WEBVTT
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Language: en
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So the next thing that i want to discuss is
schottky’s theorem, which is very easy to
00:01:06.290 --> 00:01:07.540
prove, okay.
00:01:07.540 --> 00:01:31.799
So let me write it down, schottky’s theorem,
so let alpha be a positive constant and beta
00:01:31.799 --> 00:01:47.380
be a fraction, positive real number between
0 and 1, okay, then there is a constant c,
00:01:47.380 --> 00:02:14.830
alpha, beta such that
if script f is a family of analytic functions
00:02:14.830 --> 00:02:45.769
on the open unit disc
that omits the values 0 and 1 and satisfies
00:02:45.769 --> 00:03:08.920
f of 0 is bounded above alpha for all functions
small f in the family f, then, the mod fz
00:03:08.920 --> 00:03:23.740
is less than c alpha, beta for all z with
mod z less than beta, okay. This is schottky’s
00:03:23.740 --> 00:03:29.550
theorem and the point is that it is constant
c alpha, beta, it depends only on alpha and
00:03:29.550 --> 00:03:38.280
beta and it does not have anything to do with
what the family is, it works for any family.
00:03:38.280 --> 00:03:48.390
And, such theorems when they were 1st proved,
they were pretty, they were considered pretty
00:03:48.390 --> 00:03:59.849
difficult but then because of, because we
have montel’s, you know theorem on normality
00:03:59.849 --> 00:04:07.290
and montel’s test on normality, it is easy
to deduce this theorem. Okay, so let me tell
00:04:07.290 --> 00:04:17.680
you the proof, you can see immediately that
the, you, you want the family to be arbitrary,
00:04:17.680 --> 00:04:23.150
therefore you consider the biggest possible
family, namely you take all analytic functions
00:04:23.150 --> 00:04:31.500
on the unit disc, satisfying the condition
that the, you know value at 0 is bounded by
00:04:31.500 --> 00:04:32.740
alpha okay.
00:04:32.740 --> 00:04:41.020
So you apply it to the largest possible family
that you can think of, okay. And, and you
00:04:41.020 --> 00:04:51.050
know, see the moment you given that the, these
functions omit the values 0 and 1, it means
00:04:51.050 --> 00:04:58.350
that the family is normal. See in fact, you
see what is, what is montel’s theorem on
00:04:58.350 --> 00:05:03.250
normality, otherwise it is called the fundamental
normality test. See if you want to decide
00:05:03.250 --> 00:05:10.030
a family of meromorphic functions on a domain
is normal, then you need to know that it omits
00:05:10.030 --> 00:05:18.490
3 values. But the values in the extended complex
plane, so one of them could be infinity, right.
00:05:18.490 --> 00:05:23.259
But then if you are working only with analytic
functions, you already know infinity is not
00:05:23.259 --> 00:05:31.199
going to be taken, okay, so you have to only
ensure that for a family to be normal, to
00:05:31.199 --> 00:05:38.150
be able to apply the normality test, you have
to only ensure that the family does not, every
00:05:38.150 --> 00:05:43.240
function in the family does not take 2 values.
So here it is given that the sum all the functions,
00:05:43.240 --> 00:05:49.210
they do not take the values 0 and 1. So you
know if you apply the fundamental normality
00:05:49.210 --> 00:06:00.210
test, that is montel’s theorem, it will
follow that if you take the, if you take the
00:06:00.210 --> 00:06:06.650
family of all analytic functions on the unit
disc, which omit the values 0 and 1, that
00:06:06.650 --> 00:06:08.350
will be normal, okay.
00:06:08.350 --> 00:06:14.490
To this is montel’s theorem on normality,
right. But then we also saw another care of
00:06:14.490 --> 00:06:21.550
montel, okay, which was translation or improvement
of the arzela ascoli theorem, which said that
00:06:21.550 --> 00:06:28.559
for a family of analytic functions to be normal
on a domain, you need that the family is normal
00:06:28.559 --> 00:06:33.719
uniformly bounded, that is it is uniformly
bounded on the compact subsets. So if you
00:06:33.719 --> 00:06:41.240
see mod z less then, so you know, so let me
write, let me consider mod z less than or
00:06:41.240 --> 00:06:47.029
equal to beta, okay, if you look at mod z
less than or equal to beta, so i will change
00:06:47.029 --> 00:06:53.729
this here to mod z less than or equal to beta
which is what i meant to but i did not.
00:06:53.729 --> 00:06:59.039
But if you take mod z less than or equal to
beta, that is a compact subset of the unit
00:06:59.039 --> 00:07:05.180
disc because it is closed and bounded. And
therefore by the other montel theorem, which
00:07:05.180 --> 00:07:10.379
is improved version of the arzela ascoli theorem,
the normality of the family will tell you
00:07:10.379 --> 00:07:16.629
that the family is going to be, is going to
be normal uniformly bounded, so it is uniformly
00:07:16.629 --> 00:07:20.880
bounded on any compact subset. So on this
compact subset sum all the functions should
00:07:20.880 --> 00:07:26.430
have a bound and call that bound as c alpha,
beta, it is as simple as that, okay.
00:07:26.430 --> 00:07:30.669
So what you must remember is that we have
applied 2 montel’s theorems, one montel
00:07:30.669 --> 00:07:36.599
theorem which is, which equates the normality
of a family, that is a normal sequential compactness
00:07:36.599 --> 00:07:44.849
of family with uniform boundedness of the
original functions as a family on compact
00:07:44.849 --> 00:07:51.309
subsets, normal uniform boundedness. And that
is mind you, that is an improved version of
00:07:51.309 --> 00:07:56.580
the arzela ascoli theorem and in fact it used
arzela ascoli theorem plus the diagonalization
00:07:56.580 --> 00:08:05.029
argument, okay. And then you apply the more
serious montel’s theorem on normality, the
00:08:05.029 --> 00:08:10.429
fundamental normality test or fundamental
criterion for normality which is a very deep
00:08:10.429 --> 00:08:11.429
theorem.
00:08:11.429 --> 00:08:18.219
Mind you that was the key to proving picard’s
theorem, okay. That the moment of family of
00:08:18.219 --> 00:08:25.129
metamorphic functions omits 3 values which
is normal, the moment a family of analytic
00:08:25.129 --> 00:08:30.900
functions omits 2 values, it is normal, okay.
So you apply those 2 theorems, then schottky’s
00:08:30.900 --> 00:08:40.310
theorem is simple corollary. So it happens
that there is a paper of zaltzmann in the
00:08:40.310 --> 00:08:48.170
building of the american mathematical society
where several, where he explains how several
00:08:48.170 --> 00:08:57.250
problems in functions very have easy solutions
by use of the zaltzmann lemma.
00:08:57.250 --> 00:09:04.690
So in fact there is what is called the
, there is a very deep theorem called block’s
00:09:04.690 --> 00:09:14.350
theorem and it involves, roughly it is trying
to estimate the size of, the largest size
00:09:14.350 --> 00:09:24.130
of the disc under the image of univalent
or one-to-one analytic function. You take
00:09:24.130 --> 00:09:33.650
a one-to-one analytic function, okay and then
you know
00:09:33.650 --> 00:09:40.710
you try to estimate, you take the image and
then you try to see what is the largest disc
00:09:40.710 --> 00:09:48.120
, radius of the largest disc that is contained
in the image, okay. There are theorems of
00:09:48.120 --> 00:09:51.890
this type and there is a particular theorem
called block’s theorem which is very very
00:09:51.890 --> 00:09:52.890
deep, okay.
00:09:52.890 --> 00:10:02.530
And this can be proved by using the so-called
block zaltzmann principle, which is also called
00:10:02.530 --> 00:10:08.510
the block principle, okay. And the whole point
is that zaltzmann lemma is very powerful,
00:10:08.510 --> 00:10:19.273
it gives you proofs of, easy proofs of very
deep results, right. So it is not a surprise
00:10:19.273 --> 00:10:23.930
that you get schottky’s theorem, okay, as
a simple corollary.
00:10:23.930 --> 00:10:57.320
So let me write down, proof is, so let us,
so consider the family of all analytic functions
00:10:57.320 --> 00:11:23.270
on the open unit disc that omit 0 and
1, this is normal , this is normal by
00:11:23.270 --> 00:11:46.680
montel’s theorem on normality, otherwise
it is called the fundamental normality test,
00:11:46.680 --> 00:12:12.600
sometimes it is also called fundamental normality
criterion again by another theorem of
00:12:12.600 --> 00:12:49.550
montel along the lines of arzela ascoli
the family is normally uniformly bounded,
00:12:49.550 --> 00:13:04.390
hence bounded, hence bounded uniformly by
c alpha, beta on mod z less than equal to
00:13:04.390 --> 00:13:07.240
beta, okay.
00:13:07.240 --> 00:13:14.840
See the point is i did not even use the fact
that the functions at the origin are bounded
00:13:14.840 --> 00:13:25.880
by alpha, okay. I just i know that there is
a, there is a uniform bound, all right and
00:13:25.880 --> 00:13:35.300
i simply call that uniform bound c alpha,
beta. Actually i need not put that alpha there
00:13:35.300 --> 00:13:40.230
but i can call it c alpha, beta. The point
is that i have to put in beta because i am
00:13:40.230 --> 00:13:45.070
looking at the bound on less than or equal
to beta which is sub disc of the unit disc,
00:13:45.070 --> 00:13:51.240
close sub disc of the open unit disc, okay.
Fine. So what you must understand is that
00:13:51.240 --> 00:13:55.660
this easy proof is because you have the strong
montel’s theorem on normality, which is
00:13:55.660 --> 00:14:00.640
a fundamental normality test, okay.
00:14:00.640 --> 00:14:10.660
So, all right, so this is one thing, then
i would like to, i would like to discuss the
00:14:10.660 --> 00:14:16.500
sum i would like to discuss this relation
to 1st assignment that i gave, okay. So here
00:14:16.500 --> 00:14:31.560
is the problem that i gave earlier let
d be a domain in the complex plane and f from
00:14:31.560 --> 00:15:05.170
d to c be continuous, such that for some positive
integer n, f power n is analytic, okay, then
00:15:05.170 --> 00:15:16.840
f is analytic. Of course you know i need to
take n greater than 1 because otherwise it
00:15:16.840 --> 00:15:19.590
is trivial, okay. Because if we put n equal
to1, f power n is just, f power n is just
00:15:19.590 --> 00:15:27.870
f power 1 which is f, okay. And what is the,
what is the solution to this?
00:15:27.870 --> 00:15:39.270
Well, so the 1st thing is that , so f
power n, 1st of all f power n means f of z
00:15:39.270 --> 00:15:46.260
whole power n, which means f of z into f of
z multiplied n times, okay. The 1st thing
00:15:46.260 --> 00:15:53.340
i want to tell you is that we use the fact
that the zeros of an analytic function are
00:15:53.340 --> 00:15:59.920
isolated. So since f power n is analytic,
the zeros of f power n are isolated but then
00:15:59.920 --> 00:16:04.050
the zeros of f power n are the same as the
zeros of f, therefore the zeros of f power
00:16:04.050 --> 00:16:09.750
n are isolated, okay. And therefore what we
will do is we will 1st throw away the zeros
00:16:09.750 --> 00:16:15.940
and look at the complement of zeros in the
domain, which is a sub domain, okay.
00:16:15.940 --> 00:16:22.400
And what we will do is on that subdomain we
will 1st prove that f is analytic, all right.
00:16:22.400 --> 00:16:29.690
And then we will have to worry only about
these points where f becomes 0, all right.
00:16:29.690 --> 00:16:34.779
But then we can apply riemann’s removable
singularity theorem because each of these
00:16:34.779 --> 00:16:39.650
points will be isolated points in a neighbourhood
of which f is continuous. Therefore they will
00:16:39.650 --> 00:16:50.040
be analytic even at those points and that
is the proof, okay. So let me write this down,
00:16:50.040 --> 00:17:14.380
since the zeros of an analytic function are
isolated, the zeros z of f power n, the set
00:17:14.380 --> 00:17:25.250
of zeros is isolated but the set of zeros
of f power n is same as the set of zeros of
00:17:25.250 --> 00:17:38.370
f, so the set of zeros of f in d is isolated.
00:17:38.370 --> 00:17:42.880
Of course you know, when you want to say the
zeros of an analytic function are isolated,
00:17:42.880 --> 00:17:50.780
you must make sure that the analytic function
is not identically you know constant.
00:17:50.780 --> 00:17:56.500
So the only case where this will fail even
the analytic function is identically 0. If
00:17:56.500 --> 00:18:00.660
the analytic function is identically 0, then
the 0 that is the whole domain, okay that
00:18:00.660 --> 00:18:06.500
is the only extreme case. But of course if
f power n is identically, is f is 0, then
00:18:06.500 --> 00:18:11.950
f power n is identically 0 and if f power
n is identically 0 , f is 0, so let us assume
00:18:11.950 --> 00:18:19.821
that f power n is not 0. I assume that f is
not 0, okay, so there is no need to prove
00:18:19.821 --> 00:18:23.419
if f power n is identically 0, okay.
00:18:23.419 --> 00:18:37.799
So there has to be, we assume that f is not
radically 0 on d, right. So that is the only
00:18:37.799 --> 00:18:43.559
thing that we will have to worry about. When
you, whenever, whenever you want to apply
00:18:43.559 --> 00:18:48.590
this result that zeros of an analytic function
are isolated, you better make sure that the
00:18:48.590 --> 00:18:55.330
function is not identically 0, okay. And usually
we are not interested in that function, right.
00:18:55.330 --> 00:19:09.610
Fine, so now what you do is, in any case you
look at consider d minus z f, okay,
00:19:09.610 --> 00:19:18.150
throwaway the zeros of f, it is an isolated
set of points, so that is and again. So d
00:19:18.150 --> 00:19:28.679
minus zf is also a domain, right, and it will
still be open, okay, it will still be an open
00:19:28.679 --> 00:19:41.090
set and because you are throwing away some
isolated subset and it will also be, it
00:19:41.090 --> 00:19:46.940
cannot get disconnected, okay. So because
you are just throwing isolated points away,
00:19:46.940 --> 00:19:55.389
it cannot get disconnected so d minus zf is
also a domain, okay. And now we are going
00:19:55.389 --> 00:20:02.110
to look at the domain, the advantage with
this domain is that f power n does not vanish,
00:20:02.110 --> 00:20:05.649
f does not vanish because the zeros have been
thrown away.
00:20:05.649 --> 00:20:11.269
So f power n does not vanish and f power n
is an analytic function, okay. So you have
00:20:11.269 --> 00:20:16.330
a non-vanishing analytic function on a domain,
now you know if you take any point in the
00:20:16.330 --> 00:20:20.570
domain, if you take a, there is a sufficiently
small disc around that point which is inside
00:20:20.570 --> 00:20:26.890
domain, okay. And the point is that if you
have a non-vanishing analytic function on
00:20:26.890 --> 00:20:35.850
a simply connected region, simply connected
domain, then you can find an analytic branch
00:20:35.850 --> 00:20:43.929
of the logarithm of that function and in particular
you can find nth roots of the function for
00:20:43.929 --> 00:20:46.330
any n.
00:20:46.330 --> 00:20:51.480
The point is that you can find nth roots which
are analytic, that is the whole point. So
00:20:51.480 --> 00:20:57.230
if you want to find an nth of the function
which is analytic, okay, then the functions
00:20:57.230 --> 00:21:03.940
should not vanish and the region, the set
on which you want to find it must be simply
00:21:03.940 --> 00:21:11.499
connected, all right. So the point is that
if you take f power n which is analytic and
00:21:11.499 --> 00:21:16.690
f power n does not vanish on d minus zf, so
if you take any point in d minus zf and you
00:21:16.690 --> 00:21:22.129
take a small disc surrounding that point,
in that small disc i can find nth root of
00:21:22.129 --> 00:21:27.190
f power n. And what you expect it to be, it
has to be f, okay.
00:21:27.190 --> 00:21:33.610
But this nth root is supposed to be analytic,
therefore it will prove that f is analytic,
00:21:33.610 --> 00:21:42.489
okay. But little bit of, little bit more has
to be written down, so let us do that. F power
00:21:42.489 --> 00:22:03.809
n is analytic and non-zero on d minus zf for
z0 in d minus z f, there exists a small disc,
00:22:03.809 --> 00:22:21.929
small open disc mod z minus z0 lesser than
epsilon in d minus zf, since this is simply
00:22:21.929 --> 00:22:38.470
connected
and f power n does not vanish, there exists
00:22:38.470 --> 00:22:57.129
an analytic branch
of log f power n in mod z minus z0 less than
00:22:57.129 --> 00:23:11.970
epsilon, okay. Consider, consider the analytic
function, so let me call this analytic branch
00:23:11.970 --> 00:23:13.720
as g, okay.
00:23:13.720 --> 00:23:22.399
Consider the analytic function e power 1 by
ng, okay, which is actually, see it is actually
00:23:22.399 --> 00:23:34.889
e power 1 by n, g is actually log of f power
n, okay and you know this must be f, all right,
00:23:34.889 --> 00:23:42.869
you should expect this to be equal to f, right.
Now you see, we will use the, we will use
00:23:42.869 --> 00:23:48.779
the following, we will show that this is the,
the claim is that e power 1 by ng is actually
00:23:48.779 --> 00:23:56.950
equal to f, okay. The claim is e power 1 by
ng is actually equal to f, once you, was that
00:23:56.950 --> 00:24:03.830
is true, it means f is analytic because e
power 1 by ng is already analytic. And e power
00:24:03.830 --> 00:24:08.190
1 by ng is analytic because g is analytic
and why is g analytic because g’s analytic
00:24:08.190 --> 00:24:11.610
branch of the logarithm, okay.
00:24:11.610 --> 00:24:17.389
So i just have to, we just have to prove that
e power 1 by ng is equal to f in this small
00:24:17.389 --> 00:24:26.289
disc, okay. And this will show f is analytic
in a small disc but then the point z0 was
00:24:26.289 --> 00:24:33.999
arbitrary, so it will show that f is analytic
on d minus zf, okay. And then to come at point
00:24:33.999 --> 00:24:38.779
of zf you can apply riemann’s similarity
and conclude that f is analytic on the whole
00:24:38.779 --> 00:24:44.379
of d, all right. So the only issue is that
now i will have to show that e power 1 by
00:24:44.379 --> 00:24:53.140
ng is f, okay. Now what is common to e power
1 by ng and f, they are both nth roots
00:24:53.140 --> 00:24:55.470
of f power n.
00:24:55.470 --> 00:25:01.910
F is nth root of f power n by definition,
right and e power 1 by ng is also the nth
00:25:01.910 --> 00:25:10.600
root of f power n because if i take e power
1 by ng and raise it to the power n, i will
00:25:10.600 --> 00:25:19.140
get f power n. Okay. I will get e power g,
if i take e power 1 by ng and ready to the
00:25:19.140 --> 00:25:24.940
power of n, i will get e power g but g is
log f power n, so i will get e power log f
00:25:24.940 --> 00:25:32.450
power n which is this f power in, okay. So
both e power 1 by ng and f are nth roots of
00:25:32.450 --> 00:25:46.580
f power n, all right and the point is, you
see, you take , if you see, so now we
00:25:46.580 --> 00:25:52.120
have to use the following property.
00:25:52.120 --> 00:25:59.450
If you take the, take any 2 logarithms of
a complex number, okay, the table differ by
00:25:59.450 --> 00:26:09.440
constant, there will differ by constant multiple
of 2, constant multiple of 2 pie i. See if
00:26:09.440 --> 00:26:13.919
you take, if you calculate the logarithm of
the complex number, of course it is only defined
00:26:13.919 --> 00:26:18.029
for a complex number which is different from
0, there is no logarithm for 0, okay. So if
00:26:18.029 --> 00:26:24.649
you take a nonzero complex number and calculate
its logarithm, then you know different logarithms,
00:26:24.649 --> 00:26:31.980
you know logarithm is a multivalued function,
okay and the point is that, the real point
00:26:31.980 --> 00:26:41.399
is, the, the real logarithm of the modulus
of the number which is nonzero, since the
00:26:41.399 --> 00:26:49.649
number is nonzero and imaginary part is the
argument of that number, of the complex number.
00:26:49.649 --> 00:26:59.149
And the argument can be, the argument is defined
up to a multiple of 2n pie, to a multiple
00:26:59.149 --> 00:27:06.630
of 2n pie. Therefore the imaginary part of
the logarithm can be changed by 2n pie i,
00:27:06.630 --> 00:27:16.759
i mean by 2n pie, okay. So any 2 logarithms
of a number will differ by 2 n pie i, you
00:27:16.759 --> 00:27:28.289
have to use that, for each z in mod z minus
z0 less than epsilon, we have 2 do a little
00:27:28.289 --> 00:27:46.210
bit of thinking, see you look at the function
f of z times e power minus1 by ng of z, look
00:27:46.210 --> 00:27:55.440
at this function, okay. See look at this function,
the function, if i raise this the power of
00:27:55.440 --> 00:28:04.070
n, i will get 1 because you see if i raise
this power n, f of z will give me f of z power
00:28:04.070 --> 00:28:10.070
n and e power minus1 gz, if i raise it to
the power of n, i will get e power minus gz.
00:28:10.070 --> 00:28:17.019
Okay, but e power minus gz is 1 by f power
n because g is a log of f power n, right.
00:28:17.019 --> 00:28:24.799
So this is equal to 1, so this means that
f of z into e power minus1 by ng of z is,
00:28:24.799 --> 00:28:38.590
it is an nth root of unity, okay and this
nth roots of unity but the point is an nth
00:28:38.590 --> 00:28:45.809
root of unity and this nth root of unity
in turn will change if you will change the
00:28:45.809 --> 00:28:55.119
z. I will call this omega z because it depends
on z seemingly. For every z if you take f
00:28:55.119 --> 00:29:01.669
of z times e power 1 by ng of z, it is power
n is equal to1, so it is an nth root of unity,
00:29:01.669 --> 00:29:05.629
so for every point z you are getting nth root
of unity, for that function as w of z.
00:29:05.629 --> 00:29:13.739
So w of z is that is that function, okay.
But you see what is this, so i am just calling
00:29:13.739 --> 00:29:18.669
this function as w of z. So what is w of z,
w of z is just f of z times e power minus1
00:29:18.669 --> 00:29:25.419
by n gz, okay. But notice, here is where i
will use the fact that f is continuous. We
00:29:25.419 --> 00:29:31.850
have been given that f is continuous, i have
to, i, so f is continuous and e power minus1
00:29:31.850 --> 00:29:38.620
by ng z is also continuous. So the power is
continuous, so w becomes a continuous function,
00:29:38.620 --> 00:29:43.950
w of omega of z is a continuous function.
So it is a continuous function from a disc
00:29:43.950 --> 00:29:48.749
and what is the image z, it is the nth roots
of unity, that is the discrete set, okay.
00:29:48.749 --> 00:29:59.559
Therefore the image is to be constant, okay,
the image of, the image of disc has to be
00:29:59.559 --> 00:30:06.700
connected under continuous function. We must
get a connected subset of the set of nth roots
00:30:06.700 --> 00:30:12.279
of unity, it has to be a value, it can be
only a constant, okay, it can be only a single
00:30:12.279 --> 00:30:17.979
term. So that means this omega of z is a constant,
it is what, you get the same nth roots of
00:30:17.979 --> 00:30:24.450
unity for all z, okay, you get the same nth
root of unity for z. So that is why you are
00:30:24.450 --> 00:30:36.029
using the continuity of f, okay. Since f is
continuous, w is continuous on mod z minus
00:30:36.029 --> 00:30:51.869
z0 less than epsilon which is connected, so
w of z is equal to a constant, nth root of
00:30:51.869 --> 00:30:58.990
unity, okay.
00:30:58.990 --> 00:31:08.059
So what you get is you get f of z times e
power minus1 by ng of z is equal to constant.
00:31:08.059 --> 00:31:12.679
So which means, which tells you that f of
z is equal to the constant times e power 1
00:31:12.679 --> 00:31:26.539
by ng of z, but of course right side is analytic,
so f is analytic. So which is analytic on
00:31:26.539 --> 00:31:30.379
mod z minus z0 less than epsilon, okay.
00:31:30.379 --> 00:31:36.769
So the moral of the story is that since z0
was arbitrary, you get that f is analytic
00:31:36.769 --> 00:31:51.229
on d minus z f, okay, on d minus z of f. Now
i have to only worry at points of z of f,
00:31:51.229 --> 00:31:56.749
points at which f becomes 0. You will take
a point where f is 0, that is of course an
00:31:56.749 --> 00:32:02.539
isolated point, we have already seen that.
So it is an isolated singularity for f, all
00:32:02.539 --> 00:32:07.679
right but f is continuous there, therefore
by riemann’s removable similarities f is
00:32:07.679 --> 00:32:13.480
analytic at those points as well, therefore
f is analytic on all of d, okay. So it is
00:32:13.480 --> 00:32:17.679
an application riemann’s removable singularity
theorem.
00:32:17.679 --> 00:32:46.729
At each point of zf we have an isolated singularity
of f but f is continuous there, so by riemann’s
00:32:46.729 --> 00:33:17.389
removable singularities theorem f is analytic,
this f is analytic in d, okay. So for that
00:33:17.389 --> 00:33:23.110
is the proof that f is analytic, okay. So
you should see that, the point is that you
00:33:23.110 --> 00:33:28.550
are bringing in, you are using isolatedness
of zeros of an analytic function, okay, you
00:33:28.550 --> 00:33:34.830
are using the existence of an analytic branch
logarithm, you are using riemann’s removable
00:33:34.830 --> 00:33:42.120
singularity theorem. You have a question?
00:33:42.120 --> 00:33:51.369
Look, what we proved this around that point
f is analytic you have proved, so it becomes,
00:33:51.369 --> 00:33:54.539
if around the point the function is analytic,
that point is automatically by definition
00:33:54.539 --> 00:34:01.940
it is a singularity, it is an isolated singularity.
And riemann’s removable similarities theorem
00:34:01.940 --> 00:34:16.230
applies, okay. What is a singular point of
functions? It is a point which can be approached
00:34:16.230 --> 00:34:23.060
by the points of the, where the function is
analytic. And what is an isolated singularity?
00:34:23.060 --> 00:34:28.980
It is a point where in a deleted neighbourhood
the function is analytic.
00:34:28.980 --> 00:34:35.859
So if you take any point of zf, it is an isolated
point and you can find a deleted neighbourhood
00:34:35.859 --> 00:34:39.129
of that point where the function is analytic
because i have already, we have already shown
00:34:39.129 --> 00:34:45.540
that the function is analytic outside the
zeros of f. So that point becomes an isolated
00:34:45.540 --> 00:34:50.770
singularity and then the question is what
kind of isolated singularity is it? And you
00:34:50.770 --> 00:34:55.250
know riemann’s removable singularity theorem
says that if the function is, as a limit at
00:34:55.250 --> 00:34:59.050
that point of discontinuous at that point,
or bounded in the deleted neighbourhood of
00:34:59.050 --> 00:35:02.790
that point, all these things are equivalent
to the function of being analytic at that
00:35:02.790 --> 00:35:08.541
point, you can extend the function to that
point, you can define, we defined the function
00:35:08.541 --> 00:35:13.920
value at that point if it is not already defined
and make it analytic.
00:35:13.920 --> 00:35:22.680
But in our case the function value at the
point of zf is 0 by our own definition, okay.
00:35:22.680 --> 00:35:28.210
And the point is again, it is given, you are
again using importantly the hypothesis that
00:35:28.210 --> 00:35:34.470
the function is continuous even at points
of z f, that is importantly used. You have
00:35:34.470 --> 00:35:38.250
given the function is continuous everywhere,
so in particular the function is continuous
00:35:38.250 --> 00:35:46.870
at each point of zf and you now apply riemann
sphere removable singularities theorem, okay.
00:35:46.870 --> 00:35:55.170
That is one thing and then of course i also
wanted to discuss this problem, namely that
00:35:55.170 --> 00:36:05.670
the only one-one onto maps from the complex
plane to the complex plane are the form z
00:36:05.670 --> 00:36:10.610
going to az plus b where a is not 0, okay.
00:36:10.610 --> 00:36:19.140
So these are the only automorphisms of the
complex plane, okay. So let me do that also,
00:36:19.140 --> 00:36:29.080
because it is an application of the idea of
singularity. So here is another problem, the
00:36:29.080 --> 00:37:03.700
only bijective holomorphic maps f from
c to c are those of the form f of z is equal
00:37:03.700 --> 00:37:12.420
to az plus b where a and b complex numbers
and a is not 0. And what is the solution to
00:37:12.420 --> 00:37:29.200
this? Well, the point is that if f from c
to c is bijective, then f inverse from c to
00:37:29.200 --> 00:37:52.770
c is defined and bijective, okay. And mind
you that see f is, f is analytic and
00:37:52.770 --> 00:37:55.450
that is the inverse function theorem which
will tell you that f inverse will also be
00:37:55.450 --> 00:37:58.990
analytic because f inverse will be locally
analytic, okay.
00:37:58.990 --> 00:38:18.030
So by the inverse, by the inverse function
theorem, f inverse is analytic, okay. And
00:38:18.030 --> 00:38:25.150
now we use the following thing, you know,
you treat, so the whole point is to treat
00:38:25.150 --> 00:38:31.390
infinity is an isolated singularity of f,
okay. You treat infinity is an isolated singularity
00:38:31.390 --> 00:38:39.070
of of f inverse, okay. So you look at f inverse,
okay and look at infinity, all right, infinity
00:38:39.070 --> 00:38:47.220
is an isolated singularity, okay. Or you can
also take f, actually does not matter. Now
00:38:47.220 --> 00:38:53.790
what kind of singularities is isolated singularity
is infinity? It can be either removable or
00:38:53.790 --> 00:38:56.730
it can be pole or it can be essential, okay.
00:38:56.730 --> 00:39:03.080
If it is removable , since f is entire,
it will, by louisville’s theorem f will
00:39:03.080 --> 00:39:09.090
become a constant, okay. So certainly f is
not a constant function because it is bijective,
00:39:09.090 --> 00:39:15.570
okay it is surjective, so infinity not a removable
singularity. The other possibility is infinity
00:39:15.570 --> 00:39:21.290
is a pole, if infinity is a pole then f has
to be polynomial, okay. But if it has to be
00:39:21.290 --> 00:39:27.540
bijective, in particular if it has to be injective,
it should be polynomial of degree 1, so it
00:39:27.540 --> 00:39:32.070
has to be of the form az plus b, all right.
And the only other possibility is that f is,
00:39:32.070 --> 00:39:38.650
the infinity is an isolated essential singularity
but if infinity is an isolated essential singularity,
00:39:38.650 --> 00:39:43.410
then in every neighbourhood of infinity f
will take every complex value except one,
00:39:43.410 --> 00:39:48.250
several times, in fact infinitely many times
and that will contradict the injectivity of
00:39:48.250 --> 00:39:49.250
f.
00:39:49.250 --> 00:39:54.030
So it cannot be an essential singularity,
so here you are using picard’s theorem,
00:39:54.030 --> 00:40:00.700
all right. So so the moral of the story
is that because of picard’s theorem you
00:40:00.700 --> 00:40:07.900
are forced to conclude that f is of the form,
f of z is of the form az plus b, all right
00:40:07.900 --> 00:40:13.270
and a cannot be 0. So this is an application
of picard’s theorem, that is why i wanted
00:40:13.270 --> 00:40:47.100
to mention it. By, look at z equal to infinity
as an isolated singularity of f or f inverse.
00:40:47.100 --> 00:41:08.640
It cannot be removable
by louisville, it cannot be essential by picard,
00:41:08.640 --> 00:41:34.110
so it has to be a pole, hence a polynomial.
This must be of degree 1 by injectivity and
00:41:34.110 --> 00:41:39.070
that finishes the proof, okay. So i will stop
here.