Welcome back.
Today we're going to be doing part two of the Complex Numbers section.
In this section, we will be doing a review of the complex numbers arithmetic
as well as present a couple of
useful identities for working with complex numbers.
In the previous lesson, we graphed complex numbers on the
complex plane, we did conversions from rectangular to polar format.
And we talked about complex conjugates and how to find them.
In this lesson we will adding, subtracting, multiplying, and dividing
complex numbers.
And then using Euler's Identity
to simplify working with trigonometric identities.
First of all, adding complex numbers. If I have z1 and add it to z2.
We can just basically say what we're going to do is add the real
parts and the imaginary parts individually and then put them together in this form.
And we can do this because the real part and the imaginary parts are orthogonal
to each other, so it makes it possible for us to do math this way.
There's also a graphical representation that we can look at.
If I have two vectors, here the red vector is 1 plus j2 and the blue
vector is minus j, ore minus 3 plus j, and we want to add them together.
Graphically, it's the same thing as taking one of these vectors
and sticking it on the tail end of the other vector,
like this.
Or equivalently taking the blue vector and sticking it
on the tail of the red vector like this.
And then finding where we get those, where we
end up, at the point where they kind of converge.
And, so that would be this point shown by the green.
If we look at the mathematical formula, doing the
arithmetic, we do 1 minus 3 gives us minus
2, j2 plus j gives us min, plus 3j, and so that corresponds to the value in green.
So our graphical representation matches up with our arithmetic calculation.
If you want to multiply complex numbers, first of all
we'll expand this out, it's the a plus jb formats.
If we do this in rectangular form, we need to use the distributive property.
So what we see is we get a1 times a2 plus jb1 times a2 plus a1 times jb2.
So all of these things this b1 a2, is right here.
This a1 b2 is here. And then we have, j times j, b1 times b2.
And so that j squared actually ends up being a negative 1.
So the real part becomes a1, a2 minus b1, b2.
And the imaginary part is j times a1, b2 plus b1, a2.
This is a little bit of work.
And perhaps a little bit difficult to memorize.
But let's go and see kind of what multiplication means
in complex numbers.
So if I have some vector like this and I multiply it by 2 what I'm doing is I'm
multiplying the real part and the imaginary part by
2, which means that basically I am doubling the length.
I guess I should use a different color.
The result is doubling the length.
If I did it by one half, it would be one half length.
Well now suppose that I multiply this,
this black vector by a j.
So now I'm taking the real part and multiplying
it by a j, so now it becomes an imaginary.
I'm taking the imaginary part and multiplying it by a j.
So now it becomes a negative real part. So it ends up being,
a vector of the same magnitude was the black one
but we've rotated it by 90 degrees or by pi halves.
And this is essentially what we see when we multiply complex
numbers and I take two complex numbers and multiply them together.
Is can be more easily seen in polar format.
We're scaling it by multiplying the magnitudes together.
And then we're doing a rotation by adding
the phases.
And we can see why that is if we actually go back to what this means.
This is A1 times e to the jth theta 1, times A2 e to the j theta 2.
Multiplying this together we get A1 times A2, times e to the j theta 1 times e to
the j theta 2, which if, since I'm multiplying
2 things with the same base, we can just add
the exponentials. So it's e to the j theta 1 plus j theta 2.
So I can factor out the j, either the j theta 1 plus theta
2, which if I put it into polar form it looks just like that.
So we see the multiplication basically is just stretching or, or,
compressing the, the length of the vector as well as the rotation.
Now suppose I want to subtract complex numbers.
Well subtraction is basically the inverse of multiplication or, we can take
the second term and multiply by negative 1 and just add them together.
So, to better understand what's going on here, we need
to know what it means to multiply by a negative 1.
Well if I have a vector coming out like this, you can
look at the negative, multiplying by negative one one of two ways.
We're either taking this amplitude and making it
a negative amplitude in the same direction, theta,
which brings us over like this. Or, we could look at it as multiplying it
by j twice, which means that I'm taking this and rotating around by 90 degrees.
One time and two times. And that will give us the same result.
So what, basically what we're going to be doing, is
taking this vector, and flipping it around over the axis.
So arithmetically all we're doing is taking the real part of
the first term and subtracting the real part of the second term.
And do this, and doing the same thing with the complex terms.
Graphically, if I have this vector I'm going to need to take this
vector, flip it across the axis which is multiplying it by the negative 1.
And then we'd just use the same addition where we put
the, the vectors on the ends to find where we correspond.
And so the corresponding point here is 1 minus negative 3
gives us 4 and then 2 minus
1 gives us 1.
Now, lets look at dividing complex numbers.
Again, we can do it using the rectangular format.
But what we need to typically do when we're dividing complex
numbers is to get rid of the complex terms in the denominator.
And we do this by multiplying by the complex conjugate of the bottom.
So, what we're doing is doing a2 minus jb2 divided
by a2 minus jb2. This is
just a fancy way of multiplying by one. But we're doing it in a creative way
because when we do this, this a2 plus jb2 becomes a2 squared plus b2 squared.
And since a and b are real value things, the denominator is now purely real.
And then we can find that this is the real part and this is the imaginary part
of doing this calculation.
But once again not particularly intuitive and it turns out
that again it's easier to do in the polar format.
So if I have A1, angle theta 1, and A2, angle theta 2,
we get A1 over A2 with an angle of theta 1 minus theta 2.
And again you can see this if we go back to the exponentials.
And multiply the exponentials together, we'll see that this is going to give
us a result equal to A1 over A2, e to the j theta 1 minus theta 2.
Corresponding to this polar equation right here.
So it turns out that typically what's going to
be easiest is any time you're adding or subtracting.
Put the value into rectangular format.
And then combine the real and imaginary
terms, add them, respectively, and get your result.
If you're multiplying or dividing, it's typically easier
to go through and put them into polar form, do your calculation.
And then if you want them in rectangular form, you can convert back.
Couple of useful identities.
First of all, if I take a number, a complex number, multiply it by
its complex conjugate, I'm going to get
the amplitude of that value quantity squared.
And the reason the, that occurs is if I look
at the calculation a plus jb times a minus jb.
This gives us, a squared plus b squared since
the a bj term here and the minus j times ab term here
cancel each other out, and then the minus 1 and the two j's give us a positive 1.
And if we look at this graphed on the plane,
this is the real part a, and the imaginary part b.
Well the length of this leg here c, we can calculate using
Pythagorean Theorem a squared plus b squared is equal to c squared.
Well this a squared plus b squared has to be equal to c
squared where it turns out that c is the length of the vector.
The, or the absolute
value, or the modulus of z quantity squared.
Couple of other useful things is, if I want to find the real part of z,
I can do that by taking z, adding its complex conjugate, and dividing it by 2.
Because a plus jb,
plus a minus jb divided by 2.
Well, we have these two a's, and the jb
here and the minus jb here cancel each other out.
So this gives us a.
You can do the same thing for the
imaginary terms, except now this is a minus.
So now the a's cancel each other out, this becomes minus
j2b, and so you divide it by the j2, and that
cancels out to give you b.
These can be useful in and of themselves, but one place that
they find particular use is when we want to work with trigonometric functions.
Instead of dealing with trigonometric functions
directly, we can use complex exponentials.
And this can be really handy because you might remember from learning trigonometry.
There are all of these identities and you can never remember
exactly the identities that were being used and where they came from.
Well, if I looked at
either the j theta we know that either the j theta is equal to the cosine
of theta plus j times the sine of theta, which is Spoilers identity.
If I want the real part,
it's cosine of theta.
So the cosine of theta is the real part of this.
So all we need to do is take this, add its complex conjugate and divide by 2.
So cosine of theta could be e to the j
theta plus e to the minus j theta, divided by 2.
You're
going to do the same thing to define sine of theta, just using the other side.
And to give an example of the way this can be useful,
suppose that I wanted to calculate the cosne of theta quantity squared.
This is, an identity.
It's very useful.
But a lot of people forget what this equals off the top of their heads.
Well we can actually use this to find out.
So this is going to be equal to e to the j theta plus e to the minus j theta
divided by 2, times, got again, e to the j theta,
plus e to minus j theta, divided by 2.
So using the distributive law, you get e to the j theta
times e to the j theta is e to the j 2 theta,
e to the j theta times e to the minus j theta, is equal to 0.
Which is 1, e to the minus j theta and e to the j theta again gives us 1.
And then e to the minus j theta times e to the
minus j theta is equal to e to the minus j2 theta.
And on the bottom we have 2 times 2, which is 4.
So splitting this up a little bit, we see that we have two right
here that we can put together.
So 2 over 4 is one half plus, and then here, this
e to the j2 theta and e to the j2 theta, over 4.
So that is actually equal to one half times e to
the j2 theta plus e to the j2 theta divided by 2.
And going back to here, we see that that's
just going to be the cosine of 2 times theta.
So it turns out that the cosine of theta
quantity squared is equal to one half plus one half cosine of 2 times theta.
So you might not have been able to
remember the trigonometric identity but converting it into these
exponential terms and doing it a little bit of
algebra allows you to derive them on your own.
An, and some exercises to go back and look at trigonometric
identities and see if you can derive them using this complex calculation.
So this shows another example of why complex numbers can be useful.
In summary, we demonstrated arithmetic operations with complex
numbers and even provided a couple of useful identities.
Hopefully the material covered here on complex
numbers is a useful reference for you.
You will be using it extensively in the other modules in the
course, since complex numbers are a very important part of electrical engineering.
As always, if you have any questions on the material that's covered here,
go to the forums and good luck
using complex numbers in your calculations, farewell.