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 it's this per listing
Mon 1/2 1/3 1/4 1/5 and so forth this is 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 plane and so say one is here one half is here and one-third here 1/4 1/5 so these
points seem to accumulate near the origin let's look at another example I I
over 2i over 3 I over 4 and so forth where are those numbers well eyes over
here I / tooth here i / 3 I over 4 i / 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 1/2 minus I over 3 and so forth so I again is this point right here minus 1/2 that seems to be here
minus I over 3 over here 1/4 I over 5 minus 1/6 minus iOS 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 0 informally we
say a sequence SN converges to a limit s if the sequence even
Angele lies in any ever so small disk centered at s this means that no matter how small a disco 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 disk how do you make this
mathematically precise we don't really need to have a mathematically precise
notion but I want to show you how to make this mathematically precise in case
you're interested so here is 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 and so a point in the sequence from which on s and minus s
is less than Epsilon so what is s n minus s that is how far as n is from s
so if I were to draw a picture and suppose s was not the origin necessarily
what suppose this is my limit and I draw my circle of radius epsilon around that
point then s is the center of this 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 to be 1 SN to be inside that
disk of radius epsilon far enough out along this sequence in this case we also
use the notation the limit Li M limit of SN is equal to X and we put n goes to
infinity underneath that limit sign to indicate
what's the end 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 what we would have to show that the
limit is zero which means no matter how small a disc of radius epsilon that
shoes 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 from 1
over N to be less than Epsilon this actually simply simplifies to 1 over n
less than Epsilon and I 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 1 over epsilon and then I just pick an index n that is bigger than that
1 over Epsilon and I'm guaranteed from that index on the entire sequence will
be located inside this disk
another example is the sequins one over N to some power P the peak can be
anything it can be two three four it could be one half so for example when P is equal to two the sequins we'd be
looking at would be 1 over 1 squared 1 over 2 squared 1 over 3 squared and so
forth so one 1/4 1/9 that sequence clip converges to 0 but also when P is equal
to 1/2 what does it mean to raise a number to the power 1/2 it means taking
the square root of that number so in that case the sequence would be 1 over
square root of 1 1 over square root of 2 1 over square root of 3 and so forth
that sequence would also go to 0 and we can show that very similarly to how we
showed the sequence 1 over N converges to 0 and in general for any power PE
afix that power and a sequence will go to 0 it'll take it longer to go to 0
it'd 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 1/3 then the
sequence we would be looking at is 1/3 2 to the power 1 1/3 squared which is 1/9
1/3 cubed just 1 over 27 1/3 to the fourth this is 1 over 81 and so forth so
as long as Q is a number less than 1 whenever you raise that to a higher and
higher power the numbers will get smaller and smaller and smaller and
again they approach zero if Q was the number one self you would be raising one
to the power and it would be constantly equal to one that would not convert to
zero and if he was bigger than one these numbers would blow up and definitely not
converge to zero more generally if instead of a real number Q between zero and one we plug in a complex number here
whose length is less than 1 that will still converge to zero 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 arguments maybe Z squared will be over here Z cubed or
we're tripling the argument and even closer to the origin Z to the 4th 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 end root of 10
so that sequence starts out with being the first root of 10 what's the first
root of 10 oh that's 10 the second root of 10 is the square root of 10 so that's 3 point something then the cube root of
10 so that's the number whose cube is equal to 10 so that's between 2 and 3
then the 4th root of 10 so that's a number so that when our raises the power
4 is equal 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 what really happens let's look at what
happens for n is equal to 1 that's the first root
one so it's just one put n is equal to two we get the square root of two we
know that's one point for something then we get the cubed root of three the
fourth root of four the fifth root of five and it's pretty unclear because the number says roots were taking or getting
larger and larger but also taking larger and larger roots of these numbers and
this requires some more thought but 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's our coordinate axis and
suppose we have a sequence that converges to some point s that means for
any disk would choose around this point s eventually the whole remainder of the
sequence will lie inside the disk of radius s so only a few elements of that
sequence can be outside of the disk of radius s and there's only finally many so that's one largest one and you can
just pick a disk that's large enough to contain all of these finally many
elements plus the disk around the point s and that disk 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 T and supposes 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 knew that if SN is equal to 1 over n then that
sequence converges to zero we also knew that if TN is equal to say 1 3 to the N
then T n also converges to zero so now according to this fact I can add these
two things up I can look at SN plus TN which is 1 over n plus 1 3 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 in 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 also form the quotient of two sequences
and it will converge to the quotient of the limits obviously I can't divide by
zero so I can't do that for the example we looking at above one third to the N
is no good it converges to zero and so zero in the denominator is not something
that we would want so here we would need another example something that converges
to run for example so let's apply these facts that we just learn and find some
limits let's look at the sequence n over n plus one well in itself is not a convergent sequence I just get speaker 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 an 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 the 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 converging to 1 so the
denominator converges to 1 1 divided by 1 is once the whole sequence converges
to 1 so here this denominator we know converges to 1 the numerator is constant so converges to 1 so the limit of the
quotient is equal to 1 here's another example again if we look at it in its
original form 3n squared plus 5 divided by I n squared plus 2i 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 fun words and sequences
conversion sequences are bounded but if I pull it N squared out of numerator and
denominator I need to do that correctly you need to pull it 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 it's only 1 in here so when you just pull a sect and out right here minus 1 over N squared
now I know 1 over N squared converges to 0 so five over N squared converges to zero so
the whole numerator converges the three in the denominator I have I over N squared going to zero I have to I over n
going to zero and so the denominator goes to I so the whole limit is three over I which is not how we write complex
numbers we don't like have imaginary numbers 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 one again both numerator and denominator go to
infinity so we try to pull out an N by pod one n I have another n left in the
numerator and 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 for large enough ends 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 our fact
convergent sequences are bounded this since this one is not bounded it cannot
converge and finally let's look at this example 3n plus 5 divided by n squared
plus 2 yn minus 1 again let's pull it 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 2i over n minus 1 over N squared in the numerator
3 over n goes to 0 5 over N squared goes to 0 in the denominator 1 over N squared
goes to 0 to I over N goes to 0 I is constant equal to I so denominator goes to I numerator 2 0 0 divided by I is 0
so this last sequence converges to 0 as n goes to infinity
now let's consider the following sequence on to the M divided by n so I
to the first power is just I divided by one twin is equal to two we get minus
1/2 because I squared is minus 1 when n is equal to 3 we get I cube which is
minus I over 3 for N equals 4 we get I to the fourth which is 1 over 4 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 and we know goes to 0 but the sequence I
to the power n just hops around 1 I minus 1 minus I 1 I minus 1 minus I it
hops around it never approaches just one point is 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'll be outside of that disk so how do we treat this
product the sequence it was our impression converges to zero because you
know we start at I and we have minus 1/2 minus I over 3 1/4 and so forth the
sequence seems to spiral itself towards zero so it seems to converge to zero but
how do we show that here's an 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 desist 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 convergence to 0 however the sequence of apps the values
converging is not enough for convergence of the sequence to some other number it
only works at 0 another fact is that the sequence of complex numbers converges to
a limit if they're real parts of the sequence converge
the real part of the limit and the imaginary part converge to the imaginary part of the limit so the real parts need
to converge and the imaginary parts need to cut approach 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 are n
SN and TN and suppose they're all nicely lined up so RN is always the one at the
bottom is less than or equal to s and that's the one in the middle and less
than or equal to T and the one to the right if you were to draw a number line
we would have our N and then SN and then TN now suppose that both are N and T and
converge okay so our n converges to some limit and TN converges to some limit and
suppose it's the same 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 could pick one of those two so this is about a sequence of real number so a
sequence that keeps getting bigger but you can't get go beyond then it has to
accumulate somewhere it could accumulate at that wall or maybe you the wall was too isn't too big could accumulate
somewhere before the wall and it's certain you cannot run off to infinity
but it has to converge a bounded monotone sequence of real numbers
converges
so let's apply these two facts to the sequence I to the end over and I already
showed you how to apply the first fact I to the end over and the absolute value
of that is 1 over and 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 what depends I to the N remember is
this I when n is equal to 1 its minus 1 when n is equal to 2 its minus I 1 n is equal to 3 it's 100 is equal to 4 and
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 - i cases and it's 1 or minus 1 and the other two
cases and you know i split it up here when is 0 when it's 1 minutes minus 1
you can check that and similarly the imaginary part it's 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 and then 1 over n and imaginary parts as well since both 1 over and n minus 1 over N converge to 0
the squeeze theorem then implies that the real parts and the imaginary parts
converge 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 valued function f has a limit L is the
approaches to 0 if the values of f are near l SZ goes to 0 so that means if
this is a point out and the values of f have to be near L that means if I draw a
smaller disc around out in a disc of radius epsilon then eventually all the F
of Z values are in that disc if the Z values were close enough to Z 0 so I can
find another disc around C 0 such that if my Z values are from within that disc
and the FFC values will be in the disk of radius epsilon so if you wanted to
make this precise you would say you know for every epsilon that's this disk
around that limit there exists a delta that's the size of the disk around C
zero such that whenever Z minus e0 is less than Delta the disease are in the Delta T's then F of C minus L is less
than Epsilon of course you know f of Z needs to be
defined near to zero for this definition to make any kind of sense but we don't
necessarily require F to be defined at C zero here's an example suppose F of C is
the function Z squared minus one over Z minus one that function is not defined
at Z equals 1 because I will be dividing by zero but we could ask does f of Z
approach your limit as C 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 1 factors and we are left with a limit of Z plus 1 se goes to 1
but Z plus 1 as Z approaches one 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 1
let's look at another example suppose F of C is the function argument of C we
are human is the upper case argument remember the upper case argument of the
C is the angle at 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
the approaches I I is right here and as the approaches I no matter from where Z
approaches I'd so I can draw a whole little disk around I and all those Z
values in here have an argument that's pretty close to PI over 2 and the
smaller enter the disk the closer all those arguments
we do PI over two so we say the limit of the argument of Z as Z approaches I
equals PI over two how about the argument of Z as Z
approaches one one is over here and again I can draw a small disk around one and the arguments above the real axis
are going to be a little bit bigger than zero and below the real axis the less than zero but they're all pretty close
to zero around here the smaller I draw my disk the closer these arguments will
be to zero so the limit of the argument is zero but now let's look at minus one
enter our little disk around minus one what happens for points in the upper
half of this disk the argument approaches PI it's a little bit less
than PI but its argument is pretty close to PI on the other hand for points in the bottom half of the disk we measure
negatively the argument is almost minus pi so down here the argument is really
close to minus pi so that make my disk smaller and smaller and smaller below
the x-axis the argument wants to approach minus y but above the x-axis
the argument wants 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 C 0 then f is bounded
let's just like a sequence that converges needs to be bounded if a
function f approaches a limit L and a function G approaches a limit M as Z
approaches C 0 then the sum of the two functions approaches L plus M or the product of the functions approaches the
product of the limits in the quotient approaches the quotient of the limits
provided you're actually allowed to divide by the denominator you can be
dividing by 0 finally a function is continuous if F of Z approaches f of C is 0 as Z approaches to 0 that's
continuity at C 0 this definition implies a lot of things implicitly so
we're saying that F needs to be defined at C 0 because otherwise we couldn't be
writing down F of C 0 right here we're also saying that F has a limit as Z goes
to Z 0 because we're saying F of Z converges to something as Z goes to Z 0
and finally we're saying these two need to be equal because we're saying F of C
doesn't dis converge to some limit L but that limit actually needs to be F of C 0
so f needs to be defined at C 0 it needs to have a limit as Z goes to Z 0 and
that limit needs to equal F of C 0 I've probably seen definitions of continuity
before for functions from R to R it's often described as you can draw a
function without picking up your pins 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 C 0 needs to be
defined there have a limit and that limit needs to equal F of C 0 examples
of continuous functions are constant functions the function f of Z equal C
polynomials F of C is equal to absolute value of Z and F of C is a polynomial
divided by some other polynomials but you only can look at points where the denominator is not zero because
otherwise you'll be dividing by zero 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 our continuous functions