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Money, everyone, we're on complex numbers, so just to the end of the last lecture, we were thinking about the arguments of a complex number.
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And I'd like to keep going from there a little bit. So sort of thinking about Cartesian coordinates, polar coordinates.
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And it's really important that it's convenient to pass between the two of those. So here is a useful thought.
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So we can pass between.
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Cartesian and polar coordinates.
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So here's my kind of picture of what I'm doing, so this is my all gang diagram, my complex plane.
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Here is some little complex number which I might think of as having Cartesian coordinates a comma b,
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or I might think of as having modulus off an argument theta.
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And I'd like to know if I know and b, can I find out and see if I feet to catch my flight?
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And B, that's the kind of point here. So if Z, which is a plus B or C, has modulus ah and argument beta, then we can find A and B.
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So maybe I'll throw this on here and another colour for you. So this is just a little bit of trigonometry.
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So this bit is our. This is a this is B sine Fita.
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Sorry, also in features. Try that again, which is B. So a is a closed beta and B is a sign theatre.
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So one kind of way of thinking about that then, is to think of it as being our times because it's plus I sign theatre,
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which is a representation that we find coming up in various contexts in the other direction.
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If that is a plus B and to find are in theatre, well, we know how to find oh, just by the definition,
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so are is the modulus of Z is just defined by root of a squared plus b squared.
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That's how we defined it. And also again, looking at the diagram time theatre is B over a, at least in the case when it's not zero.
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I'm very scared about the possibility of dividing by zero.
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I don't want to risk it, but as long as it's not zero, then B overall is going to give me time theatre.
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From that, I can kind of recover theatre. But as we know this, this sort of ambiguity about what theatre is,
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because if you have some argument, theatre will feature plus two pi or seats, plus it needs to.
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A multiple of two PI will also be a valid thing, and that's reflected in the fact that there's not a kind of unique inverse here.
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So I'll just put determining theatre is delicate.
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So in practise, this is not a problem. You just need to kind of have your wits about you a little bit.
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So I want to know what happens when we multiply.
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We saw that we could interpret addition of complex numbers geometrically with that kind of vector addition parallelogram diagram.
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What happens with multiplication when we know what happens to the modulus, when we multiply, that behaves nicely.
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What happens with the argument? So this is Proposition five. And this says take Z and W in the complex numbers, but not zero, then.
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So remember, the arguments are zero isn't defined. So I want to ignore that the argument of the Z Times W is equal to the arguments said.
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Plus the argument if w. So a couple of little comments here.
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One is the here we're working with arguments modulo two pi.
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So this idea that these two are equal up to some multiple integer multiple of two PI.
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So it's important that we remember that. So this is working with arguments modulates you PI.
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The other is that you've seen this kind of relationship before. This looks a bit like a logarithm, right?
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That's not a coincidence. There are no coincidences in mathematics, so maybe I'll just note here this looks a bit like log looks like log.
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That's very pleasingly illustrative. So let's think about how are we going to prove this?
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So this is not too bad actually, given this kind of thinking that we've been doing up here.
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So if I let this to be the arguments of Z and fi b the arguments, if w, I hope you're making good progress with letting the Greek alphabet.
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I'm not fantastic with every Greek letter of a feature for a really good Greek classes to know.
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So then what we've just said is that says it's going to be the modulus of Z Times because it's a plus.
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I sign Theta and W. It's going to be the modulus of W Times Code Phi.
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Plus I sign Phi so we can just multiply so that W is the modulus of Z times,
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the modulus of W times course theta plus i sign Theta Times because Phi plus I sign Phi and I can do some tidying up here.
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So one thing is that I know what happens with this multiplication for the modulus.
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So that's the modulus of that w. I'll write it that way round.
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And I guess we're using Proposition three here.
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I'm not going to record every time you use Prop. three for the rest of all time because it's just going to come up so much.
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But I'll just note it here to remind you. And if I multiply out here, I get constater to cause Phi minus sign C to sign Phi.
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That's coming from these terms. Plus all times cause it's a sign PHI plus sign theta calcify.
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And hopefully you might recognise these two will kind of compound annual formulae and trig.
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So this is where my jealousies at W times cause of C so plus Phi plus i times theta plus phi.
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And so we can read off immediately that the argument is that W is c c plus phi, which is the argument of Z Plus the arguments of W.
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Of course, it's not a coincidence that the trick here works out nicely, right? So we can use this to make sense of multiplication in a geometric way.
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So remark. We can interpret multiplication and see geometrically.
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So multiplying by a non-zero complex numbers that's multiplying by zero, it's not so exciting,
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multiplying by some non-zero complex number Z that rotates the complex plane.
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I'll move over here. Anti-Clockwise.
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Why the argument is that this is a consequence of that proposition, we just proved, and it enlarges.
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By a factor, the modulus of that and that sort of centred on the origin.
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So there is a nice way to make sense of multiplication without geometric view of complex numbers as well as algebraically.
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So we're going to find ourselves using this proposition. Lots.
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OK, I want to introduce you to some particularly nice complex numbers there called the inner circle.
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So definition the inner circle.
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And see, I'm defining unit circle. It's kind of what you expect,
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but I'm also telling you this because I want to tell you that it gets called s one and s one is defined to be the set of all z in C,
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such that the Modulus Z is equal to one.
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So geometrically, it's a circle of radius one centred on the origin.
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And what we've just noticed is that we can write that's in another way.
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So here are a few little remarks about the unit circle.
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So we've just seen essentially that S1 is also the sex of all cos theta plus I sign feature by feature as well.
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So that would be another way of thinking about it.
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Here is an amusing and really surprisingly useful facts about numbers in the unit circle, so if I take Zs and ask one, what is the inverse of Z?
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Well, we know that Z Times Z bar is the modulus of z squared.
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If you're in the inner circle, the Modulus Z squared is one.
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So Z times that bar is what in this case, so the inverse of Z is its complex conjugate kit.
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Also useful, and this is also going to lie in the unit circle.
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So the inner circle has all sorts of nice structure within itself, and in fact, S1 is a group under multiplication.
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This is not the time to tell you what a group is. There's a course on that.
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It's called groups. So in February, some person will introduce you to groups in the course.
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In groups and group actions, you'll meet as one as an example. So this thing about the inverse is relevant.
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There are some other things that go into that. The thing to remember here is the inner circle is a kind of important and interesting object.
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It has kind of structure. So here's a really super exciting results about numbers in the inner circle.
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This is a theorem, and it's called the maths theorem. And this is going to be, I imagine, familiar to quite a few of you.
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But I might not be phrasing it necessarily in the way that you're used to.
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So for Z in the unit circle and any integer and remember, this just means any integer we have.
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That the argument of Z to the N is end the times the arguments is that.
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So this proposition is a very special kind of it's kind of related to this.
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We're going to be using this proposition to feed into it. But this is kind of helpful.
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So you might be more familiar with equivalent formulation equivalently for feta in R and and in Z, we have.
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That cost data, plus I signed Visa to the end is cause of and.
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Plus I sign of the feature.
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So definitely when I first learnt about troops there in my last question, this form, but this form is a really helpful way of thinking about it.
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So I encourage you to think about why these are saying the same thing, because if you understand why they're saying same thing,
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that's a really good kind of sign that you are understanding the theorem.
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So let's prove this. So I'm going to fix some Z and S one and show that the theorem applies to the Z and four and greater than equal to zero.
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So for non-negative and I'm going to use induction on N. and I'm going to write that down because it
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helps the reader to know that we going to use induction so far and great ones are equal to zero.
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We use induction on N. So my base case is energy zero, and that's just a quick check.
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So Z to the zero is equal to one.
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So the argument is that's the zero zero is end times.
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The argument is that in this case. So that was important, although not super exciting.
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More interestingly, let's think about the inductive step, so let's suppose the result holds.
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For some and greater than or equal to zero, so we've got some fixed value then where we're supposing it holds.
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So we're saying that the all humans have said to the end is equals and times the
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arguments of Z and that we want to think about the arguments is that to the end +1.
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So let's do that over here. So then the argument of Z to the N +1.
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I want to relate to the arguments of Z, but I can do that using Proposition five,
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which tells us that that's the argument is that the N plus the arguments of Z so right down, that's why.
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Proposition five. And then we know about the occupants Z to the N, because that was our induction hypothesis.
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So this is enzymes. The argument is Z plus the arguments of Z by the induction hypothesis, and that's equal to, of course.
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And plus one times the arguments of Z. So we say that the results holds for and plus one.
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So that's an inductive proof that shows us that this relationship works for any greater than or equal to zero.
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No, I wouldn't think about negative ed and. One possibility would try to be to do a kind of induction downwards that could be a useful strategy,
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but actually I think we can just use the results for positive values so far and less than zero, we use the results for positive.
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And so this can be a nice strategy to kind of use work you've already done to save yourself a bit of effort.
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So if I fix some negative end, I'm going to let m b minus n.
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So of course M is then positive. So by our previous result, so we know that the argument of W to the M is m times the arguments.
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If W for any W in as one that's using the results for M, which is opposed to values.
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So that follows from our previous pod. So we just need to think about how are we going to apply this?
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Well, let's think about Z to the end. Is that to the end is z inverse to the power M?
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That would be another way of rewriting it. And Z Inverse I notice in my kind of fun fact over there was the complex country, this of Z.
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So we going to be able to apply this result with WB complex, conjugative z.
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I quite like to know what the arguments of the complex conjugate subset is,
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but if you ponder that for a moment, you'll see that it's minus the arguments of Z.
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Again, everything is modulo two pi here. So the argument is that the N is the argument of the Z bar to the M, which is m times.
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The argument is that bar, which is m times minus the arguments of Z, which is end times the argument of Z.
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So that proves the results for negative values using the result for positive values.
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So it's one theorem. It's just kind of somehow utterly fundamental. You get to use lots.
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Here is a specific example of a kind of classic way in which it gets used, namely trig compound angle formulae.
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So by dint of for any feature in R, we know the cause of three feature plus I sign of three Fita is close to plus I sign C to keep it.
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That's from that kind of equivalent formulation.
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My goal here is to find some nice formula for close to three three two in terms of close to four sign of three three two in terms of sine theta.
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So I'm going to expand this out. So this is a binomial theorem and kind of tidying up,
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and I think I'm going to get close cubed theta minus three cause features sine squared
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thetr plus i times three core square features science beta minus sign keeps the beta.
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So that's just by multiplying out quite a lot of practise and multiplying out keeps.
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And I'm just trying to tidy up the real and imaginary parts as I go along.
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So comparing real and imaginary parts.
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Gives that cause of three Fita is cause keeps B to minus three cost to sign squared theta.
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And then I got a formula for sine square theatre in terms of cost pizza, right?
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So I could kind of keep going and tidy up, and I could write out some kind of corresponding formula for sign three seater and again, tidy that up.
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So I'm going to let you do that tidying up while we have a short pause,
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I also have one other thing for you to do in this short pause at the end of each of your lecture courses.
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This year will give you a paper questionnaire towards the end of the course so that you could give some feedback on how you found the course,
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how you found the lectures and so on so that we can then, as a department,
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use that to improve teaching for next year, even though it's only Thursday if week one.
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This is in fact the last lecture of this course officially, which means it's question time.
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So I'm going to put some piles of these in strategic places if you could pass them along so everybody gets a questionnaire, that would be great.
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If you want to fill it in during the lecture, you can.
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And if you just need two piles, one at the top there and one of the top that, I'll collect them at the end.
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If you prefer to fill it in later, you can also return it directly to the academic admin team who are the people who process them.
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So this is your opportunity to give some feedback on the very short, complex numbers course.
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OK. The question is making their way round, so if you can if you can concentrate on keeping the momentum go passing those round,
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that would be great so that everybody gets them. I will try to remember a couple of minutes to check that everybody's got them,
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but hopefully that's giving you a little bit of a chance to think about these trick formulae,
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especially if this isn't something that you've thought about before.
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Let's keep going over here because excitingly, it's time for endless routes of unity.
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So here's a definition. So if this is a complex number and is a strictly positive integer, which you might call a natural number,
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but somehow we can't decide whether or not there is an actual number and I don't have an argument about it.
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So I'm just trying to be unambiguous. And that to the end is equal to one.
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Then we say the Z is a writ of unity or more precisely on the spirit of unity.
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What I learnt about this stuff, I think I remember thinking all routes of unity that sounds really cool,
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and then unity turned out to be one sort of slightly less dramatic than I expected.
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But reach of one is still pretty good. There are lots of interesting things to explore.
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For example, Proposition seven tells us which numbers are, in fact, routes of unity.
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So we can we can pinpoint where these things are. Without too much difficulty.
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So let's take Z in C then.
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Z is an end threat of unity.
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If and only if. And I've got two conditions here, one on the modulus is that one of the arguments is that so I need the modulus said to be one.
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And the argument of that? So we took PI over, and for some integer, okay, so this precisely describes what the merits of unity look like.
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So let's prove Proposition seven.
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And you'll notice the Proposition seven is an if and only if statement, so it's really two statements packaged as one.
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Let's prove the two separately and try to help you see what one we're doing.
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I find it helpful to write these little kind of implication arrows in brackets when I'm doing this for my own benefit.
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I do this not just for you. So we're going to see the left to right first.
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So we're going to suppose that is and through it of unity.
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So we're supposing that the end is equal to one, and then we wanted to do kind of interesting things.
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Well, we can say something about the modular straightaway. The module, as opposed to the power end, is the modular.
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So Z to the N is one and the modular is that is strictly positive.
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So the modular z better be equal to one.
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So much of this is that it's just a real number here and bytes and one of which conveniently we've just proved and come up to of comma.
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The argument is that to the end is end times the argument is dead, but we know the Z to the N is equal to one.
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So the arguments of Z is the arguments of one over N,
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and then we have to concentrate because there's a little subtlety here because we're working with arguments modulo two pi.
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So if we to say the arguments of one is zero, therefore this is zero and that wouldn't be correct.
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All we know is that the argument is zero plus an integer multiple of two PI.
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So what we know is that this is 2kay pi over n for some integer K, and that gives us what we were looking for the other direction.
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Let's get rid of these smudges, really the other direction. So now we're going to suppose that the modulus ZS is equal to one.
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And the arguments is Z is two pi over MN for some integer K.
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So then what can we say about the modulus? Is that to the end? Well, that's the modulus of Z.
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So the ED is one we like. Modulus modulus works out really nicely and also.
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The argument is that to the end is end times, the arguments have said, is to party.
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So what's telling us? It's telling us that Z to the N has the same.
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Modulus and argument as one. So in fact, to the end is equal to one.
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So is that is an end to of unity and that finishes are proof.
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So if I carry on over here, so one small observation from Proposition seven,
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so vermaak Proposition seven shows, but roots of unity lie in as one they lie in the unit circle.
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And that's the kind of handy thing to be aware of occasionally. Right?
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How are we doing on questionnaires? Could you please put your hand up if you do not have a questionnaire?
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Interesting. Could you please put your hand up if you have some spare questions?
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OK. The people who didn't have questionnaires were over here somewhere.
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Oh, thank you. My question is why if if you need a questionnaire? OK, so I got to send some this way and some this way,
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and it would be great if you could sort them out amongst yourselves and hopefully there are enough.
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So cute consequence of Proposition seven is that we can count and threats of unity,
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so I think maybe I said on Monday a corollary is a quick consequence of something you've already proved.
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So Corollary eight says four and basically are equal to one.
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There are exactly NW and three acts of unity.
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And that's quick to prove from Proposition seven because we know what the end streets of unity are.
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So by Proposition seven,
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the end roots of unity are so one way of writing them would be as cause of took pi over and plus I sign of two pi over n for any K in the integers.
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That makes it look like there are lots of them. But of course, there's lots of duplication here.
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So there are. All and such distinct values, e.g. we could choose K to come from zero one up to and minus one.
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So that was a handy consequence, it's good to know not only where they intrude c.a.r.,
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but that we've got and and threats of unity, that's a good thing to know.
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There are some end suites. You're going to see that even more exciting than others. There are some threats of unity, the zone and threats of unity.
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But there aren't smaller groups of unity aren't threats of unity for smaller ebb and that particularly important, so they get a special name.
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So this is a definition. So let that be an ensuite of unity.
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If Z to the M is not equal to one for one less than or equal to m less than equal to n minus one,
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then we say that Z is a primitive and threats of unity.
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And these kind of have Connexions with things from group theory and other other things that you will see those very strange noise,
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other things that you will see this year. So a primitive and through to you, this is like end is the first power of said where you get one.
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So here is a proposition that is useful and certainly applies to primitive roots of unity,
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although it applies a little bit more generally than that as well. So if that is an end of unity?
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For some and greater than or equal to two.
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And that is not equal to one, then Z two, the power and minus one plus that's the power and minus two plus plus z plus one is equal to zero.
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This is the kind of thing where if I were working on these lecture notes this afternoon,
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thinking about the might be thinking, Oh, why is Vicki excluded the case when Nichols won?
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What was special about that? That kind of why the hypotheses that that's the sort of question that you could be asking yourselves.
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So this is not too tricky to prove because we know that zero is said to the N minus one.
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We're assuming the Z is an each of unity. Oh, we have factories that we have minus one.
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If this factorisation is not second nature to you,
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I invite you to think about it because it's a really useful factorisation and kind of generalise this difference of two squares.
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And in fact, it's this. And that is not equal to one.
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And that's the end of the proof. So this is not super difficult, but it's kind of useful.
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It's it's a nice result. So this starts to have some nice consequences geometrically, for example, when you're thinking about routes of unity.
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OK, I want to think about another way to represent complex numbers.
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We sort of thought about much of this argument to be thought about this Cartesian coordinates kind of way.
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There's another very useful way of writing them, which comes from the following fact which gets called Euler's formula, at least sometimes.
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May, says Fifita, and oh, we have E to the I theatre is equal to cost, plus I sign the theatre.
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While you take 10 seconds to just absorb how lovely that is,
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how we do a question questionnaires, please put your hand up if you do not have a question at.
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Results forget night, you will notice that I have labelled this as facts rather than theorem or any of those kinds of things.
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This looks a lot like a thing that needs proving right and it does need proofing.
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But in order to prove it, we first need to define the exponential in the cosine in the sign.
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This is, we haven't done any of those things. We're not in a very good place to prove this.
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Defining the exponential is something you will do an analysis this year and also
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cosine and sine once you've learnt how to add up infinitely many things in a safe,
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controlled manner. It's very important to be careful when adding up infinitely many things.
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Once you have kind of got your licence that you are safe to do that,
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you can think about the exponential and then you'll be able to think about why this is true.
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So I just put a little bit more in the online note. So if you want to kind of read a little bit more about that.
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Have a look at the online notes, but we're just going to kind of accept this as a fact for our purposes.
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So here is a remark. This gives a very convenient way to represent a complex number.
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So if that has the modulus is that is ah.
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And the arguments is said is theatre, then that is r e to the I theatre.
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When I started learning about complex numbers, I definitely thought about the most in the A-plus B form.
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Then I think when I was doing A-level kind of studying a bit further,
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I thought about the more in the times cos that's a plus I sign thesis form and I probably think about the more in this form now.
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So I went to stress that this is very convenient, so I put flashing neon lights to remind you.
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So if this is relatively new, it feels a little bit uncomfortable, it feels natural to go back to your old ways of thinking about it.
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That's that's what we do when we meet something new is kind of try to see what can we keep go with the old way.
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This has lots of advantages.
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I really encourage you to look for opportunities to practise working with this way of thinking about it so you become more fluent, for example.
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Proposition 10. So Proposition 10 says Take Z a non-zero complex number and you'll see why I wanted to be known zero z in C, but not zero.
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That's a zero, then Z has exactly NW and three.
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So for any and greater than or equal to one.
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So we thought about risks of one right proposition and a corollary eight, so he said that exactly and and three to one.
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This is generalising to the exactly average of any non-zero complex number, which is kind of nice.
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So let's prove this.
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So I had to do this in three parts, and I got to kind of bullet point what the sections are to try to help you see what the structure is clearly.
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So the first part is that there's at least one and threw it right at the start.
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It's not completely clear that there is at least one end throughout. So if I let all be the modulus of Z and seem to be the arguments of Z,
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so z is r e to the fitter I can sort of stare at a set in this form and guess what an end through might be.
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So there is. A in fact, unique, positive real.
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As such that as to the act as to the power is equal to R.
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So this is since R is greater than zero. How do we know this is true?
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By analysis, so I'll just put C prelims analysis.
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You will think about why you can do this in that course.
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So that's going to be the modulus of my proposed rate. And now I'm going to let Phi equal theta over N.
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Why did I start over here? Yeah, OK.
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So if I let w be equal to the s e t i phi, then double t w to the end, when you do a quick calculation, you c is equal to Z.
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So this definitely at least one right now, I'd like to show that there are, in fact,
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at least and and it's how did you show that they're exactly and if something you show that at
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least and if something you say the most and if something that could be a useful strategy.
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So I would say that there are a list and so I'm going to take W as above.
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So W is a fixed and through it, I could get a whole bunch more and through it by multiplying by then three of one.
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So if I let Alfa be an M3 of unity.
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Then you do a quick calculation and you realise the alpha times the power and is also equal to Z.
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So Alpha W. is another Richard Z. And by corollary eight, there are and such alpha.
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Giving a well, at least, and distinct and fruits of that.
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So this says I can think of and distinct sense roots is that we haven't shown that there aren't some or maybe some more that have some other form,
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of course, their arms, that's what we're about to prove,
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but that's why we're kind of structuring it like this if I carry on over here for the third bullet point.
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The third bullet says most and and through it.
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So if I can take W as above, so it W is my fixed and three z,
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I want to show that any end through to Z must be a very tough one time W that would do it.
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So if I let you be an end through to Z, then my secret aim, if you like, is to show that you must be w times some root and threats of unity.
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Well, we know that W to the end is that.
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And that's also due to the N. That wasn't very good and we try again.
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And that is not zero. So W not zero.
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So I can do you divide it by W two power n? And that's not one, right?
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Having got to use the N and W to the end in the same, if I just rearrange this says that you over W is an m three to one.
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And by corollary eight, there are.
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And of these, so there are mice and choices of you.
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So that completes the proof. So it's a little bit fiddly and you might want to kind of think about that in your own time,
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but hopefully the structure at least makes some kind of sense.
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So that was thinking about routes of particular complex number,
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and that means that we've sort of thought about solutions to the polynomial equation W to the N equals Z,
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we've just counted solutions to that equation. We'd like to know about solutions polynomials more generally.
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I said on Monday we could think about solutions quadratic, so we we'd like to be able to generalise.
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And that's what we're going to do now. So the first thing is to show that if you've got a polynomial with complex coefficients,
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a complex polynomial of degree n that you can't have too many roots.
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So this is Proposition 11. And this says.
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A complex. Polynomial, meaning a polynomial with complex coefficients of degree n has most and root.
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I mean, maybe it has no roots, maybe it has some roots. I'm not making any claims about that in this proposition.
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All I'm saying is there are most and complex rules. So I mean and roots in C, so let's prove this on.
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My strategy is going to be to use induction on RN.
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The degree of the polynomial so and is one, for example, is a base case.
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That's a quick check. I'll let you make sure that you're happy to see that a linear polynomial with complex coefficients has almost most one root.
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I feel confident that you can handle that. So let's do the inductive step.
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So I'm going to let P be a complex polynomial with degree.
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And I'm going to suppose the results hold.
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For polynomials with degree less than or equal to and minus one.
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That's our inductive hypothesis. So I want to show that the number of roots of PE can't be too big.
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Dividing through by the leading coefficient and the coefficient of Z to the N,
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if you like by the leading coefficient X two, that maybe doesn't change the rates.
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And what that means is that we may assume.
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The P is a monarch polynomial monarch means it has leading coefficient one, so p looks like P of X is X to the N.
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It has leading coefficient one because its monarch. Plus a and minus one x minus one, plus the dot plus A1 x plus a zero for some A0, A1 dot dot dot.
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And minus one in C. I would like you to notice two things about this.
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One is this I very carefully introduced to zero up to and minus one.
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Or if you turn off as a party and you take your friend along and they don't know the people that you introduce,
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them writes Really not naughty because otherwise people can't have a kind of sensible conversation.
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We can't talk about these things without knowing what they are.
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Also, notice how I've carefully matched the coefficient with the power so that it's dead easy to do.
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I didn't want to start with A0 A1 here. Right? This is a top tip. This would make your life easier if he has no roots.
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Then we're done. If P has a roots, say Alpha in C, then P of X is equal to X minus alpha times f of X for some complex monarch polynomial f.
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With degree and minus one. If you haven't seen how to prove this or you're not.
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What's the point, you're not confident you could do yourself. I did put the details in the online notes.
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If you want to think about that, you can't. What this tells us is that any roots of PE is either alpha or at a rate of F.
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So by the inductive hypothesis. F has most and minus one roots.
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And any rate of p is alpha or a rate of F.
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So P has most.
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OK, i need to carry on over here. P has at most and root.
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And that completes the induction argument.
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So proving that a complex polynomial of degree and has at and roots is not too difficult, I mean, I've gone through that fairly swiftly.
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You've got to want to look at that in your time. But it's very doable. What we'd really like to know is, does it have any roots?
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Does it always have at least one? Does it always have any roots?
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And the answer is given by Theorem 12, which is super important, and it's called the fundamental theorem of algebra.
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You should draw some conclusions about the importance of this theorem from the fact that it's called the fundamental theorem of algebra.
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And this says any complex polynomial of degree n.
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Has exactly and returns counted with multiplicity, so repeated routes,
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you have to count the appropriate number of times which when you sort of think about it, is a relatively natural thing to do.
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So that is. If he is well, we may as well focus on a monarch complex polynomial.
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With dignity and.
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Then we can fix rise p of X is x minus alpha one, two x minus alpha n four, some alpha, one two alpha and in C, they don't have to be distinct.
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This theorem takes quite a bit more work to prove that Proposition 11.
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It's not in this course, you're going to want some extra ideas, so maybe some ideas from complex analysis or topology.
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So you will see proofs of this theorem in the next couple of years that you can be looking out for.
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For now, you may assume this theorem, but only if you really, really have to write.
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If you can avoid this theorem, you want to avoid it because you haven't seen a proof. And that means you're sort of not morally on top of this there.
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So if you possibly can avoid using this theorem but know that it's there if you need it, the history of this theorem is really interesting,
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and I put some links to where you can read a bit of an introduction to that on the online notes.
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That is the end of the complex numbers course on Monday.
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Excitingly, we will be here for linear algebra if you want to leave your questions at the top on that side or that side.
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I will collect them up. Otherwise, please return to academic opinion. See you on Monday.