GILBERT STRANG:
So complex numbers are going to come in to today's
video, and let me show you why. So I'm going to
solve the problem. The differential equation
we know-- first order, linear with a source term,
but now the source term has both the cosine
and the sine. And the sine, you notice, has
this imaginary square root of minus 1 involved. So I'll call this y
complex with a little c, because the answer
is going to come out a complex number instead
of a real number. So the previous lecture
solved it with the cosine, and now we're going to solve
it with that combination, and you may wonder why. The reason is that Euler left
behind a fantastic formula for that the cosine plus i
times the sine of an angle equals the exponential of the
angle, e to the i omega t. And the good part
about that is-- which we saw earlier
for exponentials-- that when the source
term is an exponential the solution is an exponential. So I look for a solution
of this with that same e to the i omega t, and it's
got some factor capital Y that we have to find. And we find that by substituting
this into the equation, and then we learn what
this capital Y is. So let me put that in. The derivative,
the left side, is-- so this is the still going
to be our complex solution. It's still complex. So I put this into the equation. The derivative will be
i omega-- of course, the derivative of
the exponential brings down the number
there, whatever it is. i omega y e to the i omega t. That's what we get for
the derivative of this, and now that should
equal a times the function plus the
source, e to the i omega t. Just as we did before with
a real, e to the st, now it's s is i omega. And the beauty is we
divide everything by e to the i omega t, and then
we get a simple equation. i omega-- bring the a on the
other side-- times y is what? Good. So y we now know is 1
over i omega minus a. It involves the frequency
omega in the source term, and it involves the rate
of growth, the constant, in the 0 order term. Good. That's our expression
for capital Y, and if I put that in here,
I have the complex solution, but my idea is to use this
complex solution to find two real solutions-- that's really
why I'm-- this video is about-- using this complex source term
to find the solutions for both that one and sine omega t. And the trick will
be-- the way to do it will be I will take--
here's my complex solution. I'll take the real part
of this complex solution. It will be the results. Match the cosine. That's the real part of this
expression, which I now know, and the imaginary part
of this same expression will give me the
sine, the output, the response to the sine term. So to problem solve one
method, but there's a step that I have to take. How do I find the real
part of this expression? The real part is easy
if the complex number is written a plus ib. Real part is a. The imaginary part is b, but I
don't have it a plus side ib. I have it in this form, and
in a way, it's a better form. So now I have to think
about-- and y itself has this thing in the denominator. So can I practice by looking at
this number i omega-- i omega minus a? That's really the
awkward quantity that I have to get
into a good form, and the form you want
for complex numbers when you're going to multiply
them is the polar form. I want to put this in the
form re to the i alpha. So a positive number,
the magnitude of this, and the angle-- so I have
to draw the complex plane. I guess this is the first
time in these lectures. So there is the real axis. Here's the imaginary axis,
and here's a complex number, i omega minus a. So in the imaginary
direction, I go up omega, and in the real
direction, I go minus a. So here's my number. This is up to i omega
and back to minus a, and there is the
number I omega minus a. And there is the angle
alpha, and the length of that is the r. So I have those
two things to find. That's called putting the
number into its polar form. This is its rectangular
form, real, imaginary. This is its magnitude, and now
what is the magnitude of that? Easy. | a right triangle with height
with that side is omega. This side is minus a. I use Pythagoras to
know that this r is the square root of a
squared plus omega squared. That length squared-- the
minus sign disappears. That length squared, omega
squared-- so that's the r. Now, what about
e to the i alpha? What's the angle--
e to the i alpha? The best I can say is I
know what that angle is. I can only tell you
its tangent, but let me just leave it as alpha. Alpha is the angle. The tangent of
this angle will be that number over that number. I'll leave it that way. So that's a key step, a first
step with complex numbers, and in a later video,
we'll be devoted entirely to working with complex numbers. Here you're getting
a first look at it, or you may have seen it before. So now I had to divide by that. That's why I like this form. Dividing by it-- so
now I'm ready for-- I'm ready to put in my complex. y complex was this y, which
was 1 over i omega minus a, and e to the i omega t. That's what we have,
but now i omega minus a we have in this nicer form. I can divide by that. If I divide by that, it'll
be 1 over the square root of a squared plus omega squared
times e to the i alpha-- e to the minus i alpha. When I divide by an exponent--
that exponent-- exponential-- that exponent changes sign--
times e to the i omega t. Are you good with that? So this number produced both
of those terms, the magnitude part and the angle part, and
then this part was pure angle. And that's my answer
in a nicer form, and here is even better form. I combine these two. e to the i something
times-- when I multiply two exponentials,
I can put that as e to the i omega t minus alpha. You see that it has that factor
e to the minus i alpha as well as the e to the i omega t. Now I'm ready to take the real
part and the imaginary part, and that will give
me my two solutions-- my two real solutions
to the real equation. So take the real
part, and that's why I'll concentrate
on the real part is the original y of t-- y
real coming from the cosine. The cosine of
omega t is the case I'm redoing from what
I had in an earlier lecture as a combination
of sines and cosines. Now we're going to see
it more beautifully. So I'm ready to take
the real part of that. Well, this is a real number--
1 over the square root of a squared plus omega squared. And what is the real part of
e to the i times an angle? The real part of e to the i
theta is the cosine of theta. The real part is the cosine. That's where using Euler--
always using Euler. So it's just the cosine
of omega t minus alpha, and that's exactly
what we ended up with with more work in
the previous lecture. The magnitude-- this
was called the gain. This was the G part. This is the G, the gain. The input was the size
of cosine was size one. It's increased or
decreased by this factor, and then the cosine
is still here. And alpha, you remember,
is the time lag. The input is a pure
cosine, but the output is a shifted cosine, which is
also a combination of cosine and sine. So there you've used complex
numbers-- maybe the first time, maybe not-- to get the
nice form for the answer. And of course, if the input
was the sine of omega t, then I would have
the same thing. That would be a sine. So it's really the
physically important quantity is the gain and the phase lag. Good. So that's the first
use of complex numbers to get us back to an
exponential, after which the solution is easy. I'll have more source
terms-- a few more-- and then I'll look
for a formula that applies for any source term. Thank you.
