I assume from high school you know how to add and multiply
complex numbers using the relation i squared equals
negative one. I'm a little less certain that
you remember how to divide them. I hope you read last night by
way of preparation for that, but since that's something
we're going to have to do a lot of a differential equations,
so remember that the division is done by making use of the
complex conjugate. So, if z is equal to a plus bi,
some people write a plus ib,
and sometimes I'll do that too if it's more
convenient. Then, the complex conjugate is
what you get by changing i to negative i.
And, the important thing is that the product of those two is
a real number. The product of these is a
squared minus the quantity ib all squared,
which makes a squared plus b squared because i
squared is negative one.
So, the product of those, that's what you multiply if you
want to multiply this by something to make it real.
You always multiplied by its complex conjugate.
And that's the trick that underlines the doing of the
division. So, for example,
I better hang onto these or I'll never remember all the
examples. Suppose, for example,
we wanted to calculate (two plus i) divided by (one minus 3
i). To calculate it means I want to
do the division; I want to express the answer in
the form a plus bi. What you do is multiply the top
and bottom by the complex conjugate of the denominator in
order to make it real. So, it's (one plus 3i) divided
by (one plus 3i), as they taught you in
elementary school, that is one,
in a rather odd notation; therefore, multiplying doesn't
change the value of the fraction.
And so, the denominator now becomes 1 squared plus 3
squared, which is ten.
And, the numerator is, learn to do this without
multiplying out four terms. You must be able to do this in
your head. And, you always do it by the
grouping, or post office method, whatever you want to call it,
namely, first put down the real part, which is made out of two
times one minus three times one. So, that's negative one.
And then, the imaginary part, which is i times one.
That's one, coefficient one, plus 6i.
So, that makes 7i. Now, some people feel this
still doesn't look right, if you wish,
and for some places and differential equations,
it will be useful to write that as minus one tenth plus seven
tenths i. And, now it's perfectly clear
that it's in the form a plus bi. So, learn to do that if you
don't know already. It's going to be important.
Now, the main thing today is the polar representation,
which sometimes they don't get to in high school.
And if they do, it's usually not in a grown
up-enough in a form for us to be able to use it.
So, I have to worry about that little bit.
The polar representation, of course, is nominally just
the switch to polar coordinates. If here's a plus bi,
then this is r, and that's theta.
And therefore, this can be written as,
in the polar form, that would be r cosine theta
plus i, or r cosine theta.
That's the A part.
And, the B part is, the imaginary part is r
sin(theta) times i.
Now, it would be customary, at this point,
to put the i in front, just because it looks better.
The complex numbers are commutative, satisfied to
commutative law of multiplication,
which means it doesn't matter in multiplication whether you
put i in front or behind. It's still the same answer.
So, this would be r cosine theta plus i times
r sine theta,
which, of course, will factor out,
and will make it cosine theta plus i sine theta.
Now, it was Euler who took the
decisive step and said, hey, look, I'm going to call
that e to the i theta.
Now, why did he do that? Because everything seemed to
indicate that it should. But that's certainly worth the
best color we have, which is what?
We are getting low here. Okay, nonetheless,
it's worth pink. I will even give him his due,
Euler. Sometimes it's called Euler's
formula, but it really shouldn't be.
It's not a formula. It's a definition.
So, in some sense, you can't argue with it.
If you want to call putting a complex number in a power,
and calling it that, you can.
But, one can certainly ask why he did it.
And the answer, I guess, is that all the
evidence seemed to point to the fact that it was the thing to
do. Now, I think it's important to
talk about a little bit because I think it's,
in my opinion, if you're seeing this for the
first time, even if you read about it last night,
it's a mysterious thing, and one needs to see it from
every possible point of view. It's something you get used to.
You will never see it in a sudden flash of insight.
It will just get as familiar to you as more common arithmetic,
and algebraic, and calculus processes are.
But, look. What is it we demand?
If you're going to call something an exponential,
what is it we want an exponential to do,
what gives an expression like this the right to be called e to
the i theta? The answer is I can't creep
inside Euler's mind. It must have been a very big
day of his life. He had a lot of big days,
but when he realized that that was the thing to write down as
the definition of e to the i theta.
But, what is it one wants of an exponential?
Well, the high school answer surely is you want it to satisfy
the exponential law. Now, to my shock,
I realize a lot of people don't know.
In my analysis class, these are some math majors,
or graduate engineers in various subjects,
and if I say prove such and such using the exponential law,
I'm sure to get at least half a dozen e-mails asking me,
what's the exponential law? Okay, the exponential law is a
to the x times a to the y equals a to the x plus y:
the law of exponents.
That's the most important reason why, that's the single
most important thing about exponents, are the way one uses
them. And, this is the exponential
function, called the exponential function because all this
significant stuff is in the exponents.
All right, so it should satisfy-- we want,
first of all, the exponential law to be true.
But that's not all. That's a high school answer.
An MIT answer would be, I mean, why is e to the x such
a popular function? Well, of course,
it does satisfy the exponential law, but for us,
an even more reasonable thing. It's the function,
which, when you differentiate it, you get the same thing you
started with. And, it's apart from a constant
factor, the only such function. Now, in terms of differential
equations, it means that it's the solution that e to the,
let's be a little generous, make it e to the ax.
No, better not to use x because complex numbers tend to be
called x plus iy. Let's use t as a more neutral
variable, which is standing outside the fray,
as it were. It satisfies the relationship
that it's the solution, if you like,
to the differential equation. That's a fancy way of saying
it. dy / dt equals a times y.
Now, of course,
that is not unique. We could make it unique by
putting in an initial value. So, if I want to get this
function and not a constant times it, I should make this an
initial value problem and say that y of zero should be one.
And now, I will get only the
function, e to the at. So, in other words,
that characterizes this function.
It's the only function in the whole world that has that
property. Now, if you're going to call
something e to the i theta, we want that to be true.
So, here are my questions. Is it true that e to the i
theta one, let's use that, times e to the i theta two,
see, I'm on a collision course
here, but that's easily fixed. Is that equal to e to the i
(theta one plus theta two)?
If that turns out to be so, that's a big step.
What would we like to be true here?
Well, will it be true that the derivative, with respect to t of
e to the i theta, I would like that to be equal
to i times e to the i theta.
So, question,
question. I think those are the two most
significant things. Now, the nodes do a third
thing, talk about infinite series.
Since we haven't done infinite series, anyway,
it's not officially part of the syllabus, the kind of power
series that are required. But, I will put it down for the
sake of completeness, as people like to say.
So, it should behave right. The infinite series should be
nice. The infinite series should work
out. There is no word for this,
should work out, let's say.
I mean, what's the little music?
Is that some weird music idea, or is it only me that hears it?
[LAUGHTER] Yes, Lord.
I feel I'm being watched up there.
This is terrible. So, there's one guy.
Here's another guy. And, I won't put a box around
the infinite series, since I'm not going to say
anything about it. Now, these things,
in fact, are both true. Otherwise, why would I be
saying them, and why would Euler have made the formula?
But, what's interesting to see is what's behind them.
And, that gives you little practice also in calculating
with the complex numbers. So, let's look at the first
one. What will it say?
It is asking the question. It says, please,
calculate the product of these two things.
Okay, I do it, I'm told.
I will calculate the product of cosine theta one plus i cosine
theta two-- Sine. Sine theta one.
That's e to the i
theta one, right?
So, that corresponds to this. The other factor times the
other factor, cosine theta two plus i sine
theta two.
Okay, what does that come out to be?
Well, again, we will use the method of
grouping. What's the real part of it?
The real part of it is cosine theta one cosine theta two.
And then, there's a real part,
which comes from these two factors.
It's going to occur with a minus sign because of the i
squared. And, what's left is sine theta
one sine theta two.
And then, the imaginary part, I'll factor out the i.
And then, what's left, I won't have to keep repeating
the i. So, it will have to be sine
theta one cosine theta two.
And, the other factor will be cosine theta one sine theta
two-- plus sine theta two cosine theta one.
Well, it looks like a mess,
but, again, high school to the rescue.
What is this? The top thing is nothing in
disguise, but it's a disguised form of cosine (theta one plus
theta two).
And the bottom is sine of (theta one plus theta two).
So, the product of these two
things is this, and that's exactly the formula.
In other words, this formula is a way of
writing those two trigonometric identities for the cosine of the
sum and the sine of the sum. Instead of the two identities
taking up that much space, written one after the other,
they take up as much space, and they say exactly the same
thing. Those two trigonometric
identities are exactly the same as saying that e to the i theta
satisfies the exponential law.
Now, people ask, you know, what's beautiful in
mathematics? To me, that's beautiful.
I think that's great. Something long turns into
something short, and it's just as good,
and moreover, connects with all these other
things in the world, differential equations,
infinite series, blah, blah, blah,
blah, blah. Okay, I don't have to sell
Euler. He sells himself.
Now, how about the other one? How about the other one?
Now, that's obviously, I haven't said something
because for one thing, how do you differentiate if
there's theta here, and t down there.
Okay, that's easily fixed. But, how do I differentiate
this? What kind of a guy is e to the
i theta? Well, if I write it out,
take a look at what it is. It's cosine theta plus i sine
theta.
As theta varies, it's a function.
The variable is real. Theta is a real variable.
Its angle in radians, but it runs from negative
infinity to infinity. So, if you think of functions
as a black box, what's going in is a real
number. But, what's coming out is a
complex number. So, schematically,
here is the e to the i theta box,
if you like to think that way, theta goes in,
and that's real, and a complex number,
this particular complex number goes out.
So, one, we'd call it, I'm not going to write this
down because it's sort of pompous and takes too long.
But, it is a complex valued function of a real variable.
You got that? Up to now, we studied real
functions of real variables. But now, real valued functions
of real variables, those are the kind calculus is
concerned with. But now, it's a complex-valued
function because the variable is real.
But, the output, the value of the function is a
complex number. Now, in general,
such a function, well, maybe a better say,
complex-valued, how about complex-valued
function of a real variable, let's change the name of the
variable. t is always a real variable.
I don't think we have complex time yet, although I'm sure
there will be someday. But, the next Einstein appears.
A complex-valued function of a real variable,
t, in general, would look like this.
t goes in, and what comes out? Well: a complex number,
which I would then have to write this way.
In other words, the real part depends on t,
and the imaginary part depends upon t.
So, a general function looks like this, a general
complex-valued function. This is just a special case of
it, where the variable has a different name.
But, the first function would be cosine t, and the second
function would be sine t. So, my only question is,
how do you differentiate such a thing?
Well, I'm not going to fuss over this.
The general definition is, with deltas and whatnot,
but the end result of a perfectly fine definition is,
you differentiate it by differentiating each component.
The reason you don't have to work so very hard is because
this is a real variable, and I already know what it
means to differentiate a function of a real variable.
So, I could write it this way, that the derivative of u plus
iv, I'll abbreviate it that way, this means the derivative,
with respect to whatever variable, since I didn't tell
you what the variable in these functions were,
well, I don't have to tell you what I'm differentiating with
respect to. It's whatever was there because
you can't see. And the answer is,
it would be the derivative of u plus i times the derivative of
v. You differentiate it just the
way you would if these were the components of a motion vector.
You would get the velocity by differentiating each component
separately. And, that's what you're doing
here. Okay, now, the importance of
that is that it at least tells me what it is I have to check
when I check this formula. So, let's do it now that we
know what this is. We know how to differentiate
the function. Let's actually differentiate
it. That's fortunately,
by far, the easiest part of the whole process.
So, let's do it. So, what's the derivative?
Let's go back to t, our generic variable.
I want to emphasize that these functions, when we write them as
functions, that theta will almost never be the variable
outside of these notes on complex numbers.
It will normally be time or something like that,
or x, a neutral variable like x.
So, what's the derivative of e to the i theta?
I'm hoping that it will turn out to be i e to the i theta,
and that the yellow law may be
true just as the green one was. Okay, let's calculate it.
It's the derivative, with respect to,
unfortunately I can convert t's to thetas, but not thetas to
t's. C'est la vie,
okay. Times cosine t plus i sine t,
and what's that?
Well, the derivative of cosine t, differentiating the real and
imaginary parts separately, and adding them up.
It's negative sine t, plus i times cosine t.
Now, let's factor out at the i,
because it says if I factor out the i, what do I get?
Well, now, the real part of what's left would be cosine t.
And, how about the imaginary part?
Do you see, it will be i sine t because i times i
gives me that negative one.
And, what's that? e to the it.
i times e to the i t.
So, that works too. What about the initial
condition? No problem.
What is y of zero? What's the function at zero?
Well, don't say right away, i times zero is zero,
so it must be one. That's illegal because,
why is that illegal? It's because in that formula,
you are not multiplying i times theta.
I mean, sort of, you are, but that formula is
the meaning of e to the i theta.
Now, it would be very nice if this is like,
well, anyway, you can't do that.
So, you have to do it by saying it's the cosine of zero plus i
times the sine of zero.
And, how much is that? The sine of zero is zero.
Now, it's okay to say i times zero is zero because that's the
way complex numbers multiply. What is the cosine of zero?
That's one. So, the answer,
indeed, turns out to be one. So, this checks,
really, from every conceivable standpoint down as I indicated,
also from the standpoint of infinite series.
So, we are definitely allowed to use this.
Now, the more general exponential law is true.
I'm not going to say much about it.
So, in other words, e to the a, this is really a
definition. e to the (a plus ib)
is going to be, in order for the general
exponential law to be true, this is really a definition.
It's e to the a times e to the ib.
Now, notice when I look at the-- at any complex number,
-- -- so, in terms of this,
the polar form of a complex number, to draw the little
picture again, if here is our complex number,
and here is r, and here is the angle theta,
so the nice way to write this complex number is r e to the i
theta. The e to the i theta
is, now, why is that?
What is the magnitude of this? This is r.
The length of the absolute value, I didn't talk about
magnitude in argument. I guess I should have.
But, it's in the notes. So, r is called the modulus.
Well, the fancy word is the modulus.
And, we haven't given the complex number a name.
Let's call it alpha, modulus of alpha,
and theta is called, it's the angle.
It's called the argument. I didn't make up these words.
There, from a tradition of English that has long since
vanished, when I was a kid, and you wanted to know what a
play was about, you looked in the playbill,
and it said the argument of the play, it's that old-fashioned
use of the word argument. Argument means the angle,
and sometimes that's abbreviated by arg alpha.
And, this is abbreviated,
of course, as absolute value of alpha, its length.
Okay, the notes give you a little practice changing things
to a polar form. I think we will skip that in
favor of doing a couple of other things because that's pretty
easy. But let me, you should at least
realize when you should look at polar form.
The great advantage of polar form is, particularly once
you've mastered the exponential law, the great advantage of
polar form is it's good for multiplication.
Now, of course, you know how to multiply
complex numbers, even when they are in the
Cartesian form. That's the first thing you
learn in high school, how to multiply a plus bi times
c plus di. But, as you will see,
when push comes to shove, you will see this very clearly
on Friday when we talk about trigonometric inputs to
differential equations, --
-- that the changing to complex numbers makes all sorts of
things easy to calculate, and the answers come out
extremely clear, whereas if we had to do it any
other way, it's a lot more work. And worst of all,
when you finally slog through to the end, you fear you are
none the wiser. It's good for multiplication
because the product, so here's any number in its
polar form. That's a general complex
number. It's modulus times e to the i
theta times r two e to the i theta two--
Well, you just multiply them as
ordinary numbers. So, the part out front will be
r1 r2, and the e to the i theta parts gets
multiplied by the exponential law and becomes e to the i
(theta one plus theta two) --
-- which makes very clear that the multiply geometrically two
complex numbers, you multiply the moduli,
the r's, the absolute values, how long the arrow is from zero
to the complex number, multiply the moduli,
and add the arguments. So the new number,
its modulus is the product of r1 and r2.
And, its argument, its angle, polar angle,
is the sum of the old two angles.
And, you add the angles. And, you put down in your books
angles, but I'm being photographed,
so I'm going to write arguments.
In other words, it makes the geometric content
of multiplication clear, in a sense in which this is
extremely unclear. From this law,
blah, blah, blah, blah, blah, whatever it turns
out to be, you have not the slightest intuition that this is
true about the complex numbers. That first thing is just a
formula, whereas this thing is insightful representation of
complex multiplication. Now, I'd like to use it for
something, but before we do that, let me just indicate how
just the exponential notation enables you to do things in
calculus, formulas that are impossible to remember from
calculus. It makes them very easy to
derive. A typical example of that is,
suppose you want to, for example,
integrate (e to the negative x) cosine x.
Well, number one, you spend a few minutes running
through a calculus textbook and try to find out the answer
because you know you are not going to remember how to do it.
Or, you run to a computer, and type in Matlab and
something. Or, you fish out your little
pocket calculator, which will give you a formula,
and so on. So, you have aides for doing
that. But, the way to do it if you're
on a desert island, and the way I always do it
because I never have any of these little aides around,
and I cannot trust my memory, probably a certain number of
you remember how you did it at high school, or how you did it
in 18.01, if you took it here. You have to use integration by
parts. But, it's one of the tricky
things that's not required on an exam because you had to use
integration by parts twice in the same direction,
and then suddenly by comparing the end product with the initial
product and writing an equation. Somehow, the value falls out.
Well, that's tricky. And it's not the sort of thing
you can waste time stuffing into your head, unless you are going
to be the integration bee during IAP or something like that.
Instead, using complex numbers is the way to do this.
How do I think of this, cosine x?
What I do, is I think of that e to the negative x cosine x
is the real part, the real part of what?
Well, cosine x is the real part of e to the ix.
So, this thing, this is real.
This is real, too.
But I'm thinking of it as the real part of e to the ix.
Now, if I multiply these two
together, this is going to turn out to be, therefore,
the real part of e to the minus x.
I'll write it out very pompously, and then I will fix
it. I would never write this,
you are you. Okay, it's e to the minus x
times, when I write cosine x plus i sine x,
so it is the real part of that is cosine x.
So, it's the real part of, write it this way for real part
of e to the, factor out the x, and what's up there is
(negative one plus i) times x.
Okay, and now, so, the idea is the same thing
is going to be true for the integral.
This is going to be the real part of that,
the integral of e to the (minus one plus i) times x dx.
In other words,
what you do is, this procedure is called
complexifying the integral. Instead of looking at the
original real problem, I'm going to turn it into a
complex problem by turning this thing into a complex
exponential. This is the real part of that
complex exponential. Now, what's the advantage of
doing that? Simple.
It's because nothing is easier to integrate than an
exponential. And, though you may have some
doubts as to whether the familiar laws work also with
complex exponentials, I assure you they all do.
It would be lovely to sit and prove them.
On the other hand, I think after a while,
you find it rather dull. So, I'm going to do the fun
things, and assume that they are true because they are.
So, what's the integral of e to the (minus one plus i) x dx?
Well, if there is justice in heaven, it must be e to the
(minus one plus i) times x divided by minus one plus i.
In some sense,
that's the answer. This does, in fact,
give that. That's correct.
I want the real part of this. I want the real part because
that's the way the original problem was stated.
I want the real part only. So, I want the real part of
this. Now, this is what separates the
girls from the women. [LAUGHTER] This is why you have
to know how to divide complex numbers.
So, watch how I find the real part.
I write it this way. Normally when I do the
calculations for myself, I would skip a couple of these
steps. But this time,
I will write everything out. You're going to have to do this
a lot in this course, by the way, both over the
course of the next few weeks, and especially towards the end
of the term where we get into a complex systems,
which involve complex numbers. There's a lot of this.
So, now is the time to learn to do it, and to feel skillful at
it. So, it's this times e to the
negative x times e to the ix,
which is cosine x plus i sine x.
Now, I'm not ready, yet, to do the calculation to
find the real part because I don't like the way this looks.
I want to go back to the thing I did right at the very
beginning of the hour, and turn it into an a plus bi
type of complex number.
In other words, what we have to do is the
division. So, the division is going to
be, now, I'm going to ask you to do it in your head.
I multiply the top and bottom by negative one minus I.
What does that put in the denominator?
One squared plus one squared: Two.
And in the numerator, negative one minus i.
This is the same as that.
But now, it looks at the form a + bi.
It's negative one over two minus i times one half.
So, this is multiplied by e to
the minus x and cosine x.
So, if you are doing it, and practice a little bit,
please don't put in all these steps.
Go from here; well, I would go from here to
here by myself. Maybe you shouldn't.
Practice a little before you do that.
And now, what do we do with this?
Now, this is in a form to pick out the real part.
We want the real part of this. So, you don't have to write the
whole thing out as a complex number.
In other words, you don't have to do all the
multiplications. You only have to find the real
part of it, which is what? Well, e to the negative x
will be simply a factor.
That's a real factor, which I don't have to worry
about. And, in that category,
I can include the two also. So, I only have to pick out the
real part of this times that. And, what's that?
It's negative cosine x.
And, the other real part comes from the product of these two
things. I times negative i is one.
And, what's left is sine x.
So, that's the answer to the
question. That's the integral of e to the
negative x * cosine x.
Notice, it's a completely straightforward process.
It doesn't involve any tricks, unless you call going to the
complex domain a trick. But, I don't.
As soon as you see in this course the combination of e to
ax times cosine bx or sine bx,
you should immediately think,
and you're going to get plenty of it in the couple of weeks
after the exam, you are going to get plenty of
it, and you should immediately think of passing to the complex
domain. That will be the standard way
we solve such problems. So, you're going to get lots of
practice doing this. But, this was the first time.
Now, I guess in the time remaining, I'm not going to talk
about in the notes, i, R, at all,
but I would like to talk a little bit about the extraction
of the complex roots, since you have a problem about
that and because it's another beautiful application of this
polar way of writing complex numbers.
Suppose I want to calculate. So, the basic problem is to
calculate the nth roots of one. Now, in the real domain,
of course, the answer is, sometimes there's only one of
these, one itself, and sometimes there are two,
depending on whether n is an even number or an odd number.
But, in the complex domain, there are always n answers as
complex numbers. One always has n nth roots.
Now, where are they? Well, geometrically,
it's easy to see where they are.
Here's the unit circle. Here's the unit circle.
One of the roots is right here at one.
Now, where are the others? Well, do you see that if I
place, let's take n equal five because that's a nice,
dramatic number. If I place these peptides
equally spaced points around the unit circle, so,
in other words, this angle is alpha.
Alpha should be the angle. What would be the expression
for that? If there were five such equally
spaced, it would be one fifth of all the way around the circle.
All the way around the circle is two pi.
So, it will be one fifth of two pi in radians.
Now, it's geometrically clear that those are the five fifth
roots because, how do I multiply complex
numbers? I multiply the moduli.
Well, they all have moduli one. So, if I take this guy,
let's call that complex number, oh, I hate to give you,
they are always giving you Greek notation.
All right, why not torture you? Zeta.
At least you will learn how to make a zeta in this period,
small zeta, so that's zeta. There's our fifth root of
unity. It's the first one that occurs
on the circle that isn't the trivial one, one.
Now, do you see that, how would I calculate zeta to
the fifth? Well, if I write zeta in polar
notation, what would it be? The modulus would be one,
and therefore it will be simply, the r will be one
for it because its length is one.
Its modulus is one. What's up here?
I times that angle, and that angle is two pi over
five. So, there's just,
geometrically I see where zeta is.
And, if I translate that geometry into the e to the i
theta form for the formula, I see that it must
be that number. Now, let's say somebody gives
you that number and says, hey, is this the fifth root of
one? I forbid you to draw any
pictures. What would you do?
You say, well, I'll raise it to the fifth
power. What's zeta to the fifth power?
Well, it's e to the i two pi / five,
and now, if I think of raising that to the fifth power,
by the exponential law, that's the same thing as
putting a five in front of the exponent.
So, this times five, and what's that?
That's e to the i times two pi.
And, what is that? Well, it's the angle.
If the angle is two pi, I've gone all the way around
the circle and come back here again.
I've got the number one. So, this is one.
Since the argument, two pi, is the same as an
angle, it's the same as, well, let's not write it that
way. It's wrong.
It's just wrong since two pi and zero are the same angle.
So, I could replace this by zero.
Oh dear. Well, I guess I have to stop
right in the middle of things. So, you're going to have to
read a little bit about how to find roots in order to do that
problem. And, we will go on from that
point Friday.