. So denominator goes to i, numerator to 0, o divided by i is 0. So this last sequence converts this as 0, thus n goes to infinity. Now let's consider the following sequence, i to the n divided by n. So i to the first power is just i divided by 1. For in a subsequent to two we get -1/2 because i^2 is -1. When in subsequent to three, we get i cubed, which is -i / 3. For n = 4, we get i^4, which is one and so forth. So, we would like to apply our facts, but i to the power n times the sequence 1 over n. The sequence 1 over n, we know goes to 0 but the sequence i to the power n just hops around. One I minus one minus I, one I, minus one, minus I, it hops around. It never approaches just one point it's always in different places. That sequence does not converge, even though it is bounded, it never stays within a little disk of just one point because you just keep going and it will be outside of that disk. So, how do we treat this product? The sequence, it was our impression, converges to 0 because, you know, we started i and we have minus one half, minus i over 3, one fourth and so forth. The sequence seems to spiral itself towards 0. So, it seems to converge to 0, but how do we show that? Here's some additional facts; a sequence of complex numbers converges to 0 if and only if the sequence of absolute values converges to 0. Well, if you can take absolute values here, we're golden I to the n over n, the absolute value of i to the n is just 1. So, it becomes the sequence, 1 over n and we know that converges to 0. So, that's one way to show that the sequence converges to 0. This only works for converges to 0 however, the sequence of absolute values converging is not enough for convergence other sequence to some other number it only works at 0. Another fact is that a sequence of complex numbers converges to a limit if the real parts of the sequence converges to the real part of the limit and the imaginary parts converges to the imaginary part of the limit. So the real parts need to converge and the imaginary parts need to converge and that makes the whole sequence converge. Here's another really neat fact, you may have heard about this in Calculus, it's called the squeeze theorem. Suppose you have three sequences r n, s n and t n. And suppose they're all [INAUDIBLE] lined up so, r n's always the one at the bottom is less than or equal to s n, that's the one in the middle and less than or equal to t n in the one to the right. If you were to draw a number line, we would have r n, and then s n and t n. Now suppose that both r n and t converge, okay? So r n converes to some limit and t n converges to some limit. And suppose it's the number, suppose they all converge to a limit L. Well then sn who's stuck in between has no choice but to converge to L as well and that's what the squeeze theorem says and then here's the last fact. Sort of the equivalent of a sequence running against the wall, suppose you have a sequence that's bounded and monotone. Bounded means it's not going to go off anywhere and Monotone means it keeps getting bigger or, it keeps getting smaller, but you have to pick one of those two. So, this is about a sequence of real numbers, so, a sequence that keeps getting bigger. But you can't get go beyond then has to accumulate somewhere. It could accumulate at that wall, or maybe you, the wall is chosen too big, going to accumulate somewhere before the wall, and it certainly cannot run off to infinity. But it has to converge, so down the monotone sequence of real numbers converges. So let's apply these 2 facts to the sequence i to the n over n. I already showed you how to apply the first fact, i to the n over n, the absolute value of that is 1 over n, that goes to 0 and so the first theorem told us that i to the n over n must also go to 0. But now, let's look at real and imaginary parts. What is the real part of i to the n over n? Well, it depends. I to the n remember is just i1 and it's equal to 1. It's minus 1, 1 and it's equal to 2. It's minus i1 and it's equal to 3. It's 1, 1 and it's equal to 4. Again, i minus 1 minus i, 1 and so forth. So the real part could be 0, 1, or minus 1. The real part is 0 in the i, minus i cases and it's 1 or minus 1 in the other two cases. And you know, I split it up here, when it's 0 when it's 1 and when it's minus 1, you can check that. And similarly, the imaginary part is either 1, or minus 1, or 0 and I wrote that down right here. So therefore, the real parts are always between minus 1 over n and 1 over n, and the imaginary parts as well. Since both 1 over n and minus 1 over n converge to 0, the squeeze theorem then implies that the real parts and the imaginary parts converts to 0, therefore forcing i to the n over n to converge to 0. We also need to talk about limits of complex functions, we say that a complex value function f has a limit l as z approaches to 0. If the values of f are near l. Se goes to z0. So that means, if this is a point L and the values of f have to be near L that means if I draw a small enough disc around L, and a disc of radius epsilon, then eventually, all the f of values are in that disc. If the z values were close enough to z0. So. I can find another disc around z0, such that if my z values are from within that disc, and the f of z values will be in the disc of radius epsilon. So if you wanted to make this precise, you would say, you know, for every epsilon, that's this disc around this unit, there's a delta, that's the second disc around the 0's such that, whenever z minus is 0, less than delta, the z's are in the delta disc, then f of z minus l is less than epsilon. Of course you know f of z needs to be defined near z0 for this definition to make any kind of sense but we don't necessarily require f to be defined as z 0. Use an example. Suppose f of z is the function z squared minus one over z minus 1. That function is not defined as z equals one because I will be dividing by 0. But we could ask, does f of z approach a limit as z gets close to 1? It's a little unclear well, let's look at that. We look at the limit of f of z as z approaches 1 and we notice, the numerator actually factors into z minus 1 times z plus 1 and we can cancel out one of these z minus one factors and we are left with the limit of z plus 1 as z goes to 1. But z plus 1 as z approaches 1 is basically 1 plus 1, which is 2, so this function has a limit of 2. And so it does have a limit even though the function itself is not defined at z equals one. Let's look at another example, suppose f of z is the function, argument of z, where argument is the upper case argument. Remember, the upper case argument of the z, is the angle that z forms with a positive real axis, and we have to pick the angle that is between minus pi and pi. So let's look at the argument of z as z approaches i, i is right here. And as z approaches i, no matter from where z approaches i so I can draw a whole little disc around i and all those Z values in here have an argument that's pretty close to pi over two. And the smaller I draw the disc, the closer all those arguments will be to pi over 2. So we say the limit, of the argument of z as z, approaches I equals pi over 2. How about the argument of z as z approaches one? One is over here and again, I can draw a small disc around 1. And the arguments above the real axis are going to be a little bit bigger than 0 and below the real access to the less than 0. But they're all pretty close to 0 around here. The smaller I draw my disc, the closer these arguments will be to 0. So the limit of the argument is 0. But now, let's look at minus 1, I'll draw a little disk around minus 1. What happens? For points in the upper half of this disk, the argument approaches pi. It's a little bit less than pi, but it's argument is pretty close to pi. On the other hand, for points in the bottom half of the disc, we measure negatively the argument is almost minus pi. So, down here, the argument is really close to minus pi, so if I make my disc smaller and smaller and smaller. Below the X axis the argument points to approach minus 5, but above the X axis, the argument wants to approach pi, they can't agree, so there is no limit. This limit does not exist. The previous facts about limits of sequences imply similar facts about limits of functions. I'm just going to very quickly list those and I'll remind you of those facts when we need them. If f has a limit at z0 then f is bounded that's just like a sequence that converges, needs to be bounded. If a function f approaches a limit L and a limit g approaches limit M. As z approaches to 0, then the sum of the two functions approaches L plus M, or the product of the functions approaches the product of the limits, and the quotient approaches the quotient of the limit, provided you're actually allowed to divide by the denominator. You, you can't be dividing by 0. Finally, a function is continuous if f of z approaches f of z 0, as z approaches 2 0. That's continuity at z 0, this definition implies a lot of things implicitly. So, we're saying that f needs to be defined at z 0. Because otherwise we couldn't be writing down f of z0 right here. We are also saying that f has a limit as z goes to 0, because we're saying f of z converts to something as z goes to 0. And finally, we are saying these two need to be equal because we're saying f of z doesn't just convert to some limit l. But that limit actually needs to be f of z0. So F needs to be defined at z0. It needs to have a limit as z goes to 0. And that limit needs to equal f of z0. You've probably seen definitions of continuity before, for functions from r to r that's often described as you can draw a function without picking up your pencil that's not quite accurate, there are some other functions that are continuous, but it's good enough for our understanding, you cannot have a function that makes a jump for example. That function would not be continuous because it wouldn't have a limit. And so a function that is continuous at, at a point Z0 needs to be defined there, have a limit and that limit is equal f of z 0. Examples of continuous functions are constant functions, the function f of z equals z, polynomials, f of z is equal to absolute value of z. And f of z is a polynomial divided by some other polynomials, but you only can look at points where the denominator is not 0, cause otherwise you'd be dividing by 0 and then you would not have a limit. So that's enough for today, for the next lecture we'll finally be ready to start talking about Julia sets of quadratic polynomials, which we just learned are continuous functions.