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GILBERT STRANG:
So complex numbers

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are going to come in to today's
video, and let me show you why.

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So I'm going to
solve the problem.

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The differential equation
we know-- first order,

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linear with a source term,
but now the source term

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has both the cosine
and the sine.

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And the sine, you notice, has
this imaginary square root

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of minus 1 involved.

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So I'll call this y
complex with a little c,

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because the answer
is going to come out

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a complex number instead
of a real number.

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So the previous lecture
solved it with the cosine,

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and now we're going to solve
it with that combination,

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and you may wonder why.

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The reason is that Euler left
behind a fantastic formula

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for that the cosine plus i
times the sine of an angle

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equals the exponential of the
angle, e to the i omega t.

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And the good part
about that is--

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which we saw earlier
for exponentials--

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that when the source
term is an exponential

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the solution is an exponential.

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So I look for a solution
of this with that same e

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to the i omega t, and it's
got some factor capital Y

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that we have to find.

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And we find that by substituting
this into the equation,

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and then we learn what
this capital Y is.

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So let me put that in.

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The derivative,
the left side, is--

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so this is the still going
to be our complex solution.

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It's still complex.

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So I put this into the equation.

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The derivative will be
i omega-- of course,

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the derivative of
the exponential

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brings down the number
there, whatever it is.

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i omega y e to the i omega t.

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That's what we get for
the derivative of this,

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and now that should
equal a times

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the function plus the
source, e to the i omega t.

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Just as we did before with
a real, e to the st, now

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it's s is i omega.

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And the beauty is we
divide everything by e

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to the i omega t, and then
we get a simple equation.

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i omega-- bring the a on the
other side-- times y is what?

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Good.

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So y we now know is 1
over i omega minus a.

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It involves the frequency
omega in the source term,

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and it involves the rate
of growth, the constant,

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in the 0 order term.

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Good.

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That's our expression
for capital Y,

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and if I put that in here,
I have the complex solution,

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but my idea is to use this
complex solution to find two

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real solutions-- that's really
why I'm-- this video is about--

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using this complex source term
to find the solutions for both

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that one and sine omega t.

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And the trick will
be-- the way to do

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it will be I will take--
here's my complex solution.

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I'll take the real part
of this complex solution.

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It will be the results.

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Match the cosine.

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That's the real part of this
expression, which I now know,

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and the imaginary part
of this same expression

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will give me the
sine, the output,

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the response to the sine term.

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So to problem solve one
method, but there's a step

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that I have to take.

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How do I find the real
part of this expression?

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The real part is easy
if the complex number

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is written a plus ib.

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Real part is a.

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The imaginary part is b, but I
don't have it a plus side ib.

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I have it in this form, and
in a way, it's a better form.

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So now I have to think
about-- and y itself has

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this thing in the denominator.

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So can I practice by looking at
this number i omega-- i omega

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minus a?

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That's really the
awkward quantity

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that I have to get
into a good form,

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and the form you want
for complex numbers

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when you're going to multiply
them is the polar form.

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I want to put this in the
form re to the i alpha.

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So a positive number,
the magnitude of this,

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and the angle-- so I have
to draw the complex plane.

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I guess this is the first
time in these lectures.

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So there is the real axis.

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Here's the imaginary axis,
and here's a complex number,

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i omega minus a.

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So in the imaginary
direction, I go up omega,

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and in the real
direction, I go minus a.

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So here's my number.

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This is up to i omega
and back to minus a,

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and there is the
number I omega minus a.

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And there is the angle
alpha, and the length

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of that is the r.

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So I have those
two things to find.

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That's called putting the
number into its polar form.

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This is its rectangular
form, real, imaginary.

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This is its magnitude, and now
what is the magnitude of that?

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Easy.

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| a right triangle with height
with that side is omega.

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This side is minus a.

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I use Pythagoras to
know that this r is

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the square root of a
squared plus omega squared.

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That length squared-- the
minus sign disappears.

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That length squared, omega
squared-- so that's the r.

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Now, what about
e to the i alpha?

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What's the angle--
e to the i alpha?

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The best I can say is I
know what that angle is.

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I can only tell you
its tangent, but let

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me just leave it as alpha.

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Alpha is the angle.

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The tangent of
this angle will be

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that number over that number.

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I'll leave it that way.

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So that's a key step, a first
step with complex numbers,

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and in a later video,
we'll be devoted entirely

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to working with complex numbers.

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Here you're getting
a first look at it,

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or you may have seen it before.

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So now I had to divide by that.

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That's why I like this form.

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Dividing by it-- so
now I'm ready for-- I'm

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ready to put in my complex.

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y complex was this y, which
was 1 over i omega minus a,

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and e to the i omega t.

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That's what we have,
but now i omega minus

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a we have in this nicer form.

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I can divide by that.

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If I divide by that, it'll
be 1 over the square root

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of a squared plus omega squared
times e to the i alpha--

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e to the minus i alpha.

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When I divide by an exponent--
that exponent-- exponential--

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that exponent changes sign--
times e to the i omega t.

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Are you good with that?

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So this number produced both
of those terms, the magnitude

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part and the angle part, and
then this part was pure angle.

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And that's my answer
in a nicer form,

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and here is even better form.

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I combine these two.

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e to the i something
times-- when

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I multiply two exponentials,
I can put that as e

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to the i omega t minus alpha.

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You see that it has that factor
e to the minus i alpha as well

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as the e to the i omega t.

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Now I'm ready to take the real
part and the imaginary part,

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and that will give
me my two solutions--

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my two real solutions
to the real equation.

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So take the real
part, and that's

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why I'll concentrate
on the real part

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is the original y of t-- y
real coming from the cosine.

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The cosine of
omega t is the case

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I'm redoing from what
I had in an earlier

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lecture as a combination
of sines and cosines.

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Now we're going to see
it more beautifully.

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So I'm ready to take
the real part of that.

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Well, this is a real number--
1 over the square root

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of a squared plus omega squared.

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And what is the real part of
e to the i times an angle?

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The real part of e to the i
theta is the cosine of theta.

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The real part is the cosine.

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That's where using Euler--
always using Euler.

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So it's just the cosine
of omega t minus alpha,

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and that's exactly
what we ended up

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with with more work in
the previous lecture.

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The magnitude-- this
was called the gain.

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This was the G part.

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This is the G, the gain.

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The input was the size
of cosine was size one.

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It's increased or
decreased by this factor,

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and then the cosine
is still here.

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And alpha, you remember,
is the time lag.

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The input is a pure
cosine, but the output

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is a shifted cosine, which is
also a combination of cosine

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and sine.

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So there you've used complex
numbers-- maybe the first time,

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maybe not-- to get the
nice form for the answer.

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And of course, if the input
was the sine of omega t,

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then I would have
the same thing.

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That would be a sine.

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So it's really the
physically important quantity

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is the gain and the phase lag.

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Good.

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So that's the first
use of complex numbers

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to get us back to an
exponential, after which

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the solution is easy.

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I'll have more source
terms-- a few more--

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and then I'll look
for a formula that

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applies for any source term.

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Thank you.