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.
