Hello everyone. Welcome to Week 10 of Carmen's Core Concepts, my name is Carmen Bruni.
In this video series, we talk about the weekly
topics in Math 135.
So without further ado, let's get to it.
Let us begin with the start.
Polar Multiplication of Complex Numbers, so in the last
week of lectures, we saw
the polar...
polar notation for a complex number.
So remember that cis theta means cosine
theta plus i sine theta. cis is just an abbreviation.
So z1 is equal to r1 times cis theta 1,
and z2 is equal to r2 times cis theta 2,
then z1 z2 is equal to r1 r2 cis theta 1 plus theta 2.
So...
this gives us a way to multiply two numbers that are in
polar form.
Remember that polar form is r times cosine theta plus i sine theta.
If we look at this proof, this proof is actually not too hard.
The idea is just sort of a “follow your nose” thing. Take the product of z1 z2,
plug it into the cosine theta 1 plus i sin theta 1 notation,
multiply it out, and collect the like-terms. So r1 r2
is common to everything, so we can pull that out of everything,
then if I multiply these two complex numbers, well
what's the real part? It’s cosine theta 1 times cosine theta 2,
minus sine theta 1 times sine theta 2,
that's the first term. The imaginary part is cosine theta 1 times sine theta 2,
plus sine theta 1 plus cosine theta 2.
And if you look at this, well these two terms,
these are just actually trigonometric identities.
The first one is just cosine theta 1 plus theta 2,
and the second one is just sine theta 1 plus theta 2,
and that gives us r1 r2 cis theta 1 plus theta 2.
And that's it. So very straightforward proof, you
start with the left-hand side and show that it equals the right-hand side
like we normally do with an equality proof.
It's just a matter of distributing the result -
or the...
expanding the complex numbers in polar form.
Okay.
So that’s Polar Multiplication of Complex Numbers, and
this, for us, our primary use of this is to discuss
De Moivre's theorem,
and De Moivre’s theorem, what does it say? So if theta is a real number and
n is an integer, then
cis theta to the n is equal to
cis of n theta, okay,
the middle one is just a description of what these things are, just another way to rewrite it.
So again, if you're taking nth powers of cosine plus i sin theta,
it’s the same as taking cosine times n theta plus i sine n theta.
The proof of this,
it's actually pretty clever. What we're going to do is we're actually going to use induction,
which is strange because we're over the
integers and not over the natural numbers.
Actually the induction proof I'm not going to do because it's actually pretty straightforward
and just uses Polar Multiplication of Complex Numbers.
So the idea is take off the n equals 0 case,
use induction on the positive integers,
and then what do we do for the negative integers? Well
the negative integers, we're going to use a little trick.
We're going to write -
so if n is negative, we're going to write n equals
minus m for some positive number m,
then we're going to take cis theta to the power of n is equal to cis
theta to power of negative m.
That's the same as cis theta to the m
inverse, that's what this notation means,
or the other way around cis inverse…
cis theta inverse to the power of m but this way is easier for us.
And cis theta to the m, well
by induction the positive cases are true, and m is positive,
so this is the same as cis m theta to the power of negative 1.
This, we know how to invert this. We can invert this by using
z inverse is equal to z over the mod of z squared.
This is a nice little trick when you're dealing with Properties of Modulus, right, so remember that…
this should be z bar, sorry.
So this is z inverse is equal to z bar over
the mod of z squared.
So that's why we have a minus sign here. So remember,
cis m theta is equal to cosine m theta plus [i] sine m theta, but
when we have the z bar, we change the plus to a minus.
There also should be an i here.
I apologize for the two typos.
There should be an i here as well.
And the bottom, which is cosine squared of m theta plus sine squared m theta,
that's 1. That's a trigonometric identity, that’s the Pythagorean Identity.
So what do we get? We get cosine m theta
minus i times sine m theta.
And now what happens here? So
now, this isn't really what we wanted, right? We really wanted this to be
negative m because we want this to be n theta
plus n theta. Here we have m instead of n,
but we can use the fact that cosine is an even function,
so cosine of m theta is equal to cosine of minus m theta,
and sine is an odd function. So minus sine of m theta is equal to
positive i sine of
negative m theta.
And this therefore is then equal to…
what we have here.
Again I'm missing an i everywhere, I apologize. There's lots of i’s because of sine.
But I’m missing the imaginary unit, put those in.
Since cosine is even and sine is odd, and we're done,
because now cosine of m theta that's - or negative m
theta that's just n,
and sine of negative m theta well that’s just n,
and that actually proves De Moivre’s Theorem.
Okay, so...
this is the first time we've seen induction used on a question where the
variable n ranged over the integers,
and it involved breaking it up into the positive integers, doing the induction argument there,
isolating the 0 case,
and then looking at the negative case and reducing it to the positive case.
Another common trick in mathematics which is pretty neat.
So that’s it for De Moivre’s Theorem.