Part I: Complex Variables, Lec 2: Functions of a Complex Variable Video Lecture - Engineering Mathematics

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hi I hope enough time has elapsed since
our last lecture so that you feel at
home doing arithmetic of complex numbers
because you see now what we'd like to do
today is the next phase that leads into
our calculus study of complex numbers
namely what we would like to do is study
functions of a complex variable and
again in terms of a function machine I'm
visualizing
the input of my machine being a complex
number and the output being a complex
number by the way I should mention that
there are situations and we'll cover
these in the hope in the homework
exercises where one sometimes likes the
input to be a real number and the output
a complex number or the input a complex
number and the output a real number but
for reasons that I hope will become
apparent in a moment we're going to
focus our attention on the case where
both the input and the output of complex
numbers noticing that if we do that
especially from a geometrical point of
view a vector point of view our input is
a 2-tuple
say the complex number X comma Y or X
plus iy is represented by the to topple
X comma Y the output W equals F of Z is
what u plus IV say R as a to topple u
comma V the idea being that graphically
we can view a function that map's the
complex numbers into the complex numbers
as a mapping from the XY plane into the
UV plane where the complex number Z is
identified with the pair X comma Y the
complex number W is identified with the
ordered pair u comma V the XY plane is a
domain of our function the UV plane is
the range of our function by way of
example suppose I take the function f of
Z equals Z squared recalling that Z is X
plus iy if I Square this I get as the
real part x squared minus y squared the
imaginary part is 2xy
so as a 2-tuple type mapping F Maps X
comma Y into x squared minus y squared
comma 2xy
in other words you as x squared minus y
squared V is 2x y so in summary notice
that the complex valued function of a
complex variable in other words we're
mapping complex inputs into complex
outputs but that particular mapping F of
Z equals Z squared is equivalent to the
and this is important to the real system
of equations u equals x squared minus y
squared V equals 2x Y you see this is a
system of two real equations and two
unknowns and what I'm saying is that I
cannot distinguish between this problem
and this problem in other words somehow
or other studying complex valued
functions is going to help me replace or
to view from a different vantage point
the study of systems of real functions
but again more about that as we proceed
any rate leaving further drill a further
drill we haven't met any drill leaving
drill to the exercises because I think
conceptually this is all old hat and why
do I say it's old hat well heck we've
spent the whole block of material
working on these kind of mappings also
geometrically we're viewing the complex
numbers as being vectors so that in
another manner of speaking
complex valued functions of a complex
variable are equivalent to what
functions that map two-dimensional space
into two-dimensional space
and we talked about that in our
treatment of vector functions so
whichever way we look at this thing I
think that the drill is fairly
straightforward let's go on the next
thing after we have functions be
ultimately we want to get to derivatives
and integrals and the one concept that
we need for a derivative after we know
the theory of functions is the concept
of a limit right so we want limits next
and what are we going to do letting C
denote the complex numbers just for
purposes of identification suppose we
have a mapping from C into c suppose a
is in C in other words a is some complex
number we define the limit of f of Z as
Z approaches a to BL and I'm hoping that
by this time you're way ahead of me cuz
we've done this many times what am I
going to do here
I am going to work forward mimic the
real definition that was the case for
functions of a real variable or if you
wish for vector functions see this is
complex this is complex I can think of
this as a vector this is a vector same
definition that we had before namely
given epsilon greater than 0 there
exists a delta greater than 0 such that
when the absolute value of Z minus a is
greater than 0 but less than Delta the
absolute value of F of Z minus L is less
than Epsilon
noticing that even though Z a f of Z and
L are complex magnitudes are real so all
of these symbols make sense and do they
capture intuitively what we'd like them
to mean again by way of a quick
pictorial review the answer is exactly
yes namely if we think of wanting to be
with an epsilon of L we simply now in
two dimensions take a circle of radius
epsilon senator Dell and what we're
saying is to be with an epsilon of L we
can find the circle of radius Delta
Center today such that whenever Z is in
this circle f of Z will be in this
circle which is exactly the intuitive
feeling that limit is supposed to convey
because structurally the definition of
limit is the same now as it was in the
real case and in the real case we only
use properties of a definition to prove
our theorems it means that the usual
limit theorems will hold and in
particular by the way if we decide to
write f of Z as a real plus an imaginary
part in other words f of Z is what it's
u plus IV where u and V are functions of
X and y the complex number L has a real
part and an imaginary part L is L 1 plus
il 2 where these are real you see so
this is complex these are real
and the complex number a can be written
in the form a one plus ia2 with a 1 and
a 2 are real and essentially what we're
saying is that if the limit as the
approaches a f of Z is L it must be that
the real part of F of Z approaches the
real part of L and the imaginary part of
F of Z approaches the imaginary part of
L in other words this condition here is
equivalent to a system of two limits in
real variables namely the limit of U of
XY as XY approaches a1 a2 must be the
real number l1 and the limit of V of XY
as XY approaches the real ordered pair a
1 a 2 must be l2 and again what this
means is is that we can often view
systems of limit problems as being the
limit of a complex variable function ok
again that's all we're saying from that
point of view I simply want to rush
through this because this is old hat
there is really nothing new
theoretically here all I want you to see
is that the same structure that held in
our study for real valued functions
holds for complex valued functions but
in particular when we map complex
numbers into complex numbers there are
many real interpretations identified
with the pair of real mappings you with
some function of x and y and via sub
function of x and y at any rate with
that is background we can immediately
launch in to the definition of a
derivative now we may not be able to
immediately launch into what does a
derivative mean this is one of the
beauties you see if mathematical
structure one of the beauties of
mathematical structure is is that when
I'm mimicking a structure all I have to
do is copy it that's what mimic means in
fact the beauty isn't that you can copy
it the beauty is is that since all of
the powerful results of the thing that
you're copying came logically in other
words they were valid conclusions based
on the structure you started with the
new structure that you got by mimicking
the old one will inherit the same
structure provided of course that in
mimicking the resulting expression still
make sense without being around the bush
all I'm saying is why don't we copy
as our definition of f prime of Z Sub
Zero where Z Sub Zero is some complex
number
why don't we mimic what we would have
done for F prime of X sub zero only
replacing the X's by the Z's and if we
do that
remembering what our definition for
derivative was in the case of a real
variable we simply define f prime of Z
zero to be the limit as Delta Z
approaches zero F of Z zero plus Delta Z
minus F of Z zero all over Delta Z the
questions of course that we raise our
first of all does this make sense well
remember F is defined for complex
numbers so and it's an F of a complex
number is a complex number so this is
the difference of two complex numbers
the difference of two complex numbers is
a complex number we're dividing that
difference by another complex number
which by the way can't be 0 remember
Delta Z is not 0 it approaches zero
means that Delta Z is greater than 0 but
less than Delta etc you know little
Delta the same is in the epsilon-delta
definition this says what the Delta Z
gets arbitrarily close to zero but is
never equal to 0 consequently this is
the quotient of two complex numbers the
denominator is not zero consequently as
we saw in the last lecture this is again
a complex number and this makes sense to
compute this limit and we'll actually do
this in a problem before we're through
today and we'll also have a lot of
homework problems on this but to try to
give you some feeling as to what this
thing means geometrically think of this
in terms of U and V again in other words
think of the mapping the XY plane into
the UV plane if w is f of z where f of z
is what u plus IV then f prime of Z 0 is
just the symbol DW DZ where by DW DZ we
mean the limit as Delta Z approaches 0
Delta W over Delta Z and by the way to
help you see that hopefully a little bit
more clearly if this is W Delta W
involves the change in U and a change in
V so Delta W would be Delta u plus I
Delta V Delta Z by definition is Delta X
plus I Delta Y so computing the
derivative involves computing this
and notice that with the exception of
the eye appearing here everything else
involves real numbers the very very
crucial thing to notice is that not only
must this limit exist but it must exist
independently of the direction in which
Delta Z approaches zero this is much
stronger than the corresponding notion
of a directional derivative when we were
talking about real valued functions of
several real variables you see for a
directional derivative to exist in every
direction all you require is that in
each direction the limit exists but that
the limit could be different in two
different directions notice that in our
definition this says what that this
particular limit must exist no matter
what direction Delta Z approaches zero
in now to help you see this
geometrically and for heaven's sakes I'm
not the greatest guy at visualizing
things geometrically myself so if I
either confuse you or else you're not
turned on by this ignore it because we
don't need this geometric interpretation
until next lecture but for those of you
who might be interested in pursuing this
here's what we're saying notice that F
prime of Z 0 is a complex number it's in
the uv-plane complex numbers are like
vectors consequently I can visualize the
complex number as a directed length an
arrow if I take F prime of Z 0 let me
assume it originates at W 0 where W 0 is
the image of Z 0 under F so as a vector
f prime of Z 0 is this now here's what
we're saying pick any point Z 1 near Z 0
any point at all Z 1 it maps into some
point W 1 the directed length from Z 0
to Z 1 is the vector of a complex number
Delta Z the directed distance from W 0
to W 1 is the vector the complex number
Delta W in general one would expect that
when you divide Delta W by Delta Z what
you get to depend very strongly on what
Z 1 is see different values of Z 1 would
have led to different values of W 1 the
fact that the limit of Delta W over
z is F prime of Z zero means that if z1
is very close to zero no matter where
you pick it the ratio Delta W by Delta Z
is essentially a constant in fact what
constant is it it's essentially the
vector f prime of Z zero the complex
number F prime of Z zero that contrary
to what you might suspect that in
general different Z's will lead to such
drastically different w's that the ratio
Delta W by Delta Z is bound to change
what we're saying is if a function is
differentiable that ratio doesn't change
and it's always equal to F prime of Z
zero now the question that comes up is
it's very very strong condition to
insist not only that this limit exists
in every direction but be the same in
all direction it's possible that this
imposes some very strong conditions on
Delta U and Delta V and what we're going
to do is to say look at if this must be
the same for all directions let's pick
two particularly attractive directions
to look at what directions do we like
when we deal with real functions of two
real variables what we like is in that
case we're dealing with what a function
of x and y in which case the usual
derivatives involve the partials with
respect to X and the partial with
respect to Y that means in one case
holding Delta X equal to zero and the
other case we holding X constant so
Delta X is zero the other case you hold
Y constant so Delta Y is zero let's look
at these two cases especially in other
words let's look at this first of all in
the special case that Delta Y is
identically 0 you see if Delta Y is that
identically 0 Delta Z is just Delta X in
that case and so this drops out and the
problem would be to compute the limit as
Delta X approaches 0 Delta u plus I
Delta V over Delta X and saying that
more slowly search you can see it better
if I let Delta Y be identically 0 it
means that I'm allowing Delta Z to
approach 0 horizontally in other words
I'm allowing the approach to be parallel
to the real axis that's how we're
approaching Z 0 now in that case since
Delta Y is 0 Delta Z is just
delta x this is the limit as Delta X
approaches 0 Delta u plus I Delta V over
Delta X that means I can break this down
and write it as Delta u over Delta X
plus I Delta V over Delta X the limit of
a sum is the sum of the limits etc I
take the limits here recalling that the
limit of Delta u over Delta X as Delta X
approaches 0 holding Y constant C Delta
Y is 0 is what I mean by the partial of
U with respect to X similarly the limit
of Delta V over Delta X as Delta X
approaches 0 and Delta Y is 0 in other
words Y is constant is what we mean by
Delta V over Delta X and consequently
what we're saying is if f prime of Z 0
exists it must be the partial with
respect to X plus I times the partial of
V with respect to X and I'm computing
this at Z 0 which corresponds to the
point in the XY plane say X 0 y 0 on the
other hand I know that F prime of Z 0
has to exist and be the same if Delta X
is held to be identically 0 holding
Delta X 0 in other words holding X
constant means that you're now allowing
Z to approach Z 0 Delta Z is approaching
0 parallel to the imaginary axis the Y
axis now in this case what happens if
Delta X is identically 0 this term drops
out Delta Z is just I Delta Y since I is
a constant difference from 0 the only
way at I Delta Y can approach 0 is if
Delta Y approaches 0 consequently with
Delta X 0 this reads what the limit as
delta Y approaches 0 Delta u plus I
Delta V over I Delta Y and if I now
divide each term in here by I Delta Y
you see what's going to happen I'm going
to get a Delta u over I Delta Y here and
the eyes will cancel here so that F
prime of Z 0 will turn out to be the
limit as Delta Y approaches 0 Delta u
over I Delta Y plus Delta V over Delta Y
again taking the limit as Delta Y
approaches 0 remembering that
being held constant and remembering what
the definition for the partials of U and
V with respect to Y are this limit
simply turns out to be this I would like
to write this in the standard form
something real plus something real times
I not 1 over I and so I use that usual
thing of the complex conjugate I
multiply numerator and denominator here
by minus I that gives me my new
numerator is minus I my new denominator
is minus I times I which is minus minus
1 or plus 1 so that f prime of Z 0 is
the partial of V with respect to Y minus
I partial of U with respect to Y or if
you wish it's it's plus I times minus
partial U respect to Y at any rate
without making a big issue over this
since these two must be the same and
since the only way that two complex
numbers can be equal is if they're equal
component by component in other words
the real parts must be equal and the
imaginary parts must be equal I compare
the real parts here and the imaginary
parts and conclude that they must be
equal so that in summary if F is equal
to u plus IV
and F is differentiable and by the way a
very common word that's used when a
complex valued function of a complex
variable is differentiable is that we
say it's analytic if f equals u plus IV
is analytic then the partial of U
respect to X is V sub y there was a
partial view respect to Y and the
partial of Y with respect to Y must be
minus the partial of V with respect to X
you see this equals this and this must
line up with this by the way notice that
we arrived at this result using only the
information of two special directions we
held Delta X 0 and Delta Y 0 so all this
proves is is that if the function is
differentiable this must be true it
happens that the converse is also true
and we're going to prove that as an
exercise in the unit intuitively to
suspect why this is true it's the same
reason as why the directional derivative
is determined by the partials with
respect to x and y in other words if the
partials exist and are continuous
knowing what happens in the X and y
direction is enough to tell you what's
happening in every direction and will
give you the rigorous details as a
exercise later on but to put this from a
different perspective
if either u sub X is not equal to V sub
y or u sub y is not equal to minus V sub
X then u plus IV cannot be an analytic
function and the easiest way to talk
about this is in terms of examples let's
look at this let's take three examples
but hopefully one of the time the first
example is f of Z equals Z squared and
by the way before I go any further
notice that if you had not thought of
any subtleties here and I said well is F
prime of Z the odds are overwhelming
that you would have said to Z right away
I want to show you that really is the
case and I want to do it in two
different ways one is as we've seen
before
Z squared can be written as x squared
minus y squared plus I times two X Y so
that corresponds to u equals x squared
minus y squared V equals 2x Y from which
you see we can we can compute u sub X u
sub y V sub X and V sub y and what we
see right away is that u sub X does
equal V sub y u sub y does equal minus V
sub X so in particular these conditions
are obeyed this has written up in the
text by the way the proof of what we've
just done the conditions are called the
cauchy-riemann conditions that's also
mentioned in the text so I won't take
the time to write that now but the idea
is that what because the cauchy-riemann
conditions are obeyed it means that this
function should be differentiable by the
way knowing that it should be
differentiable it's not the same as
saying that you know what the derivative
is let's now take the alternative
approach that somehow or other we don't
look at U and V but work directly from
our basic definition after all we define
what f prime of Z meant in terms of F of
Z and F of Z plus Delta Z consequently
this should have meaning regardless of
any interpretation in terms of using
these notice that if I write this
by definition of this this quotient F of
Z zero plus Delta Z minus F of Z zero
over Delta Z is just Z zero plus Delta Z
squared minus Z zero squared over Delta
Z the binomial theorem works for complex
numbers the same way as it does for real
numbers in fact that was what motivated
our definition for multiplying complex
numbers if you recall our lecture of
last time was to mimic what happens in
the real case this the same as before
turns out to be what Z 0 squared which
kills off this minus Z 0 squared plus 2
Z 0 Delta Z plus Delta Z squared all
over Delta Z I'm assuming I'm not going
to take the limit as Delta Z approaches
0 that means in particular the Delta Z
is not 0 because Delta Z is not 0 I can
divide through by it that gives me 2 Z 0
plus Delta Z and now notice if I let
Delta Z approach 0 Z 0 is a fixed number
that doesn't change as Delta Z
approaches 0 on the other hand Delta Z
approaches 0 as Delta Z approaches zero
that's a truism Delta Z approaches 0 no
matter what the direction of approach is
that is Delta Z goes to 0 it goes to 0
this term does drop out and you wind up
with the anticipated result that F prime
of Z 0 is to Z 0 notice again
structurally that in the case where
there is an analogy between the complex
variable and the real variable the steps
are verbatim because the definitions
were mimicked to be verbatim that you
cannot tell the difference between the
proofs that F prime of Z is to Z in this
case with the proof that F prime of X
was to X when f of X was equal to x
squared in the real case or as another
example one would expect that the
derivative of F of Z if f of Z is 1 over
Z should be minus 1 over Z squared
except that when Z is 0
this shouldn't be differentiable I leave
it again as a voluntary exercise for you
to apply this definition to 1 over Z 1
over Z plus Delta Z etc and show that
step by step you get the same result as
in the corresponding problem involving 1
over X
to show you how this works in terms of
using these just for the drill if Z is X
plus iy one over Z is 1 over X plus iy
multiplying numerator and denominator by
the complex conjugate X minus iy here my
denominator becomes x squared plus y
squared okay my numerator is X minus iy
that's the same as x over x squared plus
y squared plus I times minus y over x
squared plus y squared this is the U
part this is the V part
u equals x over x squared plus y squared
V equals minus y over x squared plus y
squared i again leave it as an exercise
for you to show that the cauchy-riemann
conditions are obeyed in other words
that the partial of your respect to x
equals the partial of Y with respect to
Y the partial of U with respect to Y is
minus the partial of V with respect to X
and the only place that this isn't true
turns out to be when this denominator is
0 the only time this denominator can be
0 is when x and y are both 0 in other
words the color remind conditions here
are obeyed except when x equals y equals
0 and that of course implies what that Z
is 0 the real and imaginary parts are 0
and what we're saying is that 1 over Z
is analytic except when Z is 0 and in
fact the derivative is minus 1 over Z
squared for a third example I would like
to pick one that we can't that has no
analogue in the real case I want to pick
a genuine complex valued function of a
complex variable and we'll pick a very
simple one and the reason I pick this
simple one is twofold one is the
computations are easy to do the cut the
result is easy to interpret
geometrically and thirdly and most
importantly I mentioned before that it
is not easy for a complex valued
function to have a derivative the
directional derivative must exist in
every direction and be the same that's a
very stringent condition so I picked
this third example to show you how a
relatively simple complex valued
function of a complex variable can have
no derivative and the simple functions
I've Cho
is the one which maps the complex
conjugate the number into its complex
conjugate the complex number Z is mapped
by F into f bar in terms of two tuples
if the input of the f machine is X comma
Y the output will be X comma minus y a
very simple thing right
very straightforward easy function
machine to work with here notice in this
case that U is X V is minus y the
partial of U with respect to X is 1 the
partial of V with respect to Y is minus
1 1 is not equal to minus 1 consequently
U sub X is not equal to V sub y that
means that the cauchy-riemann conditions
are not obeyed by what we just saw a few
minutes ago that in turn means that this
function does not have a derivative well
that's nice to know that the function
doesn't have a derivative it would be
even nicer if we could see why it didn't
and the best way to do this is again go
back to the definition that what we're
going to ultimately do to take the
derivative is we're going to take Delta
W where W is Z bar see Delta W over
Delta Z and take the limit as Delta Z
approaches zero along a various
direction
well Delta W you see is just Delta X
minus I Delta Y Delta Z is Delta X plus
I Delta Y to keep in mind that I'm going
to move in a particular direction in the
XY plane to approach Z 0 I'm going to
set up the slope over here namely in
this expression I'm going to assume that
Delta X is not 0 and what I'm going to
do is divide through numerator
denominator by Delta X that gives me 1
minus I times Delta Y over Delta X over
1 plus I Delta Y over Delta X and as I
approach the point Z 0 you see as I
approach the point Z 0 in the XY plane
Delta Y over Delta X approaches the
slope of the curve at Z 0 which we'll
call dy DX and I wind up with this now
to simplify this let me assume that I
approach Z 0 along some straight line
let me pick a fixed straight line I'll
pick a straight line which goes through
Z 0 and how
slope M so that in that case dy/dx is
the same as Delta X as M for all points
on the on this particular line okay so
what is dy/dx in this case it's M
consequently what is Delta W over Delta
Z in this case and by the way notice
what I'm saying here what I'm saying is
this is this line segment Z 0 maps into
W Z 0 Z 1 maps into W 1 sufficiently
close to Z 0 I can assume that the image
of this straight line is a straight line
that joins W 0 to W 1 and the ratio that
I want is this length divided by this
length where by dividing lengths I mean
the usual complex variable
interpretation dividing magnitudes and
subtracting arguments at any rate what
we're saying is look at that Delta W
over Delta Z in this case is just 1
minus I M over 1 plus I M and notice
that as delta Z approaches 0 in this
direction this limit exists in fact it's
1 there's no variable in here it's 1
minus I M over 1 plus I M but notice
that even though that limit exists see
that limit exists the value depends on M
if I change em I change this ratio and
because that limit depends on M it means
that even though the directional
derivative exists different directions
will give me see if I come in along this
line I can compute this ratio of Delta W
over Delta Z what will I get in that
case 1 minus I M 1 over 1 plus I M 1
that limit will exist but it will be
unequal to this one you see the
derivative exists in each direction but
it's not the same in each direction
that's how subtle this is I will leave
the remaining exercises to let the
remaining exercises but exercises to hit
home at the remaining fine points what I
wanted to do
in closing today's lesson was to show
you one more tremendous connection with
complex variables or between complex
variables and real problems in the
physical world you may recall from some
previous exercises in an earlier block
that we say that U of XY satisfies
Laplace's equation if the second partial
of your respect to X plus the second
partial of U respect to Y is 0 in fact
if u denotes temperature distribution at
the point the temperature at the point X
comma Y this equation is what's known as
the steady state equation we're not
going to worry about the physical
ramifications here what is important is
that to solve this partial differential
equation which comes up in the physical
world this is a real partial
differential equation meaning real in
two sense it comes up in the real world
and also in the sense that U is a real
valued function of x and y now the
interesting point is this that if u plus
IV is analytic what do we know about u
plus IV being analytic it means that the
partial of Y or respect to X is V sub y
the partial of U with respect to Y is
minus the partial of V with respect to X
if I differentiate both sides here with
respect to X and both sides here with
respect to Y I get u sub xx equals V sub
y X u sub YY is minus V sub X Y assuming
continuity so that these are equal this
says by addition that if I add these two
I get the second partial of Y or respect
to X plus the second partial of Y or
respect to Y is 0 a similar argument
would have shown that the same is true
for the imaginary part that the second
partial with respect to X plus the
second partial of V with respect to Y
also will be 0 in other words the real
and the imaginary parts which by
themselves are real valued functions of
two real variables the real and
imaginary parts each satisfy Laplace's
equation the converse by the way is also
true I'll mention that in a moment but
for the time being let me close today's
lesson with an example we already know
that f of Z equals Z squared is analytic
it's differentiable the real part of Z
squared U is x squared minus y squared
the imaginary part v is 2xy what that
tells me we are even checking is that if
I were to form U sub
xx + u sub YY or V sub xx + V sub YY
those would be 0 because of the fact
that they satisfy the Cauchy remind
conditions which is what we'll talk
about in more detail during the
exercises and we'll bring this up more
in the next lecture but for the time
being you see what I want to close on
the side from the calculus was the
application I want you to see that
solving Laplace's equation is a very
important real-world problem the
interesting point is that the solutions
of Laplace's equations have to be the
real and imaginary parts of analytic
functions and there aren't that many
analytic functions and consequently by
knowing analytic functions we have a
good hold on what solutions of Laplace's
equation in the real world look like
well I think that's enough for one
session work on the exercises and next
time we're going to pound out some other
interesting ramifications of what it
means for a complex valued function to
be differentiable at any rate then until
next time goodbye
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