So...
complex exponential function, this was another thing we talked about.
We talked about this new notation,
so instead of using cis theta, instead of using cosine theta plus i sine theta,
we can use an exponential notation, e to the i theta.
This is extremely useful and widely used.
Here we're defining it, but we can actually show there are reasons
why this notation is useful and actually is used,
and the reasons here - I'm going to give you 3 reasons here.
The first reason is that exponential laws work. This is just…
the fact that exponential laws work, that's just the statement of
Polar Multiplication of Complex Numbers.
Derivative with respect to theta makes sense, so I'll leave
this again as an exercise for you to check and you should do it.
If you take the derivative with respect to theta
of this side, treating i as just some constant number
with respect to theta,
and you take the derivative of this side, again treating i as just some number
with respect to theta, you're going to get the same answers.
So the derivatives makes sense, they agree.
More importantly, the power series agrees.
So if I take the power series of e to the i theta,
I take the power series of cosine theta, I take the power series of sine theta,
you see if they actually match up
This is in the sense of complex analysis,
so this is something you haven't seen yet.
But something that you should do, when you learn this in calculus, is actually
check out the 3 power series here,
and see if you can show that this actually works.
So take the power series for e to the x, and I plug in i theta.
Take the power series for cosine of x and plug in [theta].
I take the power series for sine of x, plug in theta and multiply by i,
do the arithmetic, you're going to see that these 2 power series actually agree.
There's very good reasons for defining this notation, it's not just cosmetic.
With this week, we went to the Complex nth Roots Theorem.
This is sort of one of the big theorems we wanted to get to.
Every complex number has exactly n distinct nth roots.
So if I have some…
which I guess we're going to talk about in a minute, okay,
so if I have some z to the n is equal to a,
then there are exactly n roots to this equation, okay,
where a is some complex number here.
The roots lie in a circle of radius the size of z,
centered at the origin, and they're spaced out evenly by angles of 2 pi over n.
This is a really, really powerful theorem, right?
So it tells you a little bit about the geometry of such solutions,
which is really neat. We'll see this in an example soon.
So concretely, if I have some complex number, a equals r to the e to the
i theta - r e to the i theta,
then the solution to z to the n are equal to a are given by
z is equal to the nth root of r
times e to the power of i times
theta plus 2 pi k, all divided by n,
and k ranges from 0 to n minus 1.
So this gives you a formulaic approach for doing it,
and this gives you a geometric visualization of what the solutions should look like.
This is called the Complex nth Roots Theorem.
So what is it saying? What is this really saying too?
This is actually says a lot, okay?
This is once we have one solution, we can find all the other ones.
So if you know that a is equal to r e to the i theta,
then we know that all the solutions are given by this, right,
so in a sense, you started with our first solution,
the nth root of r, e to the i theta over n.
That's our first solution, right, taking the nth root...
taking the nth roots of a is equal to
r e to the i theta, gives us
the case when k is 0,
and then all the other cases you just add multiples of
2 pi k over n, where k ranges from 1 to n minus 1.
That's pretty neat.
A lot easier than in the real case. Okay.
Definition, one more definition since we’re on this page,
an nth root of unity is a complex number that satisfies z to the n is equal to 1.
So before when we had z cubed equals 1, we only had one solution.
But now, if we consider z as being -
so that was over the real case, and now if we consider z to be a complex variable,
then z cubed equals 1 actually has 3 solutions according to this theorem.
And that is something that I will leave for you as an exercise.
So you can actually find all 3 solutions, right? The first one is z equals 1,
so if you factor that out of z cubed minus 1, you're going to
be left with a polynomial z squared plus z plus 1
and then you can just find the roots of that
and those are going to be the other 2 complex roots of unity…
or complex third roots of unity.
Sometimes these are denoted by zeta n.
We won't do it, we won't talk too much about roots of unity
but they are a very, very important part of mathematics.
Roots of unity example: so find all eight roots of unity in standard form
So again this is a good exercise to practice.
Pause the video, try it out, see if you can find them based on the previous theorem.
See if you can do it without looking at the theorem, and if you can’t, check out the theorem.
Hopefully you’ve paused the video and come back
with an answer, that'd be good.
Let's check out the solution. So first I'm going to
do this algebraically, and then I'm going to do this pictorially.
So we already know a solution to this. We know that z is equal to e to the…
so we already know that z equals 1 is a solution,
and from the previous theorem, we know that all other solutions are given by -
well this is 2 pi i k over 8 for k from 0 to 7, right?
Our first solution was
r equals 1 and theta equals 0
and so the nth root of 1 is 1,
and so once we have that,
we plug it in, this is what the previous theorem says.
Okay?
And all of these values give solutions,
so what are they? We can actually expand them out, so we can write them using the
cosine plus i sine notation.
We see that our solutions are 1…
oh I got these backwards. Oops.
Root 2 over 2, plus root 2 over 2i,
and the third one is i, the next one is minus root 2 over 2,
plus root 2 over 2i, I can't stop smiling,
I've goofed the fractions I don't know how I did that,
and so on and so forth. I'll fix them in the PDF file.
So okay, so here we go. So if we actually just do these computations, this is what we're going to get,
and if we look, we can draw a diagram based on these solutions,
and the diagram that we're going to get
is going something like this. We can also have solved this pictorially as well, right?
Remember that we know that once we have one solution, in this case 1,
we can find all other solutions by rotations of pi over 4,
or in other words, 2 pi over 8.
So we have our solution at 1, we rotate by
45 degrees, or pi over 4, we're going to get root over 2, plus root 2 over 2i,
then rotate by another 45 degrees, and
another, and another, and another, and another, and another, and so on and so forth.
I want to talk a little bit about this diagram because this diagram is actually pretty cool.
So we have 8th roots of unity, okay?
8 is an even number, that's not surprising,
but why is that important for this little diagram here?
So whenever I had a solution i, or
root 2 over 2 plus root 2 over 2 times i,
the negation of it was also a solution.
That's because
if I had z to the 8 is equal to 1, then negative z to the 8 equals 1.
So here we have the twofold symmetry, okay?
So all of these solutions correspond in pairs on the opposite side.
So 1 corresponds to minus 1, the solution in the second quadrant
corresponds to the solution in the fourth quadrant,
and so on and so forth, that's by negation.
By doing that, that means we only really needed to look at 4 of the solutions, okay?
Another thing to note, z to the 8
equals 1, so if z is a solution,
so z to the 8 is equal to 1, then
z bar to the 8 is also equal to 1.
So I have an answer z, then z bar is also a solution,
and that's not too hard to see either. We can pull out the conjugates and that's -
the real key to that claim is that 1 is a real number.
So because 1 is a real number, then if z is a the solution to z to the 8 is equal to 1,
then z bar is also a solution.
So that actually gives us a little bit of twofold symmetry,
so we know that i corresponds to negative i, and
this solution corresponds to this solution, and
1 is 1.
So it's actually pretty cool. You could actually just compute the solutions in the first quadrant,
and along the edge of the first quadrant,
and you get all the solutions by just using a little bit of mirror symmetry.
That's pretty neat stuff, I think it's pretty cool.
Okay.
Watch out it's a trap, what do I mean by this? So solve
z to the 5 is equal to minus 16 times z bar.
I killed the punch line unfortunately.
So what do I mean by it’s a trap?
So you might believe, when you look at this, oh well
it's z to the 5 so it probably has 5 solutions,
but it actually has 7 solutions.
So this is much different than
when you're dealing with a polynomial. So this thing, as we'll see in a couple minutes,
is not a polynomial and that's the problem that we're having here.
If it was a polynomial, then it would only have at most 5 unique solutions.
But here, since there's a z bar thing,
it's not a polynomial anymore
and that might cause it to have more solutions.
So you can get some weird things happening in Complex Number Land.
And sometimes we might have no solutions. It’s possible for this
equation to not be satisfied. For example…
you could come up with examples.
Like take like z plus z bar,
right, z plus z bar is always going to be real,
so if you had z plus z bar equals i, there's no solutions, for example
So how do we prove this? How do we show there's 7 solutions?
So one thing we note is that 0 is clearly solution, so
so we can discount that and we're just looking for 6 of the solutions, okay?
So if we do that -
so we're going to assume that now z is a non-zero solution to this equation.
If we take modulus on both sides and simplify,
then we're going to see that the size of z
to the 4 is 16, hence the size of z is 2.
So if you’re solving an equation that you don't know what to do, sometimes it helps
to look at the length of the complex number.
You can also put this into polar notation and
look at the r-value that's fine too. I like just taking the modulus on both sides.
I think that's a much easier thing to do and understand.
Now the clever thing to do is now, okay, so since z
is non-zero, I've already discounted z equals 0,
we're going to multiply this original equation by z on either side,
and we're going to note that the right-hand side is z z bar, which is the size of z squared.
So now we get z to the 6 is equal to…
what is it …negative 64.
And that we know how to solve.
z to the 6 is equal to minus 64, that was by CNRT the
Complex nth Roots Theorem,
tells us that there must be 6 solutions.
So we got 1 from 0, and then 6 from doing this
this operation.
So something to watch out for, alright? So you can write down
equations with complex things and not get a solution.
You can't write down a polynomial in the complex numbers not get a solution,
but that we'll talk about later,
but if you mix polynomials with complex conjugation, weird things can happen.