All right, so we continue with our discussion of the spherical derivative, okay. So there
are few things i wanted to point out with regards to the spherical derivative, okay.
So let me, so let me just recall.
If f of z z in a domain d, z varying over capital d which is a domain in c is a meromorphic
function, function, that is f belongs to m of d and mind you the set of meromorphic functions
on d is consider now as a subset of, you know in fact continuous maps from d to the extended
complex plane which is c union infinity, okay. So the script c denotes continuous maps and
the point is that you make the meromorphic function continuous even at the poles by defining
the function value at the pole to be infinity, okay. So you consider it like this and then
the spherical derivative of f is f hash of z, it is defined to be 2 times more f dash
of z divided by 1plus mod of fz also. This is the definition of the spherical derivative.
And mind you why, why did we need the spherical derivative is because of the following reason.
So suppose the, you have this is the complex plane with the variable z and you have this,
suppose this is your this, the area inside this dotted region is your domain d and suppose
you had an arc gamma inside d, you take the image of this arc gamma under f in the external
complex plane, okay. So it means that you know you are allowing also the value infinity,
so for example you know gamma may pass through your pole f. F is a meromorphic function,
so f is meromorphic on d means f is holomorphic, that is analytic on d, except for a subset
of isolated points of d where f has poles.
But at the poles also, the value of f has defined to be infinity. So your gamma, your
curve gamma pass through the poles and that is the technical thing that i want to explain
to you about. Now you identify this external complex plane via the stereo graphic projection
with the riemann sphere which i will briefly draw like this so this is riemann sphere,
which is s2, okay. And this, this isomorphism is actually a homeomorphism given by the
sphere graphic projection, this is the stereo graphic projection with the, with the point
infinity going to the north pole which is this point here, all right.
And the fact is that the image of gamma will, see gamma will give you, you know, if you
take the image of gamma, what will happen is that you will get some curve here on the
riemann sphere, okay. So it is a curve in the external complex plane but you know, you
are thinking of the external complex plane as the riemann sphere when you think of, you
may imagine that the image of gamma is a curve on the riemann sphere itself, okay. And what
is that curves, since this, this is just s of gamma, okay, this is f of gamma and what
is the big deal about the spherical derivative, the big deal about the spherical derivative
is that you can get the spherical length of f of gamma, okay.
You can calculate the length of that image curve, okay, and i have put subscript s for
spherical lens because it is the length you are computing the arc length on the sphere,
okay. And how do you get it? You get it in the following way, you simply integrate over
gamma with the variable, see normally if you now integrate over modern dz, if you integrate
over mod dz simply on the plane over a curve gamma, you simply get the arc length of the
curve, okay. That is what integrating over mod dz means because mod dz is infinitesimal
arc length on the euclidean plane, on c common complex plane, on the complex plane thought
of as r2, okay.
It is usual arc length, but you know if you put, if instead of doing this, suppose
i put, if i add the magnification factor given by the, suppose i add the magnification factor
given by the spherical derivative. So that means i put f hash of z here and do this.
Then what you will get is , i will get actually the length of the image curve on
the riemann sphere, okay. And so this is where the spherical derivative is used, okay, the
spherical derivative will give you, it is, so without this if i do not, see if i remove
this spherical derivative factor, okay, i will get simply integral of us gamma mod dz
and that is just length of gamma.
But if you put the spherical derivative there, okay, then i will not get the length of gamma
but i will get the length of the image of gamma under f. And mind you gamma can pass
through, it can pass through a pole, the only thing is it means that this image curve will
pass through the north pole, that is all, it is not going to create any problems. Because
if it passes through a pole, the function value there is infinity at infinity corresponds
to the north pole on the riemann sphere under the stereo graphic projection, okay. So the
point is important that the spherical derivative is that it gives you the spherical length,
okay.
But there are, there are a few technical things about this , there are a few technical
things about the spherical derivative which i just indicated towards the end of my last
lecture and i want to be more you know elaborate about that. So you see, so i want to draw
your attention to the, to this formula which is a formula for the spherical derivative,
okay. This is the formula for the spherical derivative, there is something that is a little
troublesome about this formula.
See when i have defined and you go before that, let me also tell you that here in this
formula for length, spherical length of f of gamma, you know if i replace the spherical
derivative, if i instead of putting f hash of z, suppose i put mod f dash of z, suppose
i put modulus of the derivative of f, then what i will get is actually the, and i assume
that f is you know the holomorphic function. Then i will get the image, the length of the
image of gamma under f. Then f will map only into the complex plane, if f is differentiable,
okay, everywhere on gamma, okay, then it is a meromorphic.
So the image of gamma will lie in the plane itself, it is not going to go to infinity,
okay because there are no poles, okay. And because f is, if you assume f to be an analytic
function, okay. And then if you integrate over gamma mod f dash of z into mod dz, what
you will get is the length of the image under f but this will be the euclidean length, it
will be just length on the plane. But if you integrate over gamma, mod dz with the coefficient
f hash of z which is a spherical derivative and in addition you allow also f to be meromorphic,
you will actually get the length of the image curve on the riemann sphere, part of the extended
plane, that is what you understand.
Okay, fine, you know there is a problem with this at 1st sight with this definition of
f hash because you see there is this f dash, okay. F dash is the derivative of f at the
point z but the problem is that if z is a pole, then you are in trouble. At the pole
the function is certainly not differentiable, it is a single point, it is a pucca similar
point, it is an honest singular point, it is not a removable singularity, okay, the
function is not differentiable. Alright. So you are in trouble, so when i wrote this definition
last time, you know i was only, you know i was trying to heuristically tell you things
but now i am going to tell you things more seriously, so let us worry about, let us worry
about this, this, this situation.
So here is my, here is my domain d which is interior of this dotted line, it is an open
connected site, okay. This is inside the complex plane and suppose i have a point z0 and of
course i have this map f, f is a meromorphic function of d and you know of course f is
taking values in c union infinity and z0 is a pole of f, of order let say, of order n,
okay. And of course z0 will go to f of z0, which is by definition infinity, okay, this
is our definition. Now what about the spherical derivative? Okay.
See, what is, so the question is what this f dash of z. So this is an, this is an issue,
you see because what is f dash of z. If we have to worry about this, the reason is because,
see suppose i have a gamma, suppose i have a path gamma passing through a pole, okay,
then my formula for the length of gamma, the spherical length of, the image of gamma under
f on the riemann sphere which is identified in the extended plane, what is the formula,
it is l spherical of f of gamma, you know that is what i have shown in the previous
slide , the spherical length of f of gamma is integral, you integrate over gamma.
So i put, if i put just mod dz, i will get just the length of gamma but i have put the
magnification factor given by the spherical derivative of f with respect to z. And what
is a spherical derivative of f with respect to z? It is, well it is f hash of z, you can
write it, it is 2 times mod f dash of z divided by 1plus more fz the whole square, this is
what it is. It was there in the previous slide also. Okay, now the point is, if i put z equal
to z0, z0, see now gamma passes through gamma passes through z0, okay.
So when i calculate this integral on the right side gamma i have, when you do, when you do
and integration the variable of integration will lie on the region of integration. In
this case region of integration is the path gamma. So z will pass through z0, it will
bury, at some point z will become z0. But when z become z0, there is this integrand
which is f dash, f hash of z, the spherical derivative, that is in trouble. Because you
know f hash of z depends on f dash of z0 in the numerator. F hash of z0 will be, will
involve f dash of z0 but f dash of z0 does not make sense.
Why, because z0 is a pole, i cannot differentiate at a pole, i just cannot find derivative at
a pole. So what is it, what is the big deal? So there is, so you see this formula as we
have written it last time has this issue that has to be fixed. And the reason is because,
the fact is that, as i was telling you last time, even at z0, this f dash of z0 is not
defined but this spherical derivative is defined as it finite quantity. That is the beauty,
that is the reason why this integral works, okay, that is what i want to explain to you.
So you see, so let me, so let me say that, so you see, let us assume that z0 is a pole
of order n.
So then what happens is that you know you will get a small disc surrounding z0, so let
me use a different colour, see i will get a small disc surrounding z0, okay, i can find
a small disc surrounding z0 where z0 is the only pole, okay. Because you know the poles
of an analytic function are isolated in any case. And in fact our meromorphic functions
are supposed to be having only pole singularities, okay. These are the only singularity that
are allowed, okay. So i can find a small disc surrounding z0 where z0 is the only pole and
well, you know, if you, if you call this, if you call the radius of this disc as say
epsilon, okay, then mod z minus z0 less than epsilon, which is interior of that small disc.
You see you can write f of z, you can write f of that as you know g of z divided by z
minus z0 to the power of n where g of , where g of z0 is not 0, okay. So you can,
you can write it like this. And of course g analytic in mod z minus z0 less
than epsilon, okay. So, okay, i will give a little more space, let me rewrite that.
Let me just write g analytic. I can do this because this is how the function looks near
a pole of order capital n, okay. Now, now watch carefully. Let us calculate the derivative
of f, not at z0 because at z0 you cannot calculate the derivative.
But in the deleted neighbourhood of z0, let us calculate f of z. And you also when i,
when i write this in mod z minus z0 less than epsilon, of course z should not be z0. I mean
i cannot literally plug-in z equal to z0 because z equal to z0 is a for land z minus z0, denominator
vanishes, i cannot write that. Of course we have agreed to put it as z equal to z0 and
equate it to infinity, that is when you consider f to be a function with values in c union
infinity but nevertheless if you think of it and the usual function, then you do not
plug-in z equal to z0 because you do not divide by 0, okay.
And that is the situation you must be, if you want, if you want really differentiate
this, okay. So you see, so now let us calculate, let us do this calculation. See what is, so
let me write this, in mod z minus z0 less than epsilon, z not equal to z0, what is f
dash of z. F dash of z is just d by dz of f of z which is now g of z by z minus z0 to
the power of n, okay. And you can calculate this by if you want quotient rule from basic
calculus. What will be, i will get z minus z0 to the power of 2n and i will get, what
will i get here, z minus z0 to the power of ng dash of z plus, minus g of z ,n z minus
z0 to the power of n minus1, okay, this is what i get, if i do this computation.
And you see now what you do is, see this is all right for z not equal to z0, all right.
And you will certainly have a problem if you let z tend to z0, okay. Limit, if you actually
see in the usual sense if i let z tends to z0, then what will happen is that in the numerator
the 1st term will go, all right, the 2nd term, well it will go, provided n is greater than
1, okay. If n equal to1, i will get gz0, okay but the problem will be the denominator. As
z tends to z0, i will end up with, essentially i will, what will happen is that because f
has a pole of order capital n at z0, its derivative will have a pole of capital, order capital
n plus1 at z0, okay.
That is what could happen, so it is going to only get worse, limit z tend to z0, f dash
of z will not exist. And if, in fact the worst-case you want to make it exists is you can define
it to be infinity by thinking of f dash also is a meromorphic function but now with values
in c union infinity, you can do that. But in any case it is not a finite, it is not,
it is not a proper limit in the usual sense, okay, you have to include the value infinity.
But then, so i will say limit z tends to z0, f dash of z does not, it is not a complex
number, okay. It is, if you, if you include c union infinity, then you can call it as
infinity, that is that, but that is not the case.
We do not want to include the value infinity when we are talking about derivative. But,
you see, what look at what on the other hand you look at what is f hash of z. Look at the
spherical derivative, if you look at the spherical derivative, what i will guess, i will get,
c will get 2 times modulus of f dash of z, okay, so i will get 2 times modulus of this
whole quantity, okay, divided by 1plus mod f the whole square and mod f the whole square
will be, mod f is the modulus of this quantity, so it will be 1plus modulus of g of z by z
minus z0 to the power of n the whole square, this is what i will get.
Okay, this is what i will get and now you take limit z tends to z0. Now you take limit
z tends to z0, of f hash of z you take, you calculate this limit. What will happen if
you see the, in the denominator you have 1plus mod gz the whole squared divided by mod z
minus z0 to the 2n, okay. So denominator will go through infinity, the denominator will
go to infinity faster than the numerator, so this, so the whole quantity will be bounded
as z tends to z0, that is the whole point. So you see if you calculate it, okay, if you
calculate it, what will happen, so, so let, so this exists. So if you write it out, you
know i am going to get, so i will, to simplify things i will multiply both numerator and
denominator by z minus, mod of z minus z0 to 2n, which is what is the common denominator.
So what i will get is that i will get, well let me write it here, f dash of, sorry f hash
of z is going to be 2 times, i will get the numerator of this, which is more z minus z0
to the n, so i will get this, mod z minus z0 to the power n gz minus gz, oops, i think
that must have been, that is g dash of z minus g of zn into z minus z0 to the n minus1 mod
divided by, okay, i have multiplied by, multiplied by this modulus of this quantity, mod z minus
z0 to the 2n, okay. So that is gone, so in the denominator i will get more z minus z0
to the 2n plus mod gz the whole square.
This is what i will get if i multiply it by mod z minus z0 to the 2n, okay. And mind you
the spherical derivative is an absolute derivative, so it is only absolute value, so it is a nonnegative
real valued by the way. Now you, now you do, if you take limit at z tends to z0, what is
going to happen. You see as z tends to z0, this term will vanish because z minus z0 power
n is there and of course this n is of course greater than or equal to1. It is the order
of the pole, so this whole of order 1 or higher, okay. So this is going to vanish and this
fellow here, what will happen here, depends on whether n equal to1 or n is greater than
1, okay.
See, if, if n is equal to1, what is going to happen, if n is equal to1 then this term
does not exist, okay. And let z tend to z0, i will get, i will get 2 times mod gz 0, g
is anyway mind you in analytic, discontinuous, so z tends to z0 gz, g z0 and modulus also
a continuous function, so i can push the limit inside the variable, okay, inside the argument
of the function. And then that is what i said in the numerator, so this is, this is if n
is 1, okay, this is if n is 1. And in the denominator what i am going to get, this term
is going to vanish as z tends to z0, i am going to simply get mod g, again i will get
mod gz0 the whole squared, i will get divided by mod gz 0 the whole square, which is just
able to by mod g z0, this is what i will get.
And mind you gz0 is not 0 because g z0 is , g is the, you know if you want, g is
analytic function divided by z minus z0 power n which is equal to f in the neighbourhood
of f. In fact you know g z0 is, if you check very carefully, g z0 is, if the coefficient
of the, of 1 by z minus z0 power n, if you write down the lagrange expansion, okay and
that is not supposed to be 0, okay, because g, f has a pole of capital n, right. So this
is what you will get.
And you see, and mind you in the case that n equal to1, g z0 is actually the coefficient
of 1 by z minus z0 power n which is 1 by z minus z0. But you know what is the efficient
of 1 by z minus z0 called, it is called the residue. So actually this is 2 divided by
your residue of f at z0, that is what it is. This is nothing but 2 divided by modulus of
residue of f at z0, this is what happens if you get, if f is a simple pole, capital n
equal to1, all right. And the point is that, now if n is greater than 1, everything is
gone, because you see is n is greater than 1, there is no following the denominator,
i will get mod g the whole squared, this term is anyways going to vanish.
And the numerator will also go now, numerator has z minus z0 term common, so it is going
to go. So i will get 0 if n is greater than 1. So here is the, so here is the, so of course
this is on the, i forgot to write f hash of z. So here is a nice thing, f hash of z0,
you can now call, see you can define f hash of z0 by continuity to be equal to limit z
tends to z0 f hash of z, okay. If you think of f hash as a continuous function, okay,
if you want to think of the spherical derivative as a continuous function, then it is natural
define f hash at z0 to be the limit as z tends to z0 as f hash of z, okay.
And you see, this, what this does is that it makes the spherical derivative continuous
even at z0 and mind you z0 is a pole. So what this tells you is that the spherical derivative
f hash of z is continuous at all poles. So it is continuous throughout domain and therefore
because it is continuous at all, throughout the domain, this formula is valid, okay. What
i really meant here is, what is f hash of z0, okay. Of course f dash of z0 does not
make sense, so the question is what is f hash of z0, all right. So , so now you know
f hash makes sense even at poles, so this integral is well-defined, there is no issue.
You can blindly integrate f hash, you can, you cannot blindly integrate f dash because
f dash will not exist at a pole. You can integrate f dash, mod f dash only where so long
as you are an apart which is not going through any poles. And if it is going through a pole,
you cannot integrate f dash but you can integrate f hash always, even if you are passing through
a pole, that is a big deal, that is a big deal. So that is the reason why this formula
works. And what this calculation we did just tells you is that the spherical derivative
is actually 2 divided by modulus of the residue of f at simple pole z0 if z0 is a simple pole
and it is 0 if not, this is not means, i mean pole of higher-order, all right.
So the moral the story is that you know, you are in, you are in good shape. F hash spherical
derivative is a very nice thing, okay. And therefore when you, whenever you want to find
the arc length, you can integrate mod dz over, multiplied with, with the, you know integrand
as f hash and that is pretty important. Now, and you know again, i will tell you why we
are doing all this, we are doing all this because you know somehow the kind of analysis
that is required to prove picard’s theorem is involves montel’s theorem, okay.
And this, i will tell you roughly the idea is that you know there are, there are these,
there is a, there is a very close relationship as i told you between compactness and sequential
compactness and equi-continuity and normal, normal convergence, okay and bounded mass
of the derivatives, okay. So this is the, this is a bunch of results same analysis which
is usually covered by the arzela ascoli theorem, okay. And that is a, there is a, the montel’s
theorem is something that comes out of that, okay. And why we are doing all this is because
you know you, basically you know the idea is that you want to look at the space of meromorphic
functions on a domain, okay.
So you have some domain, all right, this is a domain in extended plane, it could include
infinity also, okay, on that domain you are looking as meromorphic functions, all right.
And you are looking at, you want to think of them at least as continuous functions,
so you are allowing the value infinity at a pole. So you are looking at that pace of
meromorphic functions and you see the convergence that you are worried about is normal convergence,
okay, it is not uniform convergence everywhere, it is only uniform convergence restricted
to compact sets, which for normal convergence.
And with this convergence idea you want to study apology of this space of functions.
And explicitly what kind of topology you want to study compactness, you want to study compactness,
okay. You want to study compactness and you know if you have studied, for example in euclidean
space, compactness is as good as sequential compactness which is the same as saying that
you know every sequence, if you have an infinite sequence, there is always a convergence of
sequence, all right. And so you have compactness is somehow strong related to sequential compactness,
okay.
So basically you want, basically given, given a sequence you always want a convergence of
sequence, all right. And now you want this also to happen for meromorphic functions,
that is the, that is the central idea, the central idea is give me a bunch of, give me
a sequence of meromorphic functions on a domain and now you try to find conditions, topological
conditions, that will tell you, topological, of course includes analytics conditions. That
will tell you that i always will be able to find from this sequence i find a sub sequence
which converges.
But mind you now it is not just convergence, it is normal convergence, because in the context
of complex analysis, in the context of holomorphic functions, uniform convergence will not work.
You will get only uniform convergence on restricted subsets, namely only on compact subsets, that
is called normal convergence, okay. So you have to worry about this and in order to do
all this, see i need to, see the boundedness of the derivatives for example if something
that is strongly related to all this. So i want to be able to work with derivatives.
But the problem is that functions i am trying to work with are all what, they are meromorphic
functions.
And meromorphic functions are not differentiable, at the poles they are not differentiable.
So what do i will do at the poles with meromorphic functions? What i do is, the clever thing
is i do not look at the ordinary derivative, i look at the spherical derivative. Spherical
derivative makes sense even at pole, that is where spherical derivative comes in. Okay,
that is what you offer understands all this is required for me to do analysis on a space
of meromorphic functions. And that is a kind of, see that is the kind of you know
analysis you have to do to prove the picard’s theorem, okay.
Fine, so okay so what i want to do next is i want to tell you , well that this other
fact that i was telling you last time that you know the spherical derivative has another
important advantage. The spherical derivative allows you to, you know forget meromorphicity.
It is a very clever trick, you see on the, you introduce a spherical derivative because
you want to look at derivative of a function which is meromorphic at the pole for example,
okay. That is why you introduce a spherical derivative but i, the, it is beautiful, once
you introduce this notion, you can in most cases you can even forget the pole.
Which means you can reduce everything to just studying analytics functions. So that is the
beauty, the reason is, the spherical derivative of meromorphic functions is the same as the
spherical derivative of its reciprocal, okay. And what is the advantage of passing the reciprocal,
passing to the reciprocal, the advantage is that a pole becomes is zero, okay. And a zero
is a very nice thing, the function is differentiable there, all right, so that is the advantage.
So that is a, that is an added advantage you get free, okay. And why is this true, this
is true because the spherical length, that is invariant under inversion, okay.
So that is what i want to tell you about next. So you see, so recall that the spherical distance
d sub s is invariant under inversion, so you have the spherical distance between z1 and
z2 is the same as the spherical distance between 1 by z1 and 1 by z2 where you know z1 and
z2 are now taken to be in the external complex plane. And in the external complex plane mind
you 1 by 0 is defined to be infinity and 1 by infinity is defined to be 0, this is the
condition. So you know the spherical distance is the, is actually the spherical distance
on the riemann sphere, okay which means the distance between 2 points on the riemann sphere
is given by the length of the minor arc of the biggest, bigger circle that passes through
those 2 points and which lies on the riemann sphere, okay.
And that length, how do you get it, and that length you get it by, for example, using basic
analytic geometry, it is after all length of the arc of a circle and you can always
derive’s formula, okay. And that length you transport it via the stereo graphic projection
to the extended complex plane. So the advantage of that is that i can measure for example
the distance between the point in the complex plane and the point that infinity. Okay, i
do, which i cannot do with the usual euclidean distance because you, euclidean distance becomes
unbounded as the point, as one of the points is fixed and the other goes to infinity, all
right.
So and i told you in an earlier lecture that you the mapping z going to 1 over z which
is the inversion mapping, that is a map of c union infinity onto itself, it is a, it
is a, in fact it is a homeomorphism and the fact is that under that homeomorphism
, see what it does is that she has just maps the extended complex plane back to the extended
complex plane, if exchanges 0 and infinity. Infinity goes to 0 and 0 goes to infinity,
okay. But on the other hand since it is, it is itself homeomorphism of the extended complex
plane, it will also induce itself homeomorphism of the riemann sphere because after all the
riemann sphere is homeomorphism to the extended complex plane.
See whenever in mathematics whenever one object has an isomorphism and you take another isomorphic
object, then an isomorphism on the 1st object will automatically induce an isomorphism on
the 2nd object which is transported by this isomorphism between them. So the inversion
will also induce an, a self isomorphism, a self isomorphism, a self homeomorphism of
the, of the riemann sphere and what is it? It is nothing, i have asked you to check this,
you should do it, i hope you have done it. So it is just rotation of the riemann sphere
about the x axis about 180 degrees, that is all, that is what it is.
And you know if you if you take 2 points on a sphere, and you take the spherical distance
between them, that arc distance, now if you rotate the sphere, that is not going to change,
okay. So it is invariant under that rotation, all right. And therefore the net effect is
that the spherical distance is invariant under inversion, okay. Now this implies that the
spherical derivative of f is the same as the spherical derivative of 1 by f, okay. And
uhh so so why is this, this is for f in meromorphic, f is a meromorphic function on
d.
And why is this true? Because you see, i will tell you if you want you can try to
do direct calculations, you can do a direct calculation. So you know the what
is f hash f hash is just 2 mod f dash z by 1plus mod fz the whole square, this is what
it is, all right. Now this is okay where f dash exists, this formula is correct where
f dash exists. Let us assume that, for, for simplicity let us assume that f dash is not
0 at the point. Suppose derivative does not vanishes at the point, 1 by f also makes sense
at that point, the usual derivative of 1 by f also makes sense at that point.
So if you calculate, if you calculate, it will, what it will be, it will be 2 times,
you know i will get modulus of 1 by f dash of z divided by 1plus mod 1 by f of z the
whole squared. Now if you calculate this what will i get, i will get is well you
know 2 times, if you take the derivative of 1 by f, i am going to get minus1 by f
squared into f dash, this is what i will get, okay, if i use a chain rule. And then i have
to divide by 1plus 1 by f of z the whole square mod, okay. And now if my simplified,
you see, i will again end up with 2 times mod f dash of z divided by 1plus mod z
the whole square.
If i simply multiply numerator and denominator by mod fz the whole square, i will end up
with this which is the same as f hash, okay. So this is a very heuristic, i mean it is
a very simple calculation, the only thing, the only problem with this calculation is
that you know, so i am, i am cancelling out mod f dash, i am cancelling out more f. And
to cancel out mod f, f should not vanish, okay, otherwise i cannot cancel mod f in the
numerator and denominator. So this is okay if f is, f is, f of z is, it is okay at the
point where f is not 0, okay, so let me write that, valid if f is nonzero, that is one thing
in the 2nd thing is that f dash should exist, okay.
Derivative should exist, otherwise i cannot write, so f dash exists. So what i am saying
is that the spherical derivative of f and the spherical derivative of 1 by f, they are
the same, you can verify it at all points of f which is different from the zeros and
poles, okay. Now what you do is, you check it at a pole using the same calculation that
we did last time, all right and you will see the mobile last time i mean just some time
ago, we did this calculation to calculate the f hash z0 at a pole z0, that is by basically
writing out f locally at the point z0 which is a pole in this form, f is equal to g by
z minus z0 to the n, all right.
And you use the same calculation, if you use the same calculation, what will happen is
that you can see that when f is 0 at a point, that is at the point where f has a zero or
at a point where f has a pole, the same calculation is correct. What you do is you calculate in
the deleted neighbourhood and then you let limit z tends to z0. You do it both at the
pole and at t0 and you will see that the limit will exist and whether you calculate the limit
for 1 by f or whether you calculate the limit for f, you will get the same thing. Okay,
so that, that i leave it to you as an exercise, okay.
So one of the, one of the important things is that, see one of the important things is
that you know if you take the spherical length of f of gamma, okay, this is going to be by
definition integral over gamma f hash of z mod dz. And since f hash is 1 by f hash, this
is also integral over gamma, 1 by f hash of z mod dz and this is by definition spherical
length of 1 by f of gamma. And f of gamma and 1 by f of gamma, they only differ by an
inversion and under the inversion, the spherical length should not change, so it is correct,
okay.
So this way also you see that the, you know, you can, when you calculate the spherical
derivative, whether you calculate for f or whether you calculate for 1 by f, there is
no difference, okay. So what is the advantage, suppose you are proving something, involving
spherical derivative and suppose you have to deal with a point which is a pole, okay.
Suppose i have to deal with a function f at a point which is a pole and suppose i am working
with the spherical derivative, without loss of generality i can replace f by 1 by f. Because
by replacing f by 1 by f, my spherical derivative does not change but my pole for f becomes
is 0 for 1 by f and 1 by f becomes analytic.
So i am dealing with a nice analytic function, okay. So that is the advantage of having the
spherical derivative, okay. So now what we will do is in the forthcoming classes, we
will use all this, all this background that we have developed so far do you know in a
series of lemmas and propositions and finally theorems, we will prove the Picard’s theorems,
okay. And on the way we get the very important Montel’s theorem, okay. So i will stop here.