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HERBERT GROSS: Hi.
Today, we're going to
conclude our survey
of complex variables.
And a rather fitting
topic, I imagine,
would be what it means to
integrate a complex valued
function of a complex variable.
And I thought that that's what
we would talk about today.
And in fact, let's just call
today's lecture "Integrating
Complex Functions."
But let me emphasize,
I really do
mean by a complex
function the thing
that we've been stressing
throughout the course,
a complex valued function
of a complex variable.
The idea being that if you
have a complex function
of a real variable,
essentially, this breaks down,
as we've seen in the
previous assignment,
into studying regular functions
of a single real variable.
In other words, if you
have u of t plus iv of t,
where t is a real variable,
this really gives us no problem.
This is sort of like the vector
treatment, a vector function,
of a scalar variable, et cetera.
So what I'm saying
here, though, is
let's make sure that it's
clear that we're still
talking about mappings
that map complex numbers
into complex numbers.
Now by way of review, how did
we define the definite integral
in the case of a
single real variable?
We define the
integral from x0 to x1
f of x dx to be a certain limit.
Namely, we partitioned
the interval from x0 to x1
into n pieces, calling the
pieces delta x1 up to delta xn.
We took a number, say, c sub
k star in the k partition,
formed the sum f of c
sub k star delta xk--
xk went from 1 to n--
and took that limit as the
maximum-sized interval.
The maximum delta
x sub k went to 0.
And if that limit
existed, that was defined
to be the definite integral.
Now, we did see a few
other things along the way.
But this is the main definition.
You see, this is the
definition that we're using.
Some of the consequences
of this definition
were one, if we could
find the function capital
F, whose derivative
was little f,
then this integral
was simply capital F
of x1 minus capital F of x0.
And we also saw that if
we wished, and we didn't
have to, but if we wished,
we could view this thing
geometrically by observing that
the x-axis was the domain of f.
The y-axis was the range of f.
We could then plot
the curve y equals
f of x, look at the area of
the region R between x0 and x1,
and the area of the
region R was the value
of this definite integral.
But the important point is what?
That the definition
of the integral
is as a limit of
an infinite sum.
And now the question
is, how shall we
define the analogous
thing for a complex valued
function of a complex variable?
In other words, what
shall we mean by integral
from z0 to z1, f of z dz?
And because we've
been so successful
with this technique
in the past, it
would seem that the
simplest thing to do
would be to simply
replace x by z
every place in this definition.
In other words,
let's simply do this.
Let's define the integral
z0 to z1 f of z dz
to be the limit as the maximum
delta z sub k approaches 0.
Summation-- k goes from 1 to
n, f of c sub k star times
delta z sub k, where
all we have done
is replaced xs every place
by zs wherever they appeared.
Now, the reason I've
put question marks here
is that all of a
sudden, a problem
occurs that never
occurred in the real case.
You see, the key reason
that these problems are
going to occur, as I'll explain
in a moment, hinges on the fact
that in the real
case, the domain of f
was one-dimensional.
It was an axis, the x-axis.
In the complex variable
case, the domain of f
is two-dimensional.
It's the entire xy plane,
the Argand diagram.
You see, the problem
is simply this.
What do you mean
by delta z sub k?
Intuitively, it certainly
should seem to mean what?
The line that joins the
point in the Argand diagram z
sub k minus 1 to
the point z sub k.
In other words, remember,
we view complex numbers
as being vectors.
They have a magnitude
and a direction.
So delta z sub k can be
viewed as the vector which
joins these two points.
The trouble is this.
Let's take a look in
the Argand diagram
and locate the points z0 and z1.
Here's z0.
Here's z1.
That's all that's given.
Where shall these
intermediary points z sub--
where should these
intermediary points
that are going to give
rise to delta z sub k be?
In other words, there seem
to be many points that we
can pick for a partition.
In particular, it would
seem that the most
natural definition would be to
specify a particular curve C
and then look at delta z
sub k in terms of the curve
C. Because you see once,
the curve C is specified,
if I partition things now,
notice that the delta z sub
ks do not connect random points,
but rather points on the curve
C.
The difficult point is that
there are many different curves
that join z0 to z1.
So perhaps the first
thing that we should do
is, keeping in mind
this ambiguity, maybe
we should come back to this
definition and put in a C
to indicate that we're going
to talk about the integral of f
of z dz from the point z0 to
the point z1 along the curve
C, keeping in mind, you see,
that the reason that we have
to specify the curve here
is that z0 and z1 are
in the plane.
You see?
Going back to this example
here, notice that here,
when we said let's
go from x0 to x1,
there was only one
direction in which
you could go from x0 to x1.
Namely, since the domain of
definition of f was the x-axis,
the direction had to be in
the direction of the x-axis.
You see, in the
Argand diagram case,
I can go along any
curve in the plane,
provided that the curve connects
the two points in question.
Now, the other question that I
think I should point out here,
even though it may be bringing
up a question that you might
not have anticipated--
notice that when
I interpret the definite
integral as an area here,
notice that the lower bound
is the domain of f itself.
And the upper bound is
already from the curve y
equals f of x-- that the upper
curve brings out the function
here.
You see, the x-axis
is the domain.
The y-axis is the range.
I would like you to
observe that there
is no similar interpretation
along the curve
C. In other words, if we
were to drop perpendiculars
from z0 to z1 down to the
x-axis and talk about that area,
that would be the wrong
area to talk about.
Because you see, the point
is that that region is
determined just by our domain.
Notice that this curve--
and this is very, very crucial--
that the curve that we're
talking about here, C,
has absolutely nothing in the
world to do with this f of z.
Just like when we computed
mass using density,
the density had nothing
to do with the region
that you were integrating
with respect to.
The point to observe here is
that we're assuming that f of z
is defined every
place in the plane.
f of z is defined here.
It's defined here.
It's defined here.
It's also defined
on the curve C.
Notice that f, in this
case, is a complex number,
which means that it has a
real and imaginary part.
f is no place in this diagram.
And to plot what f
does, I would have
to use the uv plane because
f itself is two-dimensional.
In other words, I would
need two dimensions
for the domain and two
dimensions for the range.
And I hope that that part
is fairly clear to you now.
Don't confuse the path of
integration with the integrand
f of z.
They are entirely
different concepts.
At any rate, I
hope that this does
suggest the concept of a
line integral in some sense.
What I'm saying now is, OK,
I want to integrate this
from z0 to z1.
That means I want to
compute a particular limit.
I picked the curve
that joins z0 to z1.
Let's assume that in
Cartesian coordinates,
that curve parametrically
has the form
x is some function of t.
y is some function of t,
where t is a real variable.
That's important. t
is a real variable.
By the way, as we mentioned
when we were studying vectors,
if I want to introduce vector
notation, the radius vector,
and write the curve
C in vector form,
notice that the curve
C also has the form
that r is x of ti plus y of tj.
And finally, if I now want
to write the curve in terms
of complex numbers-- after
all, the Argand diagram
now is the domain of
the complex numbers.
What is the connection between
the xy plane and the Argand
diagram?
The x-axis names the real part,
the y-axis, the imaginary part.
In other words, the unit--
we think of the i
vector as representing
the real part, the j vector
as the imaginary part.
Notice that changing
this notation
into the language of the
Argand diagram results
in this equation
being replaced by z
equals x of t plus iy of t.
Where notice now that z is
a complex valued function
of the real variable t.
And t, of course, goes
from some fixed value
t0 to some fixed value t1.
Because after all,
what we're saying
is that the curve Z
equals x of t plus iy of t
is being traced
out in terms of t.
When t0 corresponds
to being at z0,
t1 corresponds to being at z1.
The idea now is this.
We go back to our definition
of what the definite integral
meant in terms of a limit.
We observe that we'll
assume that we'll
make the further restriction
that our curves are smooth,
so that dz, dt will exist.
What we now do is go
back to our infinite sum
and multiply and divide by
delta t sub k, the idea being--
notice that this is what?
A complex valued function
of a real variable.
Because I can handle complex
valued functions of a real
variable-- namely, I just
break them down into real
and imaginary parts and use
the same calculus that we used
in part one of this course--
the idea is that with
this suggestive notation,
and remembering how the
infinite sum gave rise
to the integral notation,
looking at this,
you see we arrive at the fact
that the integral from z0 to z1
along the curve C, f of z dz.
If we want to write that
in terms of an integral
involving a real variable
t, it's simply what?
The integral from t0
to t1 f of z of t.
In other words,
we're assuming that z
is a function of t
along the curve C--
we know what z looks like along
the curve C in terms of t--
times the derivative of z with
respect to t, z prime of t,
times dt.
And we'll have plenty
of exercises, including
one in this lecture,
to show you how
we use this particular
result. On the other hand,
keep in mind that
this is just what?
One way of visualizing
what the infinite sum is.
A second way to visualize
the same problem
is to recall that we
very frequently like
things in terms of u and
v components, coordinates.
The idea is that if we
now want to introduce
u and v in the usual manner,
f of z is then u plus iv.
Delta z is delta
x plus i delta y.
And coming back to our basic
definition over here, you see,
what we can now do is
rewrite f as u plus iv,
rewrite delta z, sub k as delta
x sub k plus i delta y sub k,
multiply this
thing out, pick off
the real and the
imaginary parts,
write this as two
separate sums, et cetera,
and take the limit separately.
What this leads to,
if you want to go
through the whole
rigorous process quickly,
just coming up with
the right answer,
is that essentially,
what we do is,
given this particular integral,
we write f as u plus iv.
We write dz as dx plus idy.
Y
We then multiply the
way we ordinarily
would with complex numbers.
See, udx minus vdy.
And the imaginary
part is vdx plus udy.
Break this up as two
separate integrals.
And the key point is
notice that both of these
integrals-- forgetting about
the coefficient of here being i,
this is the only
place where i appears.
Notice that both integrals are
now ordinary line integrals.
You see u as a real valued
function of x and y.
v is a real valued
function of x and y.
Along the curve C, x and
y are functions of t.
Remember, the equation
of the curve C
was x was some function of t.
y was some function of t.
In other words, again, this
is not only a line integral.
But if you want to use
this parametric form,
both of these integrals
involve real integrals
of real variables.
In other words, both
u, v, dx, and dy
are all expressible
in terms of t and dt.
And we can integrate this
expression in the usual way.
But the reason I
like the u and v form
is that we've stressed what
it means for a function
to be analytic in terms of
the real and imaginary parts,
et cetera.
Remember, keep an eye on this.
Remember this particular
expression here.
And here's the key point.
If it turns out that u and v
are the real and imaginary parts
of an analytic function--
in other words,
if u plus iv is some
analytic function f--
remember that the Cauchy-Riemann
conditions told us
that the partial of
u with respect to x
is equal to the partial
of v with respect to y.
The partial of u
with respect to y
is minus the partial
of v with respect to x.
And if you look at these
two conditions coupled
with our definition
of what it meant
for a real differential in
two variables to be exact,
notice that this condition here
tells us that this is exact.
And this condition here
tells us that this is exact.
In other words, both of
these ordinary line integrals
are exact if u plus
iv is analytic.
And again, if you want to leave
the complex number part out
of this, if all we know
is that the partial
of the real function
u with respect to x
equals the partial of
the real valued function
v with respect to y, et cetera.
And in particular then,
because this is exact,
what happened when a
differential was exact?
Remember, if the
differential was exact,
the line integral was what?
Dependent only on the
endpoints, but not
on the path that
connected the end points.
Or stated in still other words,
if the integrand was exact,
the line integral around
the closed curve was 0.
So what we're
saying in summary is
that notice that if f equals
u plus iv is analytic,
then the integral from z0 to z1
f of z dz is independent of C.
It depends only on z0 and z1.
And in that case, I do
not have to specify what
curve C is joining z0 and z1.
And in particular, if f of
z happens to be analytic,
so that this is exact, then the
integral around a closed curve
is 0 for all closed curves
C. And that means again
in particular, if I know
that f of z is analytic,
I do not have to indicate
the curve that I'm forming
the contour or integral around.
You see?
But I do have to
put this in here
if the function is not analytic.
And by the way, let me just
make one more remark in passing.
And we'll emphasize this more
in the learning exercises.
Not only is it true that
integral z0 to z1 f of z dz
depends only on the end
points if f is analytic.
But the parallel structure
that is obeyed in the real case
also is true here.
Namely, if f is analytic, not
only does the integral from z0
to z1 f of z dz not
depend on the path,
but it can be evaluated
very simply just
by computing capital F of
z1 minus capital F of z0,
where capital F is any function
whose derivative is little f--
the same structure as before.
But-- and this is extremely
important to note--
the integral around the
closed curve C, f of z dz,
need not be 0 if
f is not analytic.
And I think the
best way to show you
that is by means of an example.
Let's compute the integral
around the closed curve
C, dz over z, where C is
the circle in the Argand
diagram, centered at the
origin with radius equal to r.
And again, to emphasize
what I said to you earlier
in the lesson, do not confuse
the integrand with the path C.
Notice that the
integrand here is what?
What is the integrand?
It's 1 over z.
Notice that 1 over z is defined
at any point in the Argand
plane, all right?
Notice also from
our previous lecture
on derivatives that 1 over
z is analytic, except when z
equals 0, in which case 1
over z isn't even defined.
You see, what we're saying is
notice that on the curve C,
I can talk about f of 1 over
z for each point on the curve.
I can talk about 1 over z for
every point inside the curve.
I can talk about 1 over z for
every point outside the curve.
What I want to compute
here is the integral
as I move along this curve--
1 over z dz.
Notice, by the way, that
inside the interval,
inside the circle, f of
z is not analytic. f of z
has a bad point.
Namely, 1 over z is
trouble when z equals 0.
And notice that z equals
0 is inside my region.
In other words, notice
that my function f of z
is analytic on the circle,
but it's not analytic
every place inside the circle.
The one bad spot is when z is 0.
So I do have a case where what?
The integrand,
which is 1 over z,
is not analytic in
the entire region,
even though it does happen to
be analytic on the boundary.
You see, the boundary is made
up of points, none of which
is the origin because this is
a circle of positive radius
r centered at the origin.
And again, keep in mind that f
does not appear in this diagram
anywhere.
If I wanted to plot 1 over z,
I have to use the uv plane.
If I want to think of this in
terms of a real interpretation,
what I'm saying is
visualize a force, which
when written in the language of
complex variables, is 1 over z.
We don't have to worry
about what that means.
What it really means
is write the real
and the imaginary parts of this.
And u is the real part, and
v is the imaginary part.
It's like computing the work
as you go along this circle
under the influence of
that particular force.
And the force has two
components, you see,
an i and a j
component-- in terms
of the Argand diagram, a real
and an imaginary component.
So I can't stress
that point too much.
Separate the integrand
from the path
that you're integrating
with respect to.
Now, what I claim is, is
that if I integrate 1 over z
along the curve C from
beginning to end, let's
say for the sake
of definiteness,
I start at the point 1
over here and go around--
at the point r and
go around like this.
I claim that this
integral will not be 0.
And I'm going to do
it two ways for you,
one which emphasizes the
uv definition of integral,
and one that emphasizes the
f of z f prime dt definition.
Method one is what?
To compute this
particular integral,
you simply write the
integrand in terms of its real
and imaginary parts and
write dz as dx plus i dy.
If I do that, I mechanically get
from here to here, remembering
that to get into the
standard form of a real plus
i times a real, I must
multiply the denominator
by its complex conjugate.
I multiply the numerator and
denominator here by x minus iy.
I then get this expression.
You see x plus iy
times x minus iy
being x squared plus y squared.
I now pick off the real
and the imaginary parts.
The real part here is xdx plus
ydy because minus i times i
is plus 1.
The imaginary part is going
to be minus ydx plus xdy.
And so in terms of
u and v components,
this is the integral
that I'm evaluating.
And you see, notice that
both of the integrands
are line integrals--
real line integrals.
Now, notice that on C, how
do I describe the curve C?
Remember, I'm trying to
go so that my area always
stays to the left.
I'm starting at this point.
So why not write
this parametrically
as x equals capital R cosine
theta and y equals capital R
sine theta?
You see?
That puts me in here
when theta is 0,
brings me around in the
counterclockwise direction,
back to the same point
when theta is 2 pi.
In other words, parametrically,
the curve C is given how?
x is our cosine theta.
y is our sine theta.
I could have used t in here if
I wanted to, but why bother?
dx is just minus R
sine theta d theta.
dy is our cosine theta d theta.
x squared plus y squared
on C is just R squared.
And theta goes
continuously from 0 to 2pi.
So you see, if I now put all
of this information into here,
see, xdx is what?
It's minus R squared sine
theta cosine theta d theta.
ydy is plus R squared sine
theta cosine theta d theta.
So therefore, the sum is 0.
So therefore, this
integral will be 0
because the integrand is 0.
On the other hand, minus ydx
is minus R sine theta times
minus R sine theta d theta.
That's R squared sine
squared theta d theta.
xdy is R squared cosine
squared theta d theta.
So I add those together, and
then I divide by R squared.
And I have now converted
this in terms of theta.
So the integral goes
from 0 to 2pi, the same
as we did, you see, with
ordinary line integrals.
And to summarize this
for you, because I
may have talked kind of
fast, all we're saying
is that the integral around
the closed curve C, dz over z,
is the real part is 0.
And the imaginary part,
the coefficient of i,
is integral from
0 to 2pi R squared
sine squared theta plus cosine
squared theta d theta over R
squared.
Notice that R cannot be
0 because C is the circle
of radius R centered
at the origin.
On that circle--
I mean, R is a
positive number, here.
So it's just not 0.
I can cancel the Rs because the
radius of a circle can't be 0.
And all I'm left with
is sine squared theta
plus cosine squared
theta, which is 1.
The integral d theta from
0 to 2pi is just 2pi.
So this integral is just 2pi i.
All right?
And the second method is
to use the formula what?
Integral f of z dz is the
integral f of z of theta dz
d theta times d theta.
In this case, notice
that the circle centered
at the origin with radius R
is given in polar complex form
by z equals R e to the i theta.
Remember, in polar form,
R is the magnitude.
And theta is the angle.
To say that you're on
the circle of radius R
centered origin simply says
that the magnitude of the point
must be R, the radius.
And the angle can be any angle
whatsoever between 0 and 2pi
if you wanted to reverse
the curve C just one time.
At any rate, from this, notice
that dz d theta is simply what?
iRe to the i theta.
Therefore, dz over z around
C is just the integral
from 0 to 2pi.
f of z of t-- in this case,
that's f of z of theta--
that's 1 over z of theta.
z of theta is Re to the i theta,
dz d theta, which is this,
times d theta.
You see, I've now converted
this into a complex valued
function of a real variable.
At any rate, making all
of these substitutions
and simplifying, dz d theta
being iRe to the i theta,
and z being Re to the
i theta, these cancel,
leaving only a factor of i.
The i comes outside.
I have i integral 0 to 2pi d
theta, which is just 2pi i.
Again, notice the
answers are the same
because the techniques
are equivalent.
Which of the two ways is better?
It depends on yourself,
as to which way
you feel more comfortable with.
It depends on the particular
problem that you're
dealing with and the like.
We'll emphasize these
things in the exercises.
But for the time
being, I just hope
that you have a
feeling as to what
we mean by integrating a
complex valued function
around a particular contour.
And by the way,
the subject called
topology finds a natural
inroad to the study
of complex variables
from this point of view.
Namely, I call this "rubber
sheet" geometry simply
because of a properly
that I'll mention
for you in a few moments.
But the idea is this.
One of the very
interesting factors
about what we've just done, one
of the interesting byproducts
is this.
Let's suppose I'm integrating
f of z dz along a simple curve
C1.
And let's suppose that C2 is
another curve that encloses C1.
And let's suppose
that f is analytic
on the boundaries
C1 and C2 and also
in the region between them.
In other words, I want to
compute f of z dz along C1.
I want to compute
f of z dz along C2.
And all I know is that
f of z is analytic
on and between C1 and C2.
Then the amazing result is
that the integral around C1,
f of z dz, is equal to the
integral around C2, f of z dz.
That's where the word "rubber
sheet" geometry comes in.
What you're saying
is that if there
are no bad spots
between the two curves,
if I were to visualize this
curve as being a rubber band,
by stretching that
rubber band, I
could make it take the
form of the curve C2.
And what you're saying is
given any curve that you're
integrating around, if I
can stretch that curve out
into any position I want
without breaking it,
I stretch that curve
out in such a way
that as that curve
is being stretched,
it never goes through any
points at which f of z
fails to be analytic.
Then it turns out that the
integral around the new curve
is the same as the integral
around the old curve.
And the easiest
way to prove this
is by the method of
making cuts here.
Remember we showed in our
lecture on Green's theorem
that I could make a cut.
Actually, in the lecture
on Green's theorem,
I made two cuts so that
I could vividly show
you the two separate pieces.
All you need is one cut.
See, let me cut the
region this way.
And to emphasize
what I've done here,
I'll pull this apart
just so that we
can see what's happened here.
You see, I've made the cut.
And now, here's what I'm saying.
This S and F stands
for start and finish.
You see, I'm going to
start at this point here.
I'm going to go along the
simply connected region that I
have here.
Once I make the cut, this is
a simply connected region.
Notice that if I go along
here, I am doing what?
I am integrating a
complex valued function
over a simply connected region,
where the function is analytic.
Consequently, whatever
this integral turns out
to be-- in other
words, the integral f
of z dz along this
particular contour
must be 0 by our
previous result,
that if a function is
analytic, the integral
around the closed curve is 0.
Now, here's the point.
Notice that with
this cut in here,
when I look at
the line integral,
I traverse to cut twice--
once in one sense, once
in the opposite sense--
so that cancels.
See?
That part cancels out.
Notice that this integral
around here, this part,
is just the integral around
the closed curve of C2 f of z
dz because after all, we're
assuming that this is together.
I've just separated
it here so that we
can see what's happening.
Notice that the
inner curve is C1,
but in the opposite orientation.
You see, C1 is
traversed in this sense.
But the inner curve
here is traversed
in the opposite sense.
So this integrand is just minus
the integral C1 f of z dz.
So if we put this whole
thing together, what we have
is that this integral
with the cut along C
is, on the one hand, 0.
On the other hand, it's the
integral around C2 f of z dz
minus the integral
around C1 f of z dz.
Because this expression is 0,
it means that this equals this.
And that proves the
result that we wanted.
In other words,
this must equal this
because the difference is 0.
How is this useful in the
theory of complex integration?
Well, let me give
you an example.
Let's go back to our old
friend, integrating dz over z
along the closed curve C.
But now, the closed curve C
is going to be much
messier than before.
It's not going to
be a nice circle.
The closed curve C is going
to be something like this.
And what I'd like to do is to
find the value of this integral
along the curve C.
Now, the point is I could
get into a mess trying
to express C parametrically.
But the beauty is what?
First of all, notice that if
C doesn't enclose the origin,
the integral is just 0
because the only place
that the integrand 1 over
z fails to be analytic
is when z is 0.
Consequently, if I were to
take a region like this--
see, call this my C. If I were
to take a region like this,
notice that the
integral around here
would just be 0 because
in and on this region,
the function 1
over z is analytic.
At any rate, I know that
I'm in a bad situation
here because my function fails
to be analytic over here.
The integral around the closed
curve doesn't have to be 0.
However, what I do know is
how to find the integral
around this very
nice curve C1, where
C1 happens to be a circle
centered at the origin.
You see, the integral
around C1 was just 2pi i.
We did that in our
earlier example.
The important
point is to observe
that between the circle
C1 and the given curve
C, in that particular region,
including the two curves C
and C1, the function
1 over z is analytic.
Consequently, the integral
dz over z along the curve C
is the same as the integral
dz over z along the curve C1.
And therefore, its value
will also be 2 pi i.
Now, at any rate, all
I've tried to show you
in this particular
lecture, and I
hope this part has come
through fairly clearly,
is the fact that we can handle
complex valued functions
in terms of integration
and that this
leads to a treatment
of line integrals.
It leads to examining certain
types of line integrals
along certain regions
in the xy plane, which
we can call the Argand diagram.
We will save applications
or identifications
with the real world
for the exercises.
But all I hope by
now is that we have
had a little bit of insight
to the various aspects
of mathematical analysis as it
applies to complex variables.
This winds up our treatment
of complex variables
as far as this
course is concerned.
And next time, we will begin
a new block of material
and begin, again, a
survey-type investigation
of a subject known as ordinary
differential equations.
At any rate then, until
next time, goodbye.
Funding for the
publication of this video
was provided by the Gabriella
and Paul Rosenbaum Foundation.
Help OCW continue to provide
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