0:19

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

1:11

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.

1:43

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.

4:28

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.

5:51

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.

7:05

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,

7:55

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.

9:04

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.

10:05

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.

12:16

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.

13:07

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.

14:01

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.

14:35

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.

18:21

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?

19:04

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.

20:04

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.

21:06

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.

22:20

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.

24:00

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.

24:50

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.

26:02

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?

28:43

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.