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PROFESSOR: So today
we're going to take

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a look at sinusoidal functions.

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And specifically, we're
going to use complex numbers

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to get a handle on
sinusoidal functions.

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The reason is complex
numbers provide

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a robust way of analyzing
sinusoidal functions.

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So specifically, we're
interested in this function,

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e to the i*omega*t
divided by 2 plus 3i.

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And we're asked to
write the real part

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of this function using polar
form and then rectangular form.

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And then secondly, we're
asked several properties

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of this function, the real
part of this function.

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What's the circular frequency?

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What's the amplitude?

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And what's the phase lag?

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And then lastly, we're asked
to sketch the real part

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of this function versus time.

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So I'll let you take a look at
this and try it for yourself.

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And I'll come back in a moment.

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Hi everyone.

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Welcome back.

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OK.

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So let's take a look
at this problem.

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So we're asked to write e to
i*omega*t divided by 2 plus 3i.

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We're asked to write the
real part of this function,

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using polar form and then
also rectangular form.

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So I'll first start
off with polar form.

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And if we take a look
at this function,

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we see that the numerator is
already written in polar form.

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So just recall that polar
form is of the form r e

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to the i*theta.

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Meanwhile, the denominator
is written in Cartesian form,

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or rectangular form.

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So we need to convert
this into polar form.

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So what we can do is
we can write 2 plus 3i.

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We can combine these
two into a modulus.

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And the modulus is going
to be the square root of 2

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squared plus 3 squared.

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So this is the modulus
of the complex number.

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And then we have e to the i*phi.

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And the angle phi we can deduce
by writing a triangle, which

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is over to the
right by two units,

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and then it's up two units.

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And this is the angle phi.

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So we see from the triangle
that the tangent of phi

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is equal to 3/2.

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OK.

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So putting the pieces together,
we have e to the i*omega*t

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divided by root 13 e to i*phi.

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And the beautiful
thing about polar form

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is that when we
divide and multiply

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these complex
exponentials, it just

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turns into a simple addition
or subtraction of the phases.

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So we have 1 over root 13, e
to the i*omega*t minus phi.

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So we've successfully
combined all the terms

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into one polar term.

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And then now we're asked
to compute the real part.

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And here we use Euler's formula.

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So just to recall
Euler's formula,

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we have e to the i theta.

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Then this is equal to cosine
theta plus i sine theta.

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OK?

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So at the end of the day, we're
interested in the real part--

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and I'm going to
use this notation Re

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with a curly bracket to
denote the real part-- of 1

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over root 13.

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And I'll use Euler's formula on
e to the i*omega*t minus phi.

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So we have cosine omega*t minus
phi plus i sine omega*t minus

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phi.

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Sorry.

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1 over root 13 is multiplying
both the sine and the cosine.

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And we see that the second
term is purely imaginary.

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So when we take the real
part of a real number

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plus an imaginary
number, we just

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are left with the real
number at the end of the day.

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So the answer we're looking
for is 1 over root 13 cosine

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omega*t minus phi, where from
before phi is the arctangent

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of 3/2.

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And phi can also be thought
of-- the angle being

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between zero and pi over 2.

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OK.

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So this concludes the
polar form computation.

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Secondly, we also have
to compute this using

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a rectangular form calculation.

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So let's do the calculation
in rectangular form.

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And for this, I'm going
to use Euler's formula

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to expand out the numerator.

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So first, we want to take the
real part of the numerator is

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cosine omega*T plus
i sine omega*t.

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The denominator was 2 plus 3i.

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And typically what
we want to do is

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we want to turn the denominator
into an entirely real number.

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So what we do is we
multiply the top and bottom

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by the complex conjugate
of the denominator.

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So for example,
I have 2 plus 3i.

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So I'm going to multiply the
top and bottom by 2 minus 3i.

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OK.

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And when I do this, you'll
note that when we multiply out

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2 plus 3i and 2 minus 3i we
have a difference of squares.

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So we're left with: 2 times 2
is 4, the cross terms cancel.

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And then we have 3i
times negative 3i.

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That's going to give us negative
9 i squared, which is plus 9.

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So the denominator is
going to be 4 plus 9.

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And then the numerator-- I'll
write out all the terms just

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for completeness-- it'll be
2 times cosine omega*t plus 3

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sine omega*t.

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And then now we have
some imaginary terms,

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which are negative 3 cosine
omega*t and plus 2 sine

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omega*t.

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OK?

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But now we're going
to take the real part

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of this complex function.

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And if we take a look
at it, this first term

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is entirely real.

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This second term
is entirely complex

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because we're
multiplying by an i.

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So when we take the real
part, at the end of the day,

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the second term is
going to drop out.

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So when the dust settles,
we have 1/13 2 times cosine

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omega*t plus 3 sine omega*t.

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OK?

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So this concludes Part A.

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For Part B, we're
asked several questions

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about this sinusoidal function
when written in its real form,

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specifically, what's
the circular frequency.

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And the circular frequency
is just the frequency

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that this function
oscillates at.

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And we can get this from
looking at the rectangular form

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or the polar form.

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And it's simply just
going to be omega.

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So note how this
function oscillates

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with frequency omega.

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Secondly, we're asked
about the amplitude.

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And if we take a look
at it, the amplitude

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is a little tricky to get
from the rectangular form,

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but it's completely apparent
from the polar form.

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So if we take a look
at the polar form,

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we have 1 over root 13
cosine omega*t minus phi.

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And cosine oscillates
between plus 1 and minus 1.

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So this function takes
on a maximum value

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of plus 1 over root 13 and
a maximum negative value

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of negative 1 over root 13.

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So the amplitude
of this function

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is just going to
be 1 over root 13.

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And then lastly, the phase lag.

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This is the angle
that the cosine

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is picked up and shifted over.

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So again, we can get this
directly from the polar form.

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Meanwhile, it's a
little less obvious

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what this angle is when we take
a look at the rectangular form.

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So in our case, the
phase lag is just

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going to be phi, which
is tan inverse 3/2.

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OK?

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So this concludes Part B.

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And then lastly,
for Part C we're

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just asked to
sketch this diagram.

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And again, the most direct
way at sketching this function

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is through the polar form.

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So I'll write time on this axis.

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And if we just recall that a
cosine might look something

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like this.

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So this is a cosine function.

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And in our case, we're looking
at plotting 1 over root 13

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cosine omega*t minus phi.

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So we're interested
in a cosine which

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has an amplitude
of 1 over root 13

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and a maximum negative value
of minus 1 over root 13

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whose frequency is omega.

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So there's a quick
way on finding zeros.

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So if the zero of a regular
cosine is at pi over 2,

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then if we have an angular
frequency of omega,

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the zero here is going
to be pi over 2*omega.

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And this zero is going to
be at 3 pi over 2*omega.

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And lastly, this
is not necessarily

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the cosine of our function.

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Our function is this cosine.

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However it must be picked
up and shifted over

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by some phase lag phi.

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So in our case, the function
we're actually looking for

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is some shifted cosine,
which looks like this.

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And this distance here
is phi over omega.

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OK?

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So this just gives us a quick
sketch of this function.

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And again, we note that we
were able to plot this using

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the polar form very quickly.

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And the reason is because
the polar form gives us

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the amplitude, the circular
frequency, and the phase lag.

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Whereas if we were
to take a look

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at the equivalent
rectangular form,

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it's a little less obvious.

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The rectangular form is actually
the sum of a sine and a cosine.

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And unless you can just add
sines and cosines in your head

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very quickly, it's a
little less obvious

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on how to come up
with this picture.

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OK.

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So just to summarize, we've
used complex functions

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to analyze sinusoidal functions.

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And we've used polar
form and rectangular form

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to get a handle on
the complex functions.

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And specifically,
complex functions

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are a very good way to
represent sinusoidal functions.

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And just to
reiterate, the reason

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is that they contain
a lot of information

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that we can graphically
handle and turn into a sketch.

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So I'll just conclude there,
and I'll see you next time.