WEBVTT
Kind: captions
Language: en
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So let us continue we are looking at the key
idea of residue at infinity okay, so so you
00:01:27.770 --> 00:01:34.700
know basically the way we define residue at
infinity is exactly the way that we define
00:01:34.700 --> 00:01:43.080
residue at a point in a complex plane namely
you calculate the contour integral around
00:01:43.080 --> 00:01:52.270
the point going in a positive sense about
that point of the function and divide by 2
00:01:52.270 --> 00:02:01.790
pie i okay, so the only thing is that if the
point is a point at infinity then you see
00:02:01.790 --> 00:02:08.979
the contour you should anyway be contour a
simple closed contour on the plane but saying
00:02:08.979 --> 00:02:16.690
that it goes around infinity in the positive
sense amounts to saying that you are choosing
00:02:16.690 --> 00:02:24.660
this simple closed contour in outside a circle
of sufficiently large radius and the fact
00:02:24.660 --> 00:02:30.010
that it is going around infinity in the positive
sense means that you will have to give it
00:02:30.010 --> 00:02:36.220
the clockwise orientation okay and we saw
that using the stereographic projection that
00:02:36.220 --> 00:02:45.090
this is a this make sense okay and well
hash.
00:02:45.090 --> 00:02:51.450
Now what I want to say is that so there is
a version of the residue theorem that we want
00:02:51.450 --> 00:03:04.201
which will also work for the point at infinity
and so I was explaining last time that you
00:03:04.201 --> 00:03:09.330
see this is this version of residue theorem
will be the statement of that will be the
00:03:09.330 --> 00:03:15.600
same as the statement of the usual version
of the residue theorem but there is there
00:03:15.600 --> 00:03:19.491
are 2 significant differences the 1st thing
is that when you talk about of the residue
00:03:19.491 --> 00:03:27.030
theorem and of course you are only worried
about singular points and you are not worried
00:03:27.030 --> 00:03:28.450
about…
00:03:28.450 --> 00:03:36.800
So the residue theorem says that if you integrate
a function which is analytic except for isolated
00:03:36.800 --> 00:03:46.069
singularities around the simple closed contour
then what you pick up is the integral
00:03:46.069 --> 00:03:51.370
gives you 2 pie i times the sum of the residue
of the function at the at the finitely many
00:03:51.370 --> 00:04:01.310
points where the function has isolated singularities
okay and the the point is that you see you
00:04:01.310 --> 00:04:07.500
are really not worried about points where
the function is analytic or for example isolated
00:04:07.500 --> 00:04:11.640
singularities which are actually removable
because at such a point where if you have
00:04:11.640 --> 00:04:16.600
an isolated singularities a removable singularity
the residue will turn out to be 0 okay.
00:04:16.600 --> 00:04:24.900
So essentially what you will get is you will
in the usual residue theorem what you
00:04:24.900 --> 00:04:29.610
when you refer to some of residue or will
you refer to residues what really matters
00:04:29.610 --> 00:04:37.710
is the residues at the singularities because
residues at the removable singularities will
00:04:37.710 --> 00:04:42.220
always be 0 and that is because of Clausius
theorem because if you integrate an analytic
00:04:42.220 --> 00:04:49.150
function then you are going to get 0 okay,
so I mean of course over a closed contour.
00:04:49.150 --> 00:04:56.229
Now the point is the point is that when you
are doing this also to include the point at
00:04:56.229 --> 00:05:03.289
infinity you have to be careful, the fact
is that infinity behaves very differently
00:05:03.289 --> 00:05:09.080
in the sense that function can have a residue
at infinity even if it is analytic at infinity
00:05:09.080 --> 00:05:14.730
that is the big difference, so when you talk
about residues and also want to include the
00:05:14.730 --> 00:05:20.280
point at infinity you compulsorily include
the point at infinity okay irrespective of
00:05:20.280 --> 00:05:27.220
whether infinity is rarely singularity or
not okay so even if infinity is a removable
00:05:27.220 --> 00:05:32.120
singularity have to include the residue at
infinity is the big point okay and you would
00:05:32.120 --> 00:05:37.070
not have to do that for a point other than
infinity in the complex plane because the
00:05:37.070 --> 00:05:44.150
residue will then turn out to be 0 for a point
with the removable singularity but so the
00:05:44.150 --> 00:05:49.800
easiest simplest illustration of this is the
function 1 by w okay f of w equal to 1 by
00:05:49.800 --> 00:05:56.940
w you know that function is is continuous
at infinity because at infinity in fact it
00:05:56.940 --> 00:06:06.400
is bounded and well you know at infinity function
being bounded or continuous is good enough,
00:06:06.400 --> 00:06:12.040
it is the same as it having a limited infinity
a finite limit.
00:06:12.040 --> 00:06:16.700
A finite means a limit which is a complex
number and you know all these 3 are equivalent
00:06:16.700 --> 00:06:21.400
because of Riemann’s removable singularity
theorem rather the inspiration given by a
00:06:21.400 --> 00:06:26.849
theorem which allowed us to cleverly define
analyticity at infinity will be equal in to
00:06:26.849 --> 00:06:36.469
one of these things okay, so so if you take
one by w that is well-behaved at infinity
00:06:36.469 --> 00:06:43.610
okay in fact it is 0 at infinity okay limit
goes to 0 as w tends to infinity, so it is
00:06:43.610 --> 00:06:51.500
analytic at infinity but if you integrate
1 by w around infinity around simple closed
00:06:51.500 --> 00:06:56.169
contour that goes in the positive sense with
respect to the point at infinity then you
00:06:56.169 --> 00:07:01.250
are going to get minus 1 you are going to
get minus 1 in fact you will get minus 1 into
00:07:01.250 --> 00:07:08.779
2 pie i okay 2 pie i times minus 1 so the
residue is minus 1 okay so so the moral of
00:07:08.779 --> 00:07:14.880
the story is that here is a function 1 by
w all negative powers of w are good functions
00:07:14.880 --> 00:07:17.330
at infinity okay.
00:07:17.330 --> 00:07:23.010
So 1 by w is the simplest is good at infinity
the residue at infinity is not 0 even though
00:07:23.010 --> 00:07:29.000
it is analytic at infinity the residue at
infinity is minus 1 okay, so the big deal
00:07:29.000 --> 00:07:32.870
this is the big deal, the big deal is whenever
you talk about residue for the extended complex
00:07:32.870 --> 00:07:38.160
plane then you must always compulsorily include
the point at infinity even if the point at
00:07:38.160 --> 00:07:44.409
infinity is a point of analyticity that is
very very important, so so let me and of course
00:07:44.409 --> 00:07:50.820
this example 1 by w f of w equal to 1 by w
also illustrates another things, it illustrates
00:07:50.820 --> 00:07:55.770
the following fact that you know this 1 by
w has only 2 singular points, one is at w
00:07:55.770 --> 00:08:00.940
equal to zero that is an isolated singularity
because it is not defined at 0 of course you
00:08:00.940 --> 00:08:05.630
can define it as infinity at 0 if you want
so that you get something continuous on the
00:08:05.630 --> 00:08:11.720
extended complex plane for example what
you do with any transformation but the
00:08:11.720 --> 00:08:16.250
point is that there is also another singularity
namely the singularity at infinity that is
00:08:16.250 --> 00:08:20.000
also an isolated singularity but it is a removable
singularity and the singularity at infinity
00:08:20.000 --> 00:08:21.320
as a residue of minus 1.
00:08:21.320 --> 00:08:27.339
The singularity at the origin has a residue
of plus 1 okay and you know at then you get
00:08:27.339 --> 00:08:32.580
0 and this is actually the version of the
residue theorem for infinity says the total
00:08:32.580 --> 00:08:37.950
sum of residue is 0 okay that is the version
of residue theorem, that is one version of
00:08:37.950 --> 00:08:46.520
the residue theorem at infinity for the for
the extended complex plane okay and it tells
00:08:46.520 --> 00:08:52.110
you to it tells you also another important
thing why is that the Clausius theorem fails
00:08:52.110 --> 00:08:56.160
at infinity, see if you try to integrate the
function 1 by w which is analytic at infinity
00:08:56.160 --> 00:09:03.090
around a simple closed curve goes around the
point at infinity you not going to get 0 you
00:09:03.090 --> 00:09:07.590
are going to get of course you are going to
get minus 1 which is in fact you will get
00:09:07.590 --> 00:09:09.630
minus 2 pie i which is not 0 okay.
00:09:09.630 --> 00:09:14.600
It is 2 pie i times minus 1 which is 2 pie
i times residue at infinity you are not going
00:09:14.600 --> 00:09:18.830
to give 0 and that is the violation of Clausius
theorem because Clausius theorem by Clausius
00:09:18.830 --> 00:09:22.190
theorem we expect if a function is analytic
you expect the integral over closed curve
00:09:22.190 --> 00:09:30.490
to be 0 okay but the reason is it is not violating
Clausius theorem nothing, what is actually
00:09:30.490 --> 00:09:38.209
happening is that is nonzero integral at infinity,
the value at infinity is to compensate for
00:09:38.209 --> 00:09:42.990
the residue at 0 so that the total sum of
residues is 0 okay, so when you integrate
00:09:42.990 --> 00:09:50.930
1 by w at infinity around infinity you get
minus 2 pie i and you do not get 0 even though
00:09:50.930 --> 00:09:56.550
it is analytic at infinity but then this it
has to compensate for the plus 2 pie i that
00:09:56.550 --> 00:10:02.010
you will get if you integrate it around 0
okay and the sum so that the sum minus 2 pie
00:10:02.010 --> 00:10:05.810
i plus 2 pie i is 0 and that is the version
of the so that is the version of residue theorem.
00:10:05.810 --> 00:10:09.010
So what you must understand is that the fact
that the residue theorem works is actually
00:10:09.010 --> 00:10:14.300
equivalent to the fact that Clausius theorem
does not work this case okay and now you see
00:10:14.300 --> 00:10:18.570
that the fact that Clausius theorem does not
work is not really a sad thing because you
00:10:18.570 --> 00:10:22.190
are getting something in exchange for that
you are getting nice version of residue theorem
00:10:22.190 --> 00:10:23.190
okay.
00:10:23.190 --> 00:10:30.900
So so what I will do now is let us prove
these versions of residue theorems so let
00:10:30.900 --> 00:10:50.140
me start here suppose that suppose f of
w is defined in a deleted neighbourhood of
00:10:50.140 --> 00:10:57.910
infinity okay which means it is defined or
all w in the exterior of a sufficiently large
00:10:57.910 --> 00:11:10.530
circle of sufficiently large radius.
Now what you can do is that so if you well
00:11:10.530 --> 00:11:19.460
so the residue of f of w at infinity if you
calculate this quantity is by definition
00:11:19.460 --> 00:11:26.510
equal to you integrate over you for example
integrate you can integrate over any you
00:11:26.510 --> 00:11:33.210
can integrate over any simple closed curve
going around infinity in the positive sense.
00:11:33.210 --> 00:11:45.290
So let us take a circle and you take R sufficiently
large you take R sufficiently large so that
00:11:45.290 --> 00:11:51.920
you know there are no other singularities
outside the circle of radius R except the
00:11:51.920 --> 00:12:00.060
point at infinity okay, so that is how large
R should be chosen and of course the point
00:12:00.060 --> 00:12:08.360
is that you you give negative orientation,
so the reason you give his negative orientation
00:12:08.360 --> 00:12:16.690
this is negative orientation with respect
to the usual conventions okay and mind you
00:12:16.690 --> 00:12:35.070
this is negative orientation, so let me write
with respect to 0 is same as positive orientation
00:12:35.070 --> 00:12:43.690
with respect to infinity okay, so actually
I am taking this integral over mod w equal
00:12:43.690 --> 00:12:50.920
to R around the I mean with the clockwise
orientation okay, so I take this integral
00:12:50.920 --> 00:12:52.800
and I integrate what?
00:12:52.800 --> 00:13:01.570
I of course integrate the function fw dw and
well I divided by 2 pie i whatever I get and
00:13:01.570 --> 00:13:05.690
this is the residue at infinity alright and
the point you have to remember is that this
00:13:05.690 --> 00:13:13.780
integral is anyway defined because I am taking
am actually integrating the function over
00:13:13.780 --> 00:13:20.300
the circle and on the circle it is analytic
in fact, in fact it is analytic in a neighbourhood
00:13:20.300 --> 00:13:26.220
of the circle okay. The only problem is the
point inside the circle which is the point
00:13:26.220 --> 00:13:30.840
at infinity mind you the point at infinity
is inside the circle you heard me right that
00:13:30.840 --> 00:13:36.800
is because the interior of the circle will
be given the clockwise orientation will actually
00:13:36.800 --> 00:13:40.000
be the exterior of the circle in the usual
sense okay.
00:13:40.000 --> 00:13:49.160
So so this is the residue at infinity and
now the point is that say suppose you assume
00:13:49.160 --> 00:13:56.100
that f has of course you know I have taken
R sufficiently large so that you know f has
00:13:56.100 --> 00:14:01.920
no singularity is in mod Z greater than mod
w greater than R greater than or equal to
00:14:01.920 --> 00:14:12.880
R and you only singularities at infinity,
so let me write that down, so that So R sufficiently
00:14:12.880 --> 00:14:28.580
large chosen so that fw has no singularity
is in mod w greater than or equal to if you
00:14:28.580 --> 00:14:40.870
want R, okay there are no singularity is,
right except except so I should say except
00:14:40.870 --> 00:14:46.400
w equal to infinity because now I am also
whenever I say mod w greater than or equal
00:14:46.400 --> 00:14:51.160
to R I am actually thinking of the extended
complex plane so the point at infinity is
00:14:51.160 --> 00:14:54.760
also there it is an interior point mind you
the point at infinity is an interior point
00:14:54.760 --> 00:15:02.650
for this for this reason mod w greater than
or equal to R, okay and interestingly mind
00:15:02.650 --> 00:15:09.660
you if you look at this region on the complex
plane it is unbounded okay it is unbounded
00:15:09.660 --> 00:15:14.560
and closed okay but if you look at the same
region mod w greater than or equal to R in
00:15:14.560 --> 00:15:19.190
the extended complex plane it is compact it
is bounded, okay.
00:15:19.190 --> 00:15:23.560
So because you have added that one point at
infinity to compactify it so it becomes bounded
00:15:23.560 --> 00:15:27.360
so this is so you know mod w greater than
or equal to R whether you are looking at it
00:15:27.360 --> 00:15:30.120
in the complex plane or whether you are looking
at it in the extended complex plane makes
00:15:30.120 --> 00:15:35.360
a lot of difference topologically. In the
complex plane it is closed unbounded, in the
00:15:35.360 --> 00:15:43.851
extended complex plane it is closed and bounded
it is compact okay. Anyway so fine now you
00:15:43.851 --> 00:15:49.940
see suppose you assume that function f has
only isolated singularities on the whole plane
00:15:49.940 --> 00:15:56.600
wherever it has singularities suppose it has
only isolated singularities okay then what
00:15:56.600 --> 00:16:05.060
happens is that you know you see all those
singularities in the plane which means that
00:16:05.060 --> 00:16:10.060
I have left out the singularity at infinity
all those singularities in the plane are going
00:16:10.060 --> 00:16:18.520
to lie inside this circle mod w equal to R
even the usual positive anticlockwise orientation
00:16:18.520 --> 00:16:19.650
okay.
00:16:19.650 --> 00:16:25.470
So which is a positive orientation about the
origin a and then the usual residue theorem
00:16:25.470 --> 00:16:31.100
applies, the usual residue theorem will tell
you that this this integral will give you
00:16:31.100 --> 00:16:42.560
minus of 1 by 2 pie i the integral over the
same circle given there positive orientation
00:16:42.560 --> 00:16:47.150
okay and that will be minus 2 pie i times
sum of the residues of the function at the
00:16:47.150 --> 00:16:52.710
isolated singular points inside the circle
in the usual sense okay and then if you put
00:16:52.710 --> 00:16:57.570
these 2 together get the that the total sum
of residues and the extended plane is 0 which
00:16:57.570 --> 00:17:00.790
is the 1st version of residue theorem okay,
so let me write that down, so this is also
00:17:00.790 --> 00:17:16.169
equal to minus 1 by 2 pie i times 2 pie i
times summation of the residues of f at
00:17:16.169 --> 00:17:30.799
w i, i equal to 1 to... or rather let me use
not i let me use j wj, j equal to 1 to
00:17:30.799 --> 00:17:31.799
m.
00:17:31.799 --> 00:17:57.890
So this is what I will get and and here so
so let me say let me tell you that what I
00:17:57.890 --> 00:18:05.020
have done here is that I have simply put a
minus sign because I am evaluating around
00:18:05.020 --> 00:18:10.990
the circle in the anticlockwise sense okay
and then I am using the residue theorem okay
00:18:10.990 --> 00:18:25.400
so here is so here is by the usual
so of course you know the big deal here is
00:18:25.400 --> 00:18:38.230
that I am making an assumption I am making
this assumption that suppose f has only isolated
00:18:38.230 --> 00:18:51.049
singularities
suppose f has only isolated singularities
00:18:51.049 --> 00:19:04.500
in the complex plane okay and so let me also
say suppose it has isolated singularity in
00:19:04.500 --> 00:19:12.490
the extended plane because I want them to
be only finitely many and therefore there
00:19:12.490 --> 00:19:19.260
are only finitely many singularities mind
you an isolated set in the extended plane
00:19:19.260 --> 00:19:23.559
has to be finite because the extended plane
is compact alright.
00:19:23.559 --> 00:19:31.799
So so suppose f has only isolated singularities
the extended plane and then there are only
00:19:31.799 --> 00:19:37.570
finitely many isolated singularities for f
and infinity may or may not be a point of
00:19:37.570 --> 00:19:43.910
singularity okay but in any case if you early
without then you will get only finitely many
00:19:43.910 --> 00:19:49.710
isolated singularities in the plane and am
calling those singularities as w 1 through
00:19:49.710 --> 00:20:11.669
w m okay, so so let me write that here w 1…W
m are the singularities of f in the complex
00:20:11.669 --> 00:20:22.419
plane okay, so you know the basically what
I am saying is that if w 1 through w m are
00:20:22.419 --> 00:20:26.869
the only singular point for a function in
the complex plane okay.
00:20:26.869 --> 00:20:32.460
Then you then you can enclose them by a sufficiently
large circle okay centred at the origin and
00:20:32.460 --> 00:20:36.129
then you integrate around that circle what
you are going to get is 2 pie i times the
00:20:36.129 --> 00:20:41.789
sum of the residues of the function at those
points that is all I have used here okay and
00:20:41.789 --> 00:20:48.940
now if you put all these things together
if you look at the left side at the extreme
00:20:48.940 --> 00:20:53.669
left and you look at the extreme right and
you put them together, what you will get is
00:20:53.669 --> 00:20:58.799
that for a function which has only isolated
singularities on the extended plane sum of
00:20:58.799 --> 00:21:03.799
residue is the same and that is the that is
one version of the residue theorem for the
00:21:03.799 --> 00:21:08.950
extended plane, so let me write that down
so okay.
00:21:08.950 --> 00:21:36.110
Thus, for a function f w is analytic on the
extended plane with only isolated singularities
00:21:36.110 --> 00:22:04.369
we have residue theorem
for the extended plane summation of residues
00:22:04.369 --> 00:22:23.090
of f at the singularities in the plane plus
the residue of f at infinity is equal to 0,
00:22:23.090 --> 00:22:31.039
the total sum of residue is 0 okay, this is
one version of residue theorem at infinity,
00:22:31.039 --> 00:22:38.179
right. Now and what about the so this looks
slightly different from the usual version
00:22:38.179 --> 00:22:42.509
of the residue theorem, the usual version
of the residue theorem says that you integrate
00:22:42.509 --> 00:22:47.999
around a curve function then you are supposed
to get 2 pie i times some of the residues
00:22:47.999 --> 00:22:54.190
of the function at the single points inside
the curve okay in the interior of the curve,
00:22:54.190 --> 00:23:00.440
does this also work for the extended complex
plane? It does, okay so that is actually equivalent
00:23:00.440 --> 00:23:01.570
to as right.
00:23:01.570 --> 00:23:15.289
So let me write that down so here is another
version another version
00:23:15.289 --> 00:23:36.850
of the residue theorem for the extended plane,
let gamma be
00:23:36.850 --> 00:23:46.309
a simple closed contour in the extended plane
by which we mean actually a simple closed
00:23:46.309 --> 00:23:50.960
contour in the complex plane, okay. Whenever
we talk about simple closed contour or contour
00:23:50.960 --> 00:23:55.799
of integration you never think of a curve
going to passing through the point at infinity
00:23:55.799 --> 00:24:00.470
because it really is something that you cannot
see on the plane okay maybe you can think
00:24:00.470 --> 00:24:05.629
of that on the Riemann’s sphere and work
with that but the problem is to do that you
00:24:05.629 --> 00:24:08.870
will have 2 go to the language of Riemann
surfaces you have to convert the Riemann’s
00:24:08.870 --> 00:24:12.600
sphere into a…you must think of it as a
Riemann surface and do integration.
00:24:12.600 --> 00:24:20.090
You can actually do integration on the Riemann
sphere along a for example a circle on the
00:24:20.090 --> 00:24:24.220
sphere which passes through the North pole
you can really do that okay but to do that
00:24:24.220 --> 00:24:28.759
you will have to really use a language of
Riemann surfaces okay and for more details
00:24:28.759 --> 00:24:33.519
about that you can look at my video course
on Riemann surfaces the same NPTEL series
00:24:33.519 --> 00:24:40.139
but we are not going to do that, so I am just
going to look at the simple closed contour
00:24:40.139 --> 00:24:43.869
only in the plane okay and that is what it
will mean also for a simple closed contour
00:24:43.869 --> 00:25:05.149
in the extended plane okay in C and f analytic
in the extended plane except for isolated
00:25:05.149 --> 00:25:29.759
singularities. Then integral over gamma fw
dw is equal to 2 pie i times sum of the residues
00:25:29.759 --> 00:25:57.990
of f at the singularities inside gamma including
necessarily the residue at infinity if it
00:25:57.990 --> 00:26:03.279
lies inside, okay.
00:26:03.279 --> 00:26:15.600
So this is the so this is the extra thing
okay, so it is again the residue theorem but
00:26:15.600 --> 00:26:22.309
it works also for the extended plane the only
thing is you the integral of the function
00:26:22.309 --> 00:26:28.490
is 2 pie i use sum of the residues only thing
is you have to be careful if infinity is inside
00:26:28.490 --> 00:26:36.519
gamma okay then you have to include also the
residue at infinity irrespective of whether
00:26:36.519 --> 00:26:41.230
infinity is a point of analyticity of f or
not okay that is the big deal. The big deal
00:26:41.230 --> 00:26:46.720
is you have to include infinity absolutely
necessary you cannot omit it okay if it is
00:26:46.720 --> 00:26:52.039
a usual wind on the complex plane and if it
is a removable singularity at that point you
00:26:52.039 --> 00:26:55.850
need not included because the residue at that
point would be 0 because the function is analytic
00:26:55.850 --> 00:27:00.320
at a point where the function is analytic,
the residue is 0 okay but it is not this is
00:27:00.320 --> 00:27:06.519
not true for the point at infinity as we have
seen okay so you have to necessarily include
00:27:06.519 --> 00:27:09.289
the residue at infinity okay.
00:27:09.289 --> 00:27:18.659
So so this is the in this sense you know the
residue theorem works alright and of course
00:27:18.659 --> 00:27:24.830
it is very very important that gamma does
not pass to any of these singularities the
00:27:24.830 --> 00:27:31.030
singularities in the finite plane that is
of course always assumed okay. So so let me
00:27:31.030 --> 00:27:40.169
write that here probably so you know always
you have to keep remembering gamma of course
00:27:40.169 --> 00:27:56.529
should not pass through. Any singularity of
f this is of course this is always there okay
00:27:56.529 --> 00:28:00.950
I mean if if the curve which you are trying
to integrate passes through a singular point
00:28:00.950 --> 00:28:08.639
of the integrant then you are in trouble because
of course you know if that singular point
00:28:08.639 --> 00:28:14.590
is a removable singularity is not you really
do not have to worry about it okay and of
00:28:14.590 --> 00:28:20.509
course when we write things like this we are
really not worried about points on the plane
00:28:20.509 --> 00:28:25.679
which are removable singularity is okay, we
we simply assume that they are points where
00:28:25.679 --> 00:28:27.990
the function is analytic okay.
00:28:27.990 --> 00:28:32.570
So what really matters the points in the plane
at really matter for this statement are the
00:28:32.570 --> 00:28:37.080
points which are singularities which are either
poles or essential singularity is it is not
00:28:37.080 --> 00:28:40.400
a removable singularities okay so and I am
saying gamma should not passed through any
00:28:40.400 --> 00:28:44.619
singularity of f of course it should not passed
through a of f it should not passed through
00:28:44.619 --> 00:28:50.169
a pole or a essential singularity of f and
the reason is because at each of these singularities
00:28:50.169 --> 00:28:58.460
the function is not a continuous okay function
is not continuous at those points and you
00:28:58.460 --> 00:29:02.909
know when a function is not continuous at
a point it is in general you do not try to
00:29:02.909 --> 00:29:12.080
define its integral at that point unless you
know for example is bounded okay you know
00:29:12.080 --> 00:29:17.919
the complex contour integration is also defined
by integrating by taking part integrals of
00:29:17.919 --> 00:29:21.980
the real and imaginary parts which are real
valued functions okay and so basically it
00:29:21.980 --> 00:29:26.749
reduces to real integration and you can use
Riemann integration if you want.
00:29:26.749 --> 00:29:29.549
You can use Riemann integral at the point
is that and of course you know the Riemann
00:29:29.549 --> 00:29:34.690
integral is a limit of Riemann sums okay and
if the function become unbounded at a point
00:29:34.690 --> 00:29:41.869
then at that point you really cannot expect
Riemann sum to converge properly okay in a
00:29:41.869 --> 00:29:47.899
neighbourhood of that point, so you would
never try to in general define the integral
00:29:47.899 --> 00:29:53.489
at a point of discontinuity especially when
the discontinuity is of is not of a jump type
00:29:53.489 --> 00:29:58.239
if it is a very bad discontinuity not removable
kind of discontinuity and you do not try to
00:29:58.239 --> 00:30:04.259
integrate the function at that point, okay.
So the moral of the story is that this is
00:30:04.259 --> 00:30:08.090
usual version and how does one prove it? The
usual version of the residue theorem just
00:30:08.090 --> 00:30:12.309
follows from the other version of residue
theorem the extended plane which says that
00:30:12.309 --> 00:30:15.350
the total sum of residues including the residue
at infinity is 0.
00:30:15.350 --> 00:30:28.179
So I will I will just write it down it is
pretty easy, so let w 1 et cetera w k be the
00:30:28.179 --> 00:30:45.600
isolated singularities of f inside, so you
know okay so let me say something here, so
00:30:45.600 --> 00:30:52.649
let me go back to the statement and tell you
that you know if you are really looking at
00:30:52.649 --> 00:31:01.259
gamma being you know positively oriented in
the usual sense okay if you are looking at
00:31:01.259 --> 00:31:07.450
gamma a simple closed contour on the plane
which is positively oriented going a that
00:31:07.450 --> 00:31:10.909
is going in the anticlockwise direction then
you know the interior of the gamma is going
00:31:10.909 --> 00:31:18.359
to be a bounded domain in the plane and and
you know you are only worried about the integral
00:31:18.359 --> 00:31:24.279
there and that integral will give you 2 pie
i times sum of the residues of f inside that
00:31:24.279 --> 00:31:29.210
a bounded domain then there will be only finitely
okay and that is usual residue theorem okay
00:31:29.210 --> 00:31:33.489
and you do not include of course you do not
include the residue at infinity because infinity
00:31:33.489 --> 00:31:38.559
is not there okay because we have taken gamma
to be clockwise I mean anticlockwise okay.
00:31:38.559 --> 00:31:45.039
So actually there is nothing to prove in this
statement unless you are looking at gamma
00:31:45.039 --> 00:31:50.970
which is going clockwise so that the interior
is actually unbounded in the usual plane but
00:31:50.970 --> 00:31:58.609
of course bounded in the extended plane with
infinity as an interior point okay, so really
00:31:58.609 --> 00:32:05.940
the version of this theorem that you have
to prove is only for the case when gamma is
00:32:05.940 --> 00:32:09.559
having a clockwise orientation okay and that
is the version that is the part that I will
00:32:09.559 --> 00:32:13.600
prove. If it is anticlockwise if you taking
gamma with anticlockwise orientation then
00:32:13.600 --> 00:32:16.809
it is usual residue theorem there is nothing
to prove okay, so that is the reduction I
00:32:16.809 --> 00:32:31.739
am making obvious reduction, we only have
to look at the case when gamma has clockwise
00:32:31.739 --> 00:32:44.279
orientation because this is the only case
when infinity is inside gamma okay.
00:32:44.279 --> 00:32:57.129
So so let me write this here infinity is inside
gamma so well let w 1 w k through w k be the
00:32:57.129 --> 00:33:04.320
isolated singularities of f inside minus gamma
okay so you see gamma is taken in the clock
00:33:04.320 --> 00:33:08.350
wise sense, so minus gamma is taken in the
anticlockwise sense the usual positive sense
00:33:08.350 --> 00:33:17.450
and will contain only a piece it will only
contain a portion of the usual complex plane
00:33:17.450 --> 00:33:20.980
and it will have some singularities there,
so the picture is something like this, so
00:33:20.980 --> 00:33:30.889
here is your so here is your gamma and mind
you the orientation is clockwise okay and
00:33:30.889 --> 00:33:39.029
the reason for that is that the exterior of
gamma well the interior of gamma really the
00:33:39.029 --> 00:33:44.489
interior of gamma is actually the exterior
of gamma in the usual in the common sense.
00:33:44.489 --> 00:33:52.659
So this is the interior of gamma what I have
shaded is because it has clockwise orientation
00:33:52.659 --> 00:34:01.299
and it contains infinity in the extended plane
if you are considering this in the extended…mind
00:34:01.299 --> 00:34:10.340
you if whatever I have shaded along with the
boundary gamma if you consider it in the extended
00:34:10.340 --> 00:34:16.750
plane that is a compact set okay it is closed
end bounded so because you are actually looking
00:34:16.750 --> 00:34:22.899
at its image on the Riemann’s sphere okay
which will be like a polarized cap alright
00:34:22.899 --> 00:34:27.080
so you should remember that because sometimes
it is very difficult for people to think that
00:34:27.080 --> 00:34:33.250
this is bounded because on the plane it is
unbounded but it is but topologically you
00:34:33.250 --> 00:34:38.230
should think of it as bounded okay so when
you are thinking of Riemann sphere of the
00:34:38.230 --> 00:34:39.230
extended plane.
00:34:39.230 --> 00:34:50.870
So well so here I have w 1… w k these are
the fellows and mind you these are inside
00:34:50.870 --> 00:34:56.609
minus gamma okay minus gamma is well then
have the usual anticlockwise sense orientation
00:34:56.609 --> 00:35:06.250
okay, positive orientation with respect to
the usual plane and if you calculate the integral
00:35:06.250 --> 00:35:11.660
over gamma of f w if you calculate the integral
over gamma of f w this is going to be the
00:35:11.660 --> 00:35:16.340
same as minus of the integral over minus gamma
of f w by the very definition of the Riemann
00:35:16.340 --> 00:35:21.450
integral okay you change the orientation of
the path the sign of the integral changes
00:35:21.450 --> 00:35:28.230
but if you calculate the integral of this
over minus gamma you can apply usual residue
00:35:28.230 --> 00:35:35.290
Irom and you get that b equal to 2 pie i times
sum of the residues of f at these w i or omega
00:35:35.290 --> 00:35:40.270
i from i equal to 1 to k okay that is what
you will get okay.
00:35:40.270 --> 00:35:47.190
So on the one hand you will get a minus 2
pie i times sum of the residues and w 1 through
00:35:47.190 --> 00:35:53.040
w k alright and by the residue theorem the
other version of the residue theorem which
00:35:53.040 --> 00:35:58.100
we saw or the extended plane which says that
the total sum of residues is 0 namely sum
00:35:58.100 --> 00:36:02.890
of the residues at all finite point is the
plane does the residue at infinity that sum
00:36:02.890 --> 00:36:12.990
as 0. In this if you look at the if you look
at that you will get that this integral is
00:36:12.990 --> 00:36:20.220
also equal to 2 pie i times are some of the
residues of f outside gamma okay in the shaded
00:36:20.220 --> 00:36:25.230
region which is the diversion of residue theorem
at we want, so that gives you the proof, so
00:36:25.230 --> 00:36:32.050
let me write that here so let me just use
a different color, so let me write it here
00:36:32.050 --> 00:36:39.020
in the margin maybe I can remove some
of this and make myself a little bit more
00:36:39.020 --> 00:36:40.890
space.
00:36:40.890 --> 00:36:54.120
So integral over gamma fw dw is minus of integral
over minus gamma fw dw and this is minus of
00:36:54.120 --> 00:37:06.360
2 pie i times sum of the residues of f at
w j, j equal to 1 to k this is usual residue
00:37:06.360 --> 00:37:14.790
theorem working alright and let us assume
that of course you know I am looking at a
00:37:14.790 --> 00:37:18.340
function which has only isolated singularities
in the extended planes so there are some more
00:37:18.340 --> 00:37:23.880
singularities and they are of course lying
outside gamma. I of course as a told you I
00:37:23.880 --> 00:37:28.240
avoid the situation when gamma passes through
one of the singularities that is not allowed
00:37:28.240 --> 00:37:46.680
so there are these remaining w k plus 1, so
w k plus 1 w k plus 2 and so on and there
00:37:46.680 --> 00:37:56.260
is a w let us say w m there are m finite points
in the complex plane where f has isolated
00:37:56.260 --> 00:37:58.520
singularities and then there is a point at
infinity.
00:37:58.520 --> 00:38:08.200
So this is also equal to minus 2 pie i times
mind you now I will get you see minus of summation
00:38:08.200 --> 00:38:21.020
j equal to k plus 1 to m the residue of f
at w j plus well I will also get residue of
00:38:21.020 --> 00:38:34.930
f at infinity this is what I will get okay
I will get this and of course this minus sign
00:38:34.930 --> 00:38:44.570
is common to both okay and the reason why
I get this is because of the the earlier version
00:38:44.570 --> 00:38:48.960
of the residue theorem which says a total
sum of residues is 0 okay, so I get this now
00:38:48.960 --> 00:38:53.270
you see this minus inside the minus outside
they cancel out and what you get is 2 pie
00:38:53.270 --> 00:39:00.810
i times sum of the residues of f inside, and
mind you what are the residues of f inside
00:39:00.810 --> 00:39:06.230
gamma what are the points inside gamma? The
points inside gamma are w k plus 1, w k plus
00:39:06.230 --> 00:39:11.150
2 through w m and the point at infinity which
you have to necessarily include as a singular
00:39:11.150 --> 00:39:13.340
point okay.
00:39:13.340 --> 00:39:19.780
So it is correct so this is the proof of the
statement I gave you, so the residue theorem
00:39:19.780 --> 00:39:28.660
works finely well even for any curve, any
simple closed curve in the complex plane so
00:39:28.660 --> 00:39:32.370
long as you only thing is that the function
should be should have only finite singularity,
00:39:32.370 --> 00:39:37.300
finitely many it should have only isolated
singularities in the extended plane and your
00:39:37.300 --> 00:39:44.230
contour should be simple closed the function
should not vanish it should not have any
00:39:44.230 --> 00:39:48.460
singularities at any point on the contour
okay that is all you need okay so you can
00:39:48.460 --> 00:39:55.320
see there is a…so behaviour at infinity
you can see is quite nice I mean you get also
00:39:55.320 --> 00:39:58.050
nice version of the residue theorem okay.
00:39:58.050 --> 00:40:10.460
Now there is one more thing I want to tell
you so so let me reiterate one of the
00:40:10.460 --> 00:40:20.030
reasons well just in case this confuses, so
one of the reasons the Clausius theorem fails
00:40:20.030 --> 00:40:25.550
for a function which is analytic at infinity
because of this residue theorem okay because
00:40:25.550 --> 00:40:39.670
you get this in exchange which is good and
and of course the the advantage of this theorem
00:40:39.670 --> 00:40:46.370
is that you can do some difficult computations
okay. Some otherwise difficult computation
00:40:46.370 --> 00:40:51.580
be done using this okay is like the usual
residue theorem allows you to include things
00:40:51.580 --> 00:41:01.480
compute integrals is extended version
of the residue theorem will help you.
00:41:01.480 --> 00:41:06.040
So for example I will give you I will give
you philosophically an example suppose you
00:41:06.040 --> 00:41:11.670
have an analytic function okay which has isolated
singularities and suppose there are lot of
00:41:11.670 --> 00:41:21.540
them okay but anyway I am only looking at
chance which have only you know isolated singularities
00:41:21.540 --> 00:41:25.670
in the extended planes so there are only finitely
many at the point is that this finite number
00:41:25.670 --> 00:41:30.640
may be huge suppose I am looking at an analytic
function which has 1 million singularities
00:41:30.640 --> 00:41:34.910
okay 1 million isolated singularities and
suppose there are 1 million poles at different
00:41:34.910 --> 00:41:36.200
points alright.
00:41:36.200 --> 00:41:40.870
Now I can of course choose a huge circle which
encloses all of them because they are anyway
00:41:40.870 --> 00:41:45.710
they are finite so there is going to be a
sufficiently large circle where the which
00:41:45.710 --> 00:41:49.470
can enclosed one of them and what am I going
to get if I integrate the function around
00:41:49.470 --> 00:41:56.180
that circle? Well the usual residue theorem
will tell me that you can get the answer by
00:41:56.180 --> 00:42:02.900
you know taking a pie i times sum of residues
at each of these million poles okay so that
00:42:02.900 --> 00:42:09.000
there are millions of them you have to compute
1 million residues and that is practically
00:42:09.000 --> 00:42:12.830
you can imagine how difficult this but then
the extended version of the residue theorem
00:42:12.830 --> 00:42:17.310
says do not do all that, simply compute the
residue at infinity and put minus sign that
00:42:17.310 --> 00:42:22.480
is it okay and multiply by 2 pie i okay, so
in that way residue at infinity is very useful
00:42:22.480 --> 00:42:29.750
okay, so it is a very useful theorem okay
it allows you to compute residues when there
00:42:29.750 --> 00:42:37.490
are huge number of singularities okay that
is the advantage of this, so whenever we do
00:42:37.490 --> 00:42:41.590
something we should see some advantage in
that, so in that sense that is the advantage
00:42:41.590 --> 00:42:42.590
of this okay.
00:42:42.590 --> 00:42:47.840
So so for example you know well I will give
you couple of illustrations see suppose I
00:42:47.840 --> 00:42:59.090
write integrals model Z is equal to R, R sufficiently
large and if I ride P of Z dz by Q of
00:42:59.090 --> 00:43:13.580
Z where P and Q are polynomial
okay so suppose I do this see you would have
00:43:13.580 --> 00:43:20.660
seen in the 1st course in complex analysis
that you know if the degree of P is is
00:43:20.660 --> 00:43:26.931
less than degree Q minus less than or equal
to degree Q minus 2 okay that is the degree
00:43:26.931 --> 00:43:36.420
of the numerator is lesser than the degree
of the denominator by least 2 powers of the
00:43:36.420 --> 00:43:45.360
variable okay and this integral is actually
0 okay so actually this is equal to 0 if degree
00:43:45.360 --> 00:43:53.980
of P is less than or equal to degree of Q
minus 2 okay.
00:43:53.980 --> 00:44:00.720
Now you see so it is the way you do this in
the 1st course in complex analysis there are
00:44:00.720 --> 00:44:06.080
2 ways of doing it one way is well I mean
the easiest way which is what people normally
00:44:06.080 --> 00:44:14.510
do is use the ML formula, we use the ML inequality
which says that the integral of the modulus
00:44:14.510 --> 00:44:17.430
of an integral is less than or equal to the
integral of the modulus and that is lesser
00:44:17.430 --> 00:44:25.940
or equal to the maximum value of the integrant
M on the contour times L which is the length
00:44:25.940 --> 00:44:32.680
of the contour okay this is the ML inequality
and what you do is that you do this ML inequality
00:44:32.680 --> 00:44:40.980
estimation okay and you know if it will it
will give you immediately that this integral
00:44:40.980 --> 00:44:46.940
will go to 0 as you increase R if you make
R of course R has to be large enough so that
00:44:46.940 --> 00:44:54.840
you do not allow any zeros of Q outside R
okay they should all come inside alright.
00:44:54.840 --> 00:45:01.630
So so you make R big enough so that R includes
all the zeros of Q and mind you zeros of Q
00:45:01.630 --> 00:45:10.380
are poles of the integrant okay and you make
R sufficiently large to include all the poles
00:45:10.380 --> 00:45:15.960
of the integrant then you get a quantity which
will go to 0 as R tends to infinity okay and
00:45:15.960 --> 00:45:20.880
sense in fact this quantity is independent
of R because of Clausius theorem, it is independent
00:45:20.880 --> 00:45:24.590
of R because of Clausius theorem you can let
R tent to infinity and you can that this integral
00:45:24.590 --> 00:45:35.350
is 0 this is what people normally do okay
and well now you know try to do this using
00:45:35.350 --> 00:45:38.140
the residue at infinity.
00:45:38.140 --> 00:45:43.790
So you know integral over R integral over
mod Z equal to R, R sufficiently large is
00:45:43.790 --> 00:45:53.750
going to give you 2 pie i times minus the
residue at infinity okay and you try to calculate
00:45:53.750 --> 00:46:01.470
the residue at infinity for this meaning that
you you write this out in positive and negative
00:46:01.470 --> 00:46:06.741
powers of the variable Z if you want in this
case and you look at the coefficient of 1
00:46:06.741 --> 00:46:13.950
by Z and you will see that since the numerator
degree is less than the at least 2 less than
00:46:13.950 --> 00:46:19.340
the denominator degree you will never get
a 1 by Z term and what therefore it will tell
00:46:19.340 --> 00:46:26.790
you is that, that infinity okay at infinity
you are not going to get 1 by Z term okay
00:46:26.790 --> 00:46:33.760
Z being the variable and therefore the residue
is 0 and therefore the answer 0 okay so you
00:46:33.760 --> 00:46:39.420
can see this is 0 is like that okay, you can
see this is 0 just like that and much harder
00:46:39.420 --> 00:46:47.780
thing is supposed degree of P is actually
degree of Q minus 1 or if it is equal to degree
00:46:47.780 --> 00:46:50.440
of Q, how do you make these computations?
00:46:50.440 --> 00:46:54.970
Okay your computations will be you will see
that the computations are very easy if you
00:46:54.970 --> 00:46:58.930
really use residue at infinity, so residue
at infinity is very useful to do these kind
00:46:58.930 --> 00:47:09.420
of calculations okay, so so that is something
that you should understand, so you know for
00:47:09.420 --> 00:47:18.830
example if I write integral over mod Z equal
to R and I write d Z by Z to the 2014 plus
00:47:18.830 --> 00:47:26.790
1 okay so this is a huge polynomial in the
denominator all its zeros are simple zeros
00:47:26.790 --> 00:47:34.081
they are the 2014 roots of unit of minus 1
anyway they all lie on the unit circles, so
00:47:34.081 --> 00:47:38.530
I do not have to take R very large I just
have to take R greater than 1 alright but
00:47:38.530 --> 00:47:42.750
the point is that if you now use the usual
residue theorem and try to compute it, it
00:47:42.750 --> 00:47:48.890
is not all that easy okay you have to calculate
the residue of this at each of those simple
00:47:48.890 --> 00:47:53.741
poles and there are 2014 of them and you will
have to add all of them and then multiply
00:47:53.741 --> 00:48:00.780
by 2 pie i you would certainly not do that
okay rather what you do is take minus 2 pie
00:48:00.780 --> 00:48:07.800
i times residue at infinity and you see that,
that is 0 so you can easily see that this
00:48:07.800 --> 00:48:13.030
this is actually going to be 0 of course if
you apply the previous criterion it is 0 but
00:48:13.030 --> 00:48:15.980
even you do not have to do that okay.
00:48:15.980 --> 00:48:21.990
So let me illustrate what you would for example
do with this case okay, so here the function
00:48:21.990 --> 00:48:31.210
is f of Z I have taken the variable as Z,
so it is 1 by Z to the 2014 plus 1 okay and
00:48:31.210 --> 00:48:37.070
mind you I want to look at it at infinity
okay which means I want the Laurent expansion
00:48:37.070 --> 00:48:44.840
at infinity and mind you that should be thought
of as a Laurent expansion that is valid outside
00:48:44.840 --> 00:48:50.830
sufficiently large a circle of sufficiently
large radius. Mind you this function has all
00:48:50.830 --> 00:48:56.890
simple poles which are zeros of the denominator
and they all lie on the unit circle okay they
00:48:56.890 --> 00:49:01.990
all lie on the unit circle and therefore you
know if you calculate the Laurent expansion
00:49:01.990 --> 00:49:07.190
about the origin you will get 2 Laurent expansions,
there is one Laurent expansion that to be
00:49:07.190 --> 00:49:16.040
valid in the unit disk okay and it will actually
turn out to be a Taylor expansion, the reason
00:49:16.040 --> 00:49:20.940
is because the function is actually analytic
in the unit disk, in the unit disk there are
00:49:20.940 --> 00:49:26.300
no the denominator does not vanish right.
00:49:26.300 --> 00:49:32.110
So if you write the 1st Laurent expansion
centred at the at Z equal to 0 what you will
00:49:32.110 --> 00:49:36.401
get is it will be valid in the unit disk and
it will actually be a Taylor expansion. Then
00:49:36.401 --> 00:49:42.140
you will get another Laurent expansion which
is valid outside the unit disk okay and that
00:49:42.140 --> 00:49:47.740
is a Lauren expansion at infinity okay, so
but if I go outside the unit disk what happens?
00:49:47.740 --> 00:49:52.700
Mod Z is greater than 1 so I should write
an expansion in terms of for the situation
00:49:52.700 --> 00:49:57.410
and mod Z is greater than 1 and you know if
I am trying to use geometric series I always
00:49:57.410 --> 00:50:02.420
look at the situation when the variable has
modulus less than 1 so this tells you that
00:50:02.420 --> 00:50:08.540
I will have to write it in terms of 1 by okay,
so basically what I will do is I will take
00:50:08.540 --> 00:50:27.720
this Z to the 2014 out and then I will write
it as one by one plus Z to the 2014 to
00:50:27.720 --> 00:50:41.180
the minus 1, so I write it like this and now
you know.
00:50:41.180 --> 00:50:47.200
So this is going to give me one by Z to the
2014 and here I am going to get one plus Z
00:50:47.200 --> 00:50:54.880
to the 2014 to the minus 1 whole to the minus
1, now if I expanded using a geometric series
00:50:54.880 --> 00:51:00.350
you will see that the coefficient of 1 by
Z is 0 that will tell you the residue at infinity
00:51:00.350 --> 00:51:08.360
is 0 okay and you see immediately that this
integral is 0 okay, so so I am just trying
00:51:08.360 --> 00:51:13.250
to tell you that whenever you see problems
go back to those problems that you did when
00:51:13.250 --> 00:51:18.130
you were when you took the 1st course in complex
analysis try to apply residue at infinity
00:51:18.130 --> 00:51:21.740
and you see many of the problems are easier
okay, so I will stop here.