WEBVTT
Kind: captions
Language: en
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Alright so what we are going to discuss now
is about dealing with the residue at infinity
00:01:19.040 --> 00:01:27.650
which will make sense since we have been studying
about a point at infinity okay and so you
00:01:27.650 --> 00:01:34.970
know let me tell you at the outset something
that you have to remember which are distinguishes
00:01:34.970 --> 00:01:43.280
between the residue at infinity and residue
at a finite point in the complex plane. The
00:01:43.280 --> 00:01:49.140
point is that the residue at infinity for
a function which is analytic at infinity can
00:01:49.140 --> 00:01:56.340
be nonzero whereas the residue at a finite
point in the complex plane there is a point
00:01:56.340 --> 00:02:01.610
in the usual complex plane or an analytic
function the residue is 0 okay.
00:02:01.610 --> 00:02:09.490
So this is a very very important okay. Of
course if the residue is 0 at a point it does
00:02:09.490 --> 00:02:16.500
not of course mean that the function is
analytic but the fact is that if you have
00:02:16.500 --> 00:02:20.840
an analytic function at a point in the residue
at that point is 0, so long as that point
00:02:20.840 --> 00:02:25.860
is a point of the usual complex plane but
if it is a point at infinity okay it may be
00:02:25.860 --> 00:02:32.840
analytic at infinity but yet the residue may
not be 0, so so we are going to talk about
00:02:32.840 --> 00:02:38.260
residue at infinity and we are going to talk
about you version of the residue theorem at
00:02:38.260 --> 00:02:40.230
on the on the extended complex plane okay.
00:02:40.230 --> 00:03:00.370
So so let me write this down residue at infinity
and the residue theorem or the extended
00:03:00.370 --> 00:03:15.760
complex plane, so this is what we are going
to talk about, so so you know let me begin
00:03:15.760 --> 00:03:29.030
by what usual idea of residue is? Okay, so
so recall if Z naught is a point of the
00:03:29.030 --> 00:03:48.000
complex plane okay and f of Z is analytic
in a deleted neighbourhood of Z naught okay,
00:03:48.000 --> 00:03:56.170
so when I say this I am saying that Z naught
is actually a singular point okay, so it could
00:03:56.170 --> 00:03:58.900
be removable singularity which means that
it is not really a singular point it could
00:03:58.900 --> 00:04:04.420
be analytic point but on the other hand it
could be singularity it may be a pole or an
00:04:04.420 --> 00:04:23.190
essential singularity and then the residue
the residue of f at Z naught is given by well
00:04:23.190 --> 00:04:41.710
residue of f of Z at Z equal to Z naught is
equal to well 1 by 2 pi i so, so you integrate
00:04:41.710 --> 00:04:54.180
you integrate the function fz dz around a
simple closed contour you a simple closed
00:04:54.180 --> 00:04:58.050
contour which goes around the point the positive
sense okay.
00:04:58.050 --> 00:05:04.530
So so basically so Z naught is this point
and of course there is a deleted neighbourhood
00:05:04.530 --> 00:05:10.580
of this point sufficiently small deleted neighbourhood
of this point so in particular there is a
00:05:10.580 --> 00:05:16.280
there is a disk of sufficiently small radius
open disk surrounding that point where the
00:05:16.280 --> 00:05:23.250
function is analytic and then I am just taking
a simple closed curve gamma okay of course
00:05:23.250 --> 00:05:30.729
this is simple closed contour, so simple means
of course that it does not intersect itself.
00:05:30.729 --> 00:05:35.839
It has positive orientations which means that
it is going anticlockwise around that point
00:05:35.839 --> 00:05:40.899
so in other words the interior of the contour
contains the point okay.
00:05:40.899 --> 00:05:46.710
So the point lies to the left of the contour
as you walk along the contour and the region
00:05:46.710 --> 00:05:50.260
to the left of the contour as you walk along
the contour in the direction specified by
00:05:50.260 --> 00:05:54.940
the contour this is called the interior of
the contour okay and of course when I say
00:05:54.940 --> 00:06:00.330
contour you must remember that gamma has to
be piece wise smooth, so gamma is whenever
00:06:00.330 --> 00:06:08.169
you parameterise gamma mind you gamma basically
is continues image of of an interval okay
00:06:08.169 --> 00:06:14.320
that is what the part this but it can always
it can also be a piece wise continues image
00:06:14.320 --> 00:06:26.800
okay and well well in fact we always take
gamma to be a continuous path so but the thing
00:06:26.800 --> 00:06:31.550
is that it is also continuously differentiable
at least piecewise okay that is the condition
00:06:31.550 --> 00:06:36.890
for a contour and this much is required or
this integral to be defined as a Riemann integral
00:06:36.890 --> 00:06:37.890
alright.
00:06:37.890 --> 00:06:53.340
So so this is the the residue and well I am
wondering I am wondering if this that is right,
00:06:53.340 --> 00:07:00.100
so I was just wondering whether the fraction
1 by 2 pi i is correct, it is so for example
00:07:00.100 --> 00:07:06.400
if I plug-in 1 by Z minus Z naught if I take
the function f of Z to be 1 by Z minus Z naught
00:07:06.400 --> 00:07:14.529
if I integrate 1 by Z minus Z naught over
simple closed contour sufficiently small contour
00:07:14.529 --> 00:07:22.520
then I am going to get 2 pi i so the residue
is one and well see of course it is very important
00:07:22.520 --> 00:07:31.409
that I choose this gamma in such a way that
there are no other singular points of f inside
00:07:31.409 --> 00:07:37.760
other than Z naught and that is of course
true because I am choosing gamma inside a
00:07:37.760 --> 00:07:40.930
deleted neighbourhood of Z naught where f
is analytic so there are no other points where
00:07:40.930 --> 00:07:44.700
f is singular other than Z naught okay.
00:07:44.700 --> 00:07:48.219
Now the point is…so this is the this is
the definition of residue this is one definition
00:07:48.219 --> 00:07:54.510
of residue and as you can see the importance
of this definition is that if you multiply
00:07:54.510 --> 00:07:59.729
out the 2 pi i on on the right side if you
multiply both sides by 2 pi i what you will
00:07:59.729 --> 00:08:05.300
get is 2 pi i times the residue is the integral
of the function over gamma, so what it tells
00:08:05.300 --> 00:08:10.229
you is that it tells you how to integrate
a function around a singularity okay that
00:08:10.229 --> 00:08:15.960
is the important thing, so so the fact is
that why is this definition interesting? It
00:08:15.960 --> 00:08:22.010
is interesting because you get to know what
the integral of a function is around singularity
00:08:22.010 --> 00:08:27.530
at the important thing is that you should
be able to compute left side without having
00:08:27.530 --> 00:08:31.700
to do the integral okay and that is the method
of residue that you have learned in the 1st
00:08:31.700 --> 00:08:34.029
course and so there is.
00:08:34.029 --> 00:08:38.630
So you also can recall he you know the usual
definition of the other usual definition of
00:08:38.630 --> 00:08:43.839
residue in terms of Laurent expansion, so
the point is that since f of Z is assumed
00:08:43.839 --> 00:08:47.930
to be analytic in a deleted neighbourhood
of Z naught there is a Laurent expansion or
00:08:47.930 --> 00:08:53.750
f centred at Z naught then you write out this
Laurent expansion it will contain both positive
00:08:53.750 --> 00:08:59.310
and 0 and negative powers of Z minus Z naught
and the residue is precisely the coefficient
00:08:59.310 --> 00:09:06.310
of the of 1 by Z minus Z naught namely we
Z minus Z naught to be minus 1 okay.
00:09:06.310 --> 00:09:18.760
So that is another so let me write that down
recall also that if f of Z is equal to Sigma
00:09:18.760 --> 00:09:28.890
n equal to minus infinity to infinity a n
times a n times Z minus Z naught to the power
00:09:28.890 --> 00:09:42.430
of n is the Laurent expansion of f around
Z naught or f centred at Z naught or f about
00:09:42.430 --> 00:09:48.180
Z naught mind you that that exist because
of Laurent’s theorem. Laurent expansion
00:09:48.180 --> 00:09:52.140
exists because of Laurent’s theorem and
you know what is the big deal about Laurent’s
00:09:52.140 --> 00:09:56.320
theorem, it is the it is the analog of Taylor’s
theorem, so Taylor’s theorem tells you that
00:09:56.320 --> 00:10:00.410
a function is analytic at a point Z naught
then you can expand it as a power series in
00:10:00.410 --> 00:10:05.690
Z minus Z naught namely you can expand it
as a series which involves only 0 and positive
00:10:05.690 --> 00:10:06.990
powers of Z minus Z naught.
00:10:06.990 --> 00:10:13.090
Laurent expansion tells you that you can do
something and similarly something as good
00:10:13.090 --> 00:10:19.100
if even if Z naught was a bad point if
Z naught was singular point and the only singular
00:10:19.100 --> 00:10:27.380
point in that neighbourhood in its neighbourhood
then you can expand f and both positive and
00:10:27.380 --> 00:10:33.790
negative powers of Z minus Z naught okay,
so the the fact is that if you if you are
00:10:33.790 --> 00:10:37.310
dealing with a singular point you must allow
negative powers of Z minus Z naught also in
00:10:37.310 --> 00:10:43.160
the series expansion and of course if Z naught
is a removable singularity you know roughly
00:10:43.160 --> 00:10:46.860
I mean this is exactly what Riemann’s removable
singularity theorem is that if you write out
00:10:46.860 --> 00:10:50.680
the Laurent expansion you will only get a
Taylor expansion in Z naught is actually removable
00:10:50.680 --> 00:10:53.820
singularity okay which is equivalent to saying
that f is analytic at that point.
00:10:53.820 --> 00:11:04.450
So anyway so the point is that if f has this
Laurent expansion then then the residue of
00:11:04.450 --> 00:11:14.170
f of Z at Z equal to Z naught is none other
than a minus 1 and so you know so if I write
00:11:14.170 --> 00:11:19.430
this down this is also equal to as I wrote
earlier it is 1 by 2 pi i integral over gamma
00:11:19.430 --> 00:11:33.550
fz dz okay and so the advantage of this formula
is that you know in many cases if I can write
00:11:33.550 --> 00:11:39.230
out the Laurent expansion at Z naught then
I can explicitly and out what this a minus
00:11:39.230 --> 00:11:45.020
1 is and that a minus 1 will give you that
multiplied by 2 pile i is going to give me
00:11:45.020 --> 00:11:49.610
give you the integral of the function around
that point okay so that is the advantage of
00:11:49.610 --> 00:11:53.820
this you know and you have used this is a
1st course in complex analysis to a integral
00:11:53.820 --> 00:11:56.760
and so on and so forth, so it is a very useful
thing.
00:11:56.760 --> 00:12:02.660
Now the point is that you know and then of
course what is the generalisation of this,
00:12:02.660 --> 00:12:07.710
the generalisation of this is that suppose
you are going to integrate a function around
00:12:07.710 --> 00:12:13.110
a simple closed contour and assume that inside
that region the function has only isolated
00:12:13.110 --> 00:12:23.930
singularities okay and you know they are going
to be finitely many isolated singularities
00:12:23.930 --> 00:12:29.550
and then the the formula reads the formula
that you get is essentially the residue theorem
00:12:29.550 --> 00:12:34.490
says that if you integrate the function around
a bunch of isolated singularities what you
00:12:34.490 --> 00:12:38.420
are going to get is 2 pi i times the sum of
the residue of the function at each of those
00:12:38.420 --> 00:12:44.170
similar points okay and that is the residue
theorem and what we have written down here
00:12:44.170 --> 00:12:48.030
is that the integral of the function is 2
pi i times the residue at Z naught because
00:12:48.030 --> 00:12:52.350
Z naught is the only singularity and the point
is that this extends to 2 pi i times some
00:12:52.350 --> 00:12:57.000
of the residues at various points which are
the isolated singularities of f inside the
00:12:57.000 --> 00:12:58.070
contour okay.
00:12:58.070 --> 00:13:04.020
So that is the residue theorem essentially
so the point is that you know in a way it
00:13:04.020 --> 00:13:11.010
is very easy the residue theorem is very easy
if you look at it like this. It is rather
00:13:11.010 --> 00:13:16.570
not a theorem it is more part of the definition
except that if you want to really prove it
00:13:16.570 --> 00:13:21.540
you will actually be applying Clausius theorem
okay by surrounding each of the singular points
00:13:21.540 --> 00:13:26.600
by a sufficiently small disk and noting that
outside this disk and inside your contour
00:13:26.600 --> 00:13:31.000
the function is actually analytic okay and
you will have to apply the version of Clausius
00:13:31.000 --> 00:13:35.170
theorem or multiply connected regions and
you have to use this definition of residue
00:13:35.170 --> 00:13:43.540
to get the residue theorem, so so let me write
that down more generally more generally
00:13:43.540 --> 00:14:10.160
if if f of Z has only isolated singularities
inside simple closed contour
00:14:10.160 --> 00:14:16.770
gamma and of course I am assuming that on
the contour there are no singularities okay
00:14:16.770 --> 00:14:23.290
and it is analytic on gamma.
00:14:23.290 --> 00:14:30.029
So this means that it is analytic on a small
neighbourhood an open set which contains the
00:14:30.029 --> 00:14:35.529
contour gamma, so in particular I am avoiding
the situation that there are singular points
00:14:35.529 --> 00:14:39.560
of the function on the contour okay and of
course you know if there are similar points
00:14:39.560 --> 00:14:43.500
of the function on the contour then you know
at those similar points the function will
00:14:43.500 --> 00:14:48.250
fail to continuous if they are on a single
points and once the function fails to be continuous
00:14:48.250 --> 00:14:51.690
it is very difficult to define the Riemann
integral of the function over the contour,
00:14:51.690 --> 00:14:55.720
so you know you must understand that the moment
you define the Riemann integral more or less
00:14:55.720 --> 00:15:01.550
you are assuming that the function has nothing
wrong going on the contour is only what happens
00:15:01.550 --> 00:15:03.279
inside that matters okay.
00:15:03.279 --> 00:15:19.980
So well then there are there are only finitely
many
00:15:19.980 --> 00:15:30.980
and integral over gamma f of Z dz is equal
to 2 pi i times the sum of the the sum of
00:15:30.980 --> 00:15:49.430
the residues of a residues of f at the single
points. So this is essentially the residue
00:15:49.430 --> 00:15:55.100
theorem okay and of course that there are
only finitely many follows from the fact that
00:15:55.100 --> 00:16:00.770
if you take the if you take the contour along
with the…of course whenever I say contour
00:16:00.770 --> 00:16:05.650
mind you it is always an oriented contour
and without if nothing is mentioned always
00:16:05.650 --> 00:16:10.860
the contour is oriented and it is given the
positive orientation which means that you
00:16:10.860 --> 00:16:18.440
go around the point in the anti-clock wise
sense okay and that is always there is always
00:16:18.440 --> 00:16:24.270
taken for granted unless something else is
mentioned, so whenever integral over contour
00:16:24.270 --> 00:16:29.029
is mentioned you must understand that the
contours is already oriented and the orientation
00:16:29.029 --> 00:16:30.029
is positive okay.
00:16:30.029 --> 00:16:41.790
So well so this is essentially the residue
theorem. So let me write that down so this
00:16:41.790 --> 00:16:51.490
is this is just the residue theorem and well
as I told you it is very useful to evaluate
00:16:51.490 --> 00:16:59.980
integrals which integrals of function around
contour which have only isolated singularities
00:16:59.980 --> 00:17:06.829
inside the contour okay so so this is it.
Now the point is that…what we want to do
00:17:06.829 --> 00:17:10.800
is that? We want to do this for the point
at infinity and we want the residue theorem
00:17:10.800 --> 00:17:16.100
for a domain the extended complex plane so
how we are going to do it, so so the idea
00:17:16.100 --> 00:17:23.350
is very very simple we start off by using
this by adapting a little using the same
00:17:23.350 --> 00:17:31.890
philosophy namely the residue should be you
know you integrate the function okay around
00:17:31.890 --> 00:17:38.250
the point okay and then divide by 2 pi i okay
and then that should give you the residue,
00:17:38.250 --> 00:17:45.419
so if you want to get the residue of a function
at the point at infinity 1st of all it should
00:17:45.419 --> 00:17:51.539
be analytic in a deleted neighbourhood of
the point at infinity okay and then I must
00:17:51.539 --> 00:17:57.740
take a contour that goes around infinity in
the positive sense okay and I have to integrate
00:17:57.740 --> 00:18:05.559
the function around that contour okay and
divide by 2 pi i and I must get the residue
00:18:05.559 --> 00:18:06.559
at infinity, okay.
00:18:06.559 --> 00:18:13.120
So this is so the definition for residue remains
the same namely you just integrate the function
00:18:13.120 --> 00:18:19.550
around by going over a contour that goes around
the point in the positive sense as far as
00:18:19.550 --> 00:18:24.330
the point is concern and then you divide by
2 pi i okay that is the definition alright
00:18:24.330 --> 00:18:31.090
and let us see what that that brings up, so
I incidentally I wanted to point out something
00:18:31.090 --> 00:18:42.170
here which I just remembered, so let me tell
you so for a moment in a way once you once
00:18:42.170 --> 00:18:43.880
you believe Laurent’s theorem.
00:18:43.880 --> 00:18:52.630
Once you believe Laurent’s theorem that
this is exactly I mean that this formula is
00:18:52.630 --> 00:18:57.280
correct or something that you can more or
less see because you know so you know I have
00:18:57.280 --> 00:19:02.040
this expression I have this expression of
the Laurent series okay and what I need to
00:19:02.040 --> 00:19:08.270
compute is that I need to compute integral
of f over gamma okay and that is the integral
00:19:08.270 --> 00:19:13.320
that is the same as integral over gamma of
the series on the right side and now the point
00:19:13.320 --> 00:19:18.380
is that I can push the integral inside some
because the fact is that the Laurent series
00:19:18.380 --> 00:19:22.350
like a Taylor series you know wherever it
converges, it converges normally that means
00:19:22.350 --> 00:19:28.080
it converges uniformly on compact subset and
since I am going to integrate over gamma,
00:19:28.080 --> 00:19:33.650
gamma is of course mind you any simple closed
contour is a compact set because it is both
00:19:33.650 --> 00:19:34.650
closed end bounded.
00:19:34.650 --> 00:19:40.060
So therefore the integral of the sum is same
as some of the integrals so I can push the
00:19:40.060 --> 00:19:43.800
integrals across the some and when I push
the integral across the sum what is going
00:19:43.800 --> 00:19:50.370
to happen is that you know you will see that
if I take positive powers of Z minus Z naught
00:19:50.370 --> 00:19:53.740
the integrals are going to vanish because
the positive power of Z minus Z naught are
00:19:53.740 --> 00:19:57.590
analytic functions then Clausius theorem is
going to tell you that whenever you whenever
00:19:57.590 --> 00:20:01.520
you integrate analytic function around a simple
closed contour you are going to get 0 in fact
00:20:01.520 --> 00:20:05.960
the positive parts are all entire functions
okay they are just polynomials and the same
00:20:05.960 --> 00:20:11.520
thing is going to happen the constant term
okay and then if you take the if you take
00:20:11.520 --> 00:20:16.680
the negative powers of Z minus Z naught greater
than 2 okay.
00:20:16.680 --> 00:20:21.550
So I mean what I mean by that is if you take
terms involving 1 by Z minus Z naught the
00:20:21.550 --> 00:20:26.480
whole square, 1 by Z minus Z naught the whole
cube and so on the integrals of those sums
00:20:26.480 --> 00:20:33.560
will also vanish that is because they all
have anti-derivatives, so you know it is something
00:20:33.560 --> 00:20:41.510
very basic that I want you to understand,
the fact that an integral vanishes is you
00:20:41.510 --> 00:20:46.460
know basically it is equivalent to saying
that the integral is independent of the path
00:20:46.460 --> 00:20:51.500
and especially when the integral when the
integral has an anti-derivatives then you
00:20:51.500 --> 00:20:57.770
know that the integral is actually the final
value minus the initial value okay of the
00:20:57.770 --> 00:21:02.240
anti-derivative and if the final value is
equal to the initial value is what will happen
00:21:02.240 --> 00:21:07.700
if you go around a closed path you are going
to get the integral be 0 okay and so you know
00:21:07.700 --> 00:21:12.480
all the integrals which involves Z minus Z
naught is the power of n where n is minus
00:21:12.480 --> 00:21:15.290
2, minus 3, minus 4 and so on they are all
going to vanish.
00:21:15.290 --> 00:21:21.620
So the only thing that is going to survive
is going to be integral over gamma a minus
00:21:21.620 --> 00:21:27.940
1 Z minus Z naught power minus one dz okay
and you know the integral over gamma Z minus
00:21:27.940 --> 00:21:34.330
Z naught to the minus 1 dz is just going to
be 2 pi i and that is just because of Clausius
00:21:34.330 --> 00:21:37.370
theorem because the integral over gamma is
not going to really depend on the shape of
00:21:37.370 --> 00:21:44.470
gamma you can replace gamma by a small circle
okay going around Z naught and then you actually
00:21:44.470 --> 00:21:49.230
parameterise a circle as usual a write it
as Z equal to to the i Theta, Theta varying
00:21:49.230 --> 00:21:54.060
from 0 to 2 pi and compute the integral you
will get 2 pi i okay so therefore what I want
00:21:54.060 --> 00:21:58.270
to tell you is that once you know Laurent’s
theorem okay once you believe Laurent’s
00:21:58.270 --> 00:22:04.740
theorem is formula for the residue that a
minus 1 gives you the residue is more or less
00:22:04.740 --> 00:22:10.140
direct okay, so that is something that I want
you to recall.
00:22:10.140 --> 00:22:19.730
Now let me go back and define the residue
at infinity, so suppose f of Z is defined
00:22:19.730 --> 00:22:29.130
in a neighbourhood of infinity okay so mind
you that what this means is that f of Z is
00:22:29.130 --> 00:22:33.480
defined on a circle on the exterior of a circle
of sufficiently large radius that is what
00:22:33.480 --> 00:22:37.920
it means. Neighbourhood of infinity is by
definition you know by the stereographic projection
00:22:37.920 --> 00:22:49.640
the same as the exterior of a sufficiently
large circle okay and so at this point
00:22:49.640 --> 00:22:53.530
what we do is that, we define the residue
of f at infinity will be just the integral
00:22:53.530 --> 00:23:00.470
over gamma 1 by 2 pi i integral over gamma
fz dz where gamma is simple closed contour
00:23:00.470 --> 00:23:03.980
that goes around infinity in the positive
sense okay this is exactly the way we define
00:23:03.980 --> 00:23:09.570
it as we defined it for the point in the usual
complex plane, so let me write that down.
00:23:09.570 --> 00:23:19.570
Then residue of f of Z at Z equal to infinity
is defined to be so I put a colon and equal
00:23:19.570 --> 00:23:30.440
to 1 by 2 pi i integral over gamma so
you know let me as let me change notification
00:23:30.440 --> 00:23:41.020
to f w because I will need to appeal to something
I need to appeal to change in the variable
00:23:41.020 --> 00:23:54.330
to 1 by Z, so let me do that, so f w, w equal
to infinity 1 by fw dw okay 1 by 2 pi i integral
00:23:54.330 --> 00:23:59.630
over gamma, so let me write that is is very
very important where gamma is a simple closed
00:23:59.630 --> 00:24:16.210
contour going around a point at infinity in
the positive sense, so you know this is the
00:24:16.210 --> 00:24:23.630
this is definition and well you know so this
is exactly the definition that you would have
00:24:23.630 --> 00:24:27.290
made for a point in the complex plane and
this is the same definition I am using for
00:24:27.290 --> 00:24:31.230
a point for the point at infinity but then
there are 2 or 3 things that one has to be
00:24:31.230 --> 00:24:37.010
careful about the 1st thing is that what do
you mean by contour that is going around infinity
00:24:37.010 --> 00:24:38.150
the positive sense.
00:24:38.150 --> 00:24:47.580
So the the fact is that you know you must
think of a contour in so here is where you
00:24:47.580 --> 00:24:54.990
again appealed to the stereographic projection
okay. You know that sufficiently small neighbourhood
00:24:54.990 --> 00:25:01.220
of infinity is given by the exterior of a
sufficiently large circle centred at the origin
00:25:01.220 --> 00:25:08.760
okay, so in some sense a sufficiently large
circle centred at the origin should be a contour
00:25:08.760 --> 00:25:16.669
which goes around the point at infinity okay
and you can you can see this you can see this
00:25:16.669 --> 00:25:23.610
so more generally if you take a sufficiently
large contour which goes around the origin
00:25:23.610 --> 00:25:29.299
okay simple close contour which goes around
the origin sufficiently large means it encloses
00:25:29.299 --> 00:25:33.370
a sufficiently large area okay.
00:25:33.370 --> 00:25:39.390
Then for that matter I mean it lies in the
exterior of I should rather say it lies in
00:25:39.390 --> 00:25:45.890
the exterior of a circle of sufficiently large
radius okay then that itself is an example
00:25:45.890 --> 00:25:50.210
of a simple closed contour that goes around
the point at infinity and that is because
00:25:50.210 --> 00:25:55.100
of the that is because of the stereographic
projection and the only thing is that what
00:25:55.100 --> 00:25:58.771
is this, what is this business of the positive
orientation of that contour with respect to
00:25:58.771 --> 00:26:04.320
infinity mind you the positive orientation
means at infinity should lie in the interior
00:26:04.320 --> 00:26:11.490
of that contour okay so you should orient
the contour in such a way such that the interior
00:26:11.490 --> 00:26:17.360
of the contour contains a point at infinity.
Now if I take a circle sufficiently large
00:26:17.360 --> 00:26:22.420
circle centred at the origin and oriented
in the usual way added which we do namely
00:26:22.420 --> 00:26:27.200
give it the anti-clock wise orientation in
the origin becomes comes into the interior,
00:26:27.200 --> 00:26:33.520
the interior is just the interior of that
circle okay and the exterior will be the exterior
00:26:33.520 --> 00:26:39.110
of the circle and that will contain the point
at infinity, so you can see from this argument
00:26:39.110 --> 00:26:41.720
that will have 2 orient it clockwise okay.
00:26:41.720 --> 00:26:50.190
So the gamma must be a simple closed contour
lying in a sufficiently small neighbourhood
00:26:50.190 --> 00:26:57.350
of infinity, so it should be line in the exterior
of a circle of a sufficiently large areas
00:26:57.350 --> 00:27:01.660
and it should be given the clockwise orientation
that is what it means okay. So let me write
00:27:01.660 --> 00:27:18.610
that down this means that gamma should be
a simple closed contour
00:27:18.610 --> 00:27:33.160
the complex plane lying outside a circle of
sufficiently
00:27:33.160 --> 00:27:57.900
large radius but given the clockwise or negative
orientation. So here you see you have to be
00:27:57.900 --> 00:28:04.809
careful I am saying that gamma should have
negative orientation here but in the statement
00:28:04.809 --> 00:28:10.340
before that I am saying it should have positive
sense, so mind you in the statement preceding
00:28:10.340 --> 00:28:14.030
it was positive sense with respect to the
point at infinity and positive sense with
00:28:14.030 --> 00:28:19.179
respect to point at infinity is negative sense
with respect to the origin okay. So these
00:28:19.179 --> 00:28:26.500
are not contradictory statements you have
to understand the subtlety and well of course
00:28:26.500 --> 00:28:29.790
you can also see this pictorially more or
less.
00:28:29.790 --> 00:28:36.720
See you know if you draw the stereographic
projection, so here is the x y plane which
00:28:36.720 --> 00:28:42.700
is a complex plane Z plane and then and
then you know we draw this 3rd axis which
00:28:42.700 --> 00:28:47.950
which we call it as u and then you take the
unit sphere, surface of the unit sphere in
00:28:47.950 --> 00:29:02.030
3 space I am going to get something like this
this is the stereographic projection so you
00:29:02.030 --> 00:29:14.169
know if I take now you know if I take sufficiently
if I take a circle in the complex plane of
00:29:14.169 --> 00:29:24.500
sufficiently large radius okay and then you
know well you know the point at infinity corresponds
00:29:24.500 --> 00:29:32.380
to do this point here on the on the Riemann
sphere which corresponds to the North pole
00:29:32.380 --> 00:29:36.110
so this is the this corresponds to point at
infinity okay.
00:29:36.110 --> 00:29:42.980
So I put this triple line to tell you that
it corresponds to the point at infinity under
00:29:42.980 --> 00:29:55.690
this geographic projection and well what happens
to the well you know the exterior of the circle
00:29:55.690 --> 00:30:01.650
the exterior of the circle in the complex
plane and that is going to correspond to well
00:30:01.650 --> 00:30:10.309
on the Riemann sphere is going to correspond
to the small disk like region though it is
00:30:10.309 --> 00:30:18.460
a curved surface it is a small disk like region,
when I say disk like I mean topologically
00:30:18.460 --> 00:30:23.289
you can flatten it to look like a disk topologically
and it is a disk like neighbourhood of the
00:30:23.289 --> 00:30:30.710
North pole on the Riemann sphere okay and
that is how the shaded region on the plane
00:30:30.710 --> 00:30:41.780
namely the exterior of the circle and the
small cap on surrounding the arrest like cap
00:30:41.780 --> 00:30:43.490
okay.
00:30:43.490 --> 00:30:50.230
If you imagine the Earth it is North at the
North pole so now the point is that you see
00:30:50.230 --> 00:31:00.840
what you must understand is that if I now
give this so in particular you know if I now
00:31:00.840 --> 00:31:11.850
take a sufficiently if I take a contour which
lies in outside the circle okay if I take
00:31:11.850 --> 00:31:19.590
a contour like this. Now that contour is going
to correspond to a contour here on the Riemann
00:31:19.590 --> 00:31:24.650
sphere that goes around the point at infinity
okay and as I make this circle bigger and
00:31:24.650 --> 00:31:29.120
bigger that is going to give a smaller and
smaller contour simple close contour that
00:31:29.120 --> 00:31:32.110
goes around the point at infinity the only
thing that you have to worry about is the
00:31:32.110 --> 00:31:36.780
orientation, the orientation the way I have
drawn it the orientation should be like this
00:31:36.780 --> 00:31:42.309
mind you it is the it is actually it is actually
clockwise about the origin and the reason
00:31:42.309 --> 00:31:46.669
for that is that if you define it like that
which is what you should do if you are
00:31:46.669 --> 00:31:52.750
dealing with the point at infinity in the
interior of that contour is the exterior.
00:31:52.750 --> 00:32:03.429
So this region this thing outside this
is the interior of that contour okay and well
00:32:03.429 --> 00:32:07.460
and it is actually the region that lies to
the left of the contour as you walk along
00:32:07.460 --> 00:32:13.429
the contour okay and you can see that this
region corresponds to well this region on
00:32:13.429 --> 00:32:18.630
the on the Riemann sphere and that of course
contains the point at infinity, so the point
00:32:18.630 --> 00:32:23.500
at infinity is an interior point for the contour
oriented in the clockwise direction okay,
00:32:23.500 --> 00:32:33.490
so you can see this diagrammatically, fine.
So very well now that we have defined what
00:32:33.490 --> 00:32:40.840
the residue at infinity is so of course this
is gamma right so let us compute it.
00:32:40.840 --> 00:32:48.770
Let us compute what the residue at infinity
is, so f of Z is well f of w let me write
00:32:48.770 --> 00:32:56.559
f of W, f of w is going to be Sigma n equal
to minus infinity to infinity a n w to the
00:32:56.559 --> 00:33:05.190
n which is the Laurent series of f this is
the Laurent series of f at in a neighbourhood
00:33:05.190 --> 00:33:11.520
of infinity okay and mind you in principle
it is actually also the Laurent series of
00:33:11.520 --> 00:33:18.520
f at the origin in some sense centred at the
origin but in a domain that is a neighbourhood
00:33:18.520 --> 00:33:26.080
of infinity and the only thing that you distinguish
is that when you say it is the Laurent series
00:33:26.080 --> 00:33:33.179
at infinity you see the singular part is the
one that involves the positive powers of w
00:33:33.179 --> 00:33:38.270
and the analytic part is the one that involves
0 and negative powers of w because it is the
00:33:38.270 --> 00:33:41.150
negative powers of w that behave well at infinity,
okay.
00:33:41.150 --> 00:33:44.720
So that is the only difference but what you
are actually looking at is actually the Laurent
00:33:44.720 --> 00:33:51.000
series of f at the origin okay. Now but anyway
the fact for the same reason that I told you
00:33:51.000 --> 00:33:58.710
earlier if you compute if you if you now
compute integral over gamma fw dw if I do
00:33:58.710 --> 00:34:06.890
this what I am going to get is and you know
of course I think it is 1 by 2 pie i this
00:34:06.890 --> 00:34:19.240
is what it is, well you know mind you
if I if I compute this integral the way I
00:34:19.240 --> 00:34:24.230
normally would compute the integral on the
plane, what I first do is I always compute
00:34:24.230 --> 00:34:31.440
integrals with with my contour being positively
oriented with respect to the usual plane that
00:34:31.440 --> 00:34:32.440
is with respect to the origin.
00:34:32.440 --> 00:34:36.859
So what I will do is I will 1st write this
as minus 1 by 2 pie i integral over minus
00:34:36.859 --> 00:34:43.639
gamma where minus gamma is the gamma oriented
in the anticlockwise sense and that is the
00:34:43.639 --> 00:34:50.760
positive sense for the plane okay with respect
to the origin if you want okay and then I
00:34:50.760 --> 00:34:59.369
will get this and you know now if I plug in
the series here okay and member that I can
00:34:59.369 --> 00:35:03.900
do integration term by term because of the
same reasons I told you earlier because the
00:35:03.900 --> 00:35:08.010
the Laurent series always converges normally
it converges uniformly and compact sets and
00:35:08.010 --> 00:35:14.369
gamma or minus gamma for that matter they
are as sets their compact sets so if I do
00:35:14.369 --> 00:35:20.049
that again what is going to happen is that,
what is going to survive is only the coefficient
00:35:20.049 --> 00:35:27.849
of 1 by w okay and that is going to give me
that the coefficient when n minus 1 so I am
00:35:27.849 --> 00:35:36.049
going to get a minus 1 the only thing is that
I am going to get I am going to get a
00:35:36.049 --> 00:35:38.069
minus a minus 1 okay.
00:35:38.069 --> 00:35:43.759
So what you must see is that I mean what you
will see is that I will get 1 by 2 pie i times
00:35:43.759 --> 00:35:57.759
2 pie i times a minus 1 this is as before
and I get minus a minus 1 okay so so the moral
00:35:57.759 --> 00:36:06.579
of the story is that when you are looking
at the residue at infinity okay what you do
00:36:06.579 --> 00:36:16.579
is you literally get minus of a minus 1 okay
which is with an extra minus sign added to
00:36:16.579 --> 00:36:24.160
it okay whereas if you look at the usual Laurent
series of a point around a point in the complex
00:36:24.160 --> 00:36:28.369
plane then the residue is actually a minus
1 which is just the coefficient of the 1st
00:36:28.369 --> 00:36:37.680
negative power of the variable okay whereas
in this case it is the minus of that and so
00:36:37.680 --> 00:36:44.369
you know you can now believe that you can
see right from here you can see something
00:36:44.369 --> 00:36:51.769
happening suppose this suppose this function
f had only singularities at the origin and
00:36:51.769 --> 00:36:58.219
at infinity okay which means that this Laurent
series has in finite radius of convergence
00:36:58.219 --> 00:36:59.219
okay.
00:36:59.219 --> 00:37:06.319
So by that I mean the Laurent series is valid
for all w naught 0 and not infinity so it
00:37:06.319 --> 00:37:12.789
is valid on C star C minus the origin okay.
If that is the case then you see this function
00:37:12.789 --> 00:37:18.880
if you calculate the residue at the origin
you are going to get plus a minus 1 okay which
00:37:18.880 --> 00:37:23.739
is usual definition of residue if I if I take
this function suppose it is also analytic
00:37:23.739 --> 00:37:29.579
in the neighbourhood of the origin and suppose
the only singularities are at the origin and
00:37:29.579 --> 00:37:35.239
at infinity okay then you see you notice that
the residue of the function and the origin
00:37:35.239 --> 00:37:40.620
is plus a minus 1 and the residue of the function
at infinity is minus a minus 1, what is the
00:37:40.620 --> 00:37:48.289
sum of residues? It is 0 and that is exactly
what the residue theorem is says for the for
00:37:48.289 --> 00:37:53.329
any function which has only isolated singularities
in the extended planes.
00:37:53.329 --> 00:38:00.490
So you know so the statement is that you take
a function which has only isolated singularities
00:38:00.490 --> 00:38:05.529
in the extended planes okay which means that
know there are only mind you it means whenever
00:38:05.529 --> 00:38:11.710
you say isolated singularities in the extended
plane there are to be only finitely many that
00:38:11.710 --> 00:38:15.940
is because the extended plane is compact and
any subset of a compact isolated set subset
00:38:15.940 --> 00:38:25.380
of a compact set is finite in this case okay.
So therefore you know what the residue theorem
00:38:25.380 --> 00:38:30.799
will say is that if you take the residues
at all the points at the points at the finite
00:38:30.799 --> 00:38:34.000
complex plane and you take the residue at
infinity and you add them up you will get
00:38:34.000 --> 00:38:42.890
0 and here that is exactly what happens if
if f where having a residue if f was having
00:38:42.890 --> 00:38:48.450
a singularity at the origin only and the singularity
at infinity okay then you see that from this
00:38:48.450 --> 00:38:52.700
computation the residue at 0 is plus a minus
1 the residue at infinity is minus a minus
00:38:52.700 --> 00:38:54.269
1 the sum is 0 okay.
00:38:54.269 --> 00:39:00.380
So anyway so the moral of the story is that
you know it gives you a very easy way of computing
00:39:00.380 --> 00:39:05.339
residue at infinity it is very very simple,
what you do is that you simply write the Laurent
00:39:05.339 --> 00:39:15.499
expansion of the function of the origin but)
so that it is valid in the exterior of a sufficiently
00:39:15.499 --> 00:39:22.589
large circle okay I knew you must always remember
this that so this is probably maybe I should
00:39:22.589 --> 00:39:28.650
spend a few minutes on this. See in the 1st
course in complex analysis when you study
00:39:28.650 --> 00:39:33.200
about Laurent expansion at a point you must
remember that there are several Laurent expansion
00:39:33.200 --> 00:39:38.469
there could be several Laurent expansion at
a point that is because the Laurent expansion,
00:39:38.469 --> 00:39:44.009
the domains of the Laurent expansions are
actually annuli which are whose a boundary
00:39:44.009 --> 00:39:45.539
contains the singular points okay.
00:39:45.539 --> 00:39:53.489
So if I say if I talk about the Laurent expansion
of function at a point okay it could be even
00:39:53.489 --> 00:39:58.469
analytic point does not matter what the problem
is that because there are singularities there
00:39:58.469 --> 00:40:03.961
are other singularities the Laurent expansions
will be different expansions course you will
00:40:03.961 --> 00:40:10.309
get different annuli okay and therefore when
you write the Laurent expansion at the origin
00:40:10.309 --> 00:40:14.609
you should not look at the Laurent expansion
at the origin at may be valid in a deleted
00:40:14.609 --> 00:40:20.160
neighbourhood of the origin which at host
boundary there is a finite singularity we
00:40:20.160 --> 00:40:24.960
should not look at that, you should rather
look at a Laurent expansion that is valid
00:40:24.960 --> 00:40:29.979
outside circle of a sufficiently large radius
okay write that Laurent expansion that is
00:40:29.979 --> 00:40:34.310
the Laurent expansion at you need to work
with infinity okay and in that Laurent expansion
00:40:34.310 --> 00:40:42.359
okay at again look at the coefficient of 1
by the variable and take it to the minus sign
00:40:42.359 --> 00:40:49.079
that is residue at infinity okay and address
how you can very easily write out you can
00:40:49.079 --> 00:40:53.240
compute what the residue at infinity is an
that combined with the residue theorem is
00:40:53.240 --> 00:40:56.979
another powerful tool or computing lot of
integrals as we will see in the next lecture.