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