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Hello everyone. Welcome to Week 10 of Carmen's 
Core Concepts, my name is Carmen Bruni.

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In this video series, we 
talk about the weekly

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topics in Math 135.

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So without further ado, let's get to it.

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Let us begin with the start. 

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Polar Multiplication of Complex 
Numbers, so in the last

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week of lectures, we saw

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the polar...

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polar notation for a complex number.

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So remember that cis 
theta means cosine

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theta plus i sine theta. 
cis is just an abbreviation.

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So z1 is equal to r1 times cis theta 1,

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and z2 is equal to r2 times cis theta 2,

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then z1 z2 is equal to r1 
r2 cis theta 1 plus theta 2.

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So...

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this gives us a way to multiply 
two numbers that are in

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polar form.

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Remember that polar form is r 
times cosine theta plus i sine theta. 

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If we look at this proof, this 
proof is actually not too hard.

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The idea is just sort of a “follow your 
nose” thing. Take the product of z1 z2,

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plug it into the cosine theta 
1 plus i sin theta 1 notation,

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multiply it out, and collect 
the like-terms. So r1 r2

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is common to everything, so we 
can pull that out of everything,

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then if I multiply these two 
complex numbers, well

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what's the real part? It’s cosine 
theta 1 times cosine theta 2,

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minus sine theta 1 times sine theta 2,

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that's the first term. The imaginary part
is cosine theta 1 times sine theta 2,

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plus sine theta 1 plus cosine theta 2.

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And if you look at this, 
well these two terms,

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these are just actually
trigonometric identities.

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The first one is just cosine 
theta 1 plus theta 2,

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and the second one is just 
sine theta 1 plus theta 2, 

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and that gives us r1 r2 
cis theta 1 plus theta 2.

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And that's it. So very 
straightforward proof, you

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start with the left-hand side and show 
that it equals the right-hand side

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like we normally do with an equality proof.

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It's just a matter of distributing the result -

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or the...

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expanding the complex 
numbers in polar form.

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Okay.

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So that’s Polar Multiplication 
of Complex Numbers, and

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this, for us, our primary 
use of this is to discuss

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De Moivre's theorem,

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and De Moivre’s theorem, what does it 
say? So if theta is a real number and

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n is an integer, then

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cis theta to the n is equal to

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cis of n theta, okay,

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the middle one is just a description of what 
these things are, just another way to rewrite it.

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So again, if you're taking nth 
powers of cosine plus i sin theta,

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it’s the same as taking cosine 
times n theta plus i sine n theta.

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The proof of this,

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it's actually pretty clever. What we're going to 
do is we're actually going to use induction,

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which is strange because we're over the

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integers and not over the natural numbers.

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Actually the induction proof I'm not going to 
do because it's actually pretty straightforward

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and just uses Polar Multiplication 
of Complex Numbers.

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So the idea is take off the n equals 0 case,

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use induction on the positive integers,

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and then what do we do for 
the negative integers? Well

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the negative integers, we're 
going to use a little trick.

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We're going to write - 

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so if n is negative, we're 
going to write n equals

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minus m for some positive number m,

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then we're going to take cis theta 
to the power of n is equal to cis

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theta to power of negative m.

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That's the same as cis theta to the m

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inverse, that's what this notation means,

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or the other way around cis inverse…

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cis theta inverse to the power 
of m but this way is easier for us.

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And cis theta to the m, well

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by induction the positive cases 
are true, and m is positive, 

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so this is the same as cis m 
theta to the power of negative 1.

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This, we know how to invert 
this. We can invert this by using

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z inverse is equal to z over 
the mod of z squared.

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This is a nice little trick when you're dealing with
Properties of Modulus, right, so remember that…

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this should be z bar, sorry.

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So this is z inverse is 
equal to z bar over

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the mod of z squared.

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So that's why we have a 
minus sign here. So remember,

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cis m theta is equal to cosine 
m theta plus [i] sine m theta, but

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when we have the z bar, 
we change the plus to a minus.

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There also should be an i here.

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I apologize for the two typos.

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There should be an i here as well.

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And the bottom, which is cosine squared 
of m theta plus sine squared m theta,

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that's 1. That's a trigonometric identity, 
that’s the Pythagorean Identity.

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So what do we get? We get cosine m theta

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minus i times sine m theta.

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And now what happens here? So

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now, this isn't really what we wanted, 
right? We really wanted this to be

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negative m because we 
want this to be n theta

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plus n theta. Here we
have m instead of n,

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but we can use the fact that 
cosine is an even function,

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so cosine of m theta is equal 
to cosine of minus m theta,

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and sine is an odd function. So 
minus sine of m theta is equal to

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positive i sine of

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negative m theta.

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And this therefore is then equal to…

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what we have here.

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Again I'm missing an i everywhere, I 
apologize. There's lots of i’s because of sine.

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But I’m missing the 
imaginary unit, put those in.

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Since cosine is even and 
sine is odd, and we're done,

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because now cosine of m 
theta that's - or negative m

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theta that's just n,

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and sine of negative m
theta well that’s just n,

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and that actually proves 
De Moivre’s Theorem.

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Okay, so...

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this is the first time we've seen induction 
used on a question where the

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variable n ranged over the integers,

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and it involved breaking it up into the positive 
integers, doing the induction argument there,

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isolating the 0 case,

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and then looking at the negative case 
and reducing it to the positive case.

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Another common trick in 
mathematics which is pretty neat.

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So that’s it for De Moivre’s Theorem.