5

Problem (5) According to the result of the previous problem, it appears thatIIs = (-1)"+1 j=0where the Gj are the nth roots of unity. Show this directly:...

Question

Problem (5) According to the result of the previous problem, it appears thatIIs = (-1)"+1 j=0where the Gj are the nth roots of unity. Show this directly:

Problem (5) According to the result of the previous problem, it appears that IIs = (-1)"+1 j=0 where the Gj are the nth roots of unity. Show this directly:



Answers

Find the indicated roots of unity and express your answers in the form $a+b i$. Fourth roots of unity

We're trying to find the sixth roots of unity. So if we start, start by writing one as cysts of 360 degrees and where n is an integer. This is true because co sign and sign have a period of 360 degrees. Now we're looking to solve the values of Z where this is true. This, by definition, is the sixth roots of unity. So what we're going to do is we're gonna write Z to the sixth is equal to sis. Um co sign. Plus I sine up 360 degrees on. And now we're going to apply the mob serum. OK, do Mobs theorem is going to tell us that Z must then be equal to sis of 360 n divided by six. Okay, So that cysts of 60 end And now there are six sixth roots of unity. So we need to calculate this thing for when and is equal to zero when it is equal to one. When that is equal to two. Musical 34 and five. Okay, so we're looking to calculate consists of zero cysts of 60 just of 1 20 cysts of 1 80 sifs of 2 40 insists of 300. Okay, and so you get one 1/2 plus square three divided by two I native 1/2 plus square three divided by two I negative one negative 1/2 minus squared three divided by two I and then finally 1/2 minus were three divided by two by and these six values are the sixth roots and unity.

So to find the some of the in flu tive and, well, let's say we have a complex numbers and using go to one So we want to find me the first route event will. Then our roots will be one. And so let's say we wanted to find the B square roots of one. Well, remember, this would be plus or minus one. So this would be negative one and positive one and z cube if we work this out. And if we used the if we used the root formula for on complex numbers in polar form, the solution we would get is one. Remember, we would get one and then and we would also get this negative 1/2 plus Route three over to I. And then we would also get negative 1/2 minus minus Route three over to I. Well, we wanted to find the fourth root. We wanted to find the fourth root of one. Well, this would be the fourth roots of one we would get again. We would probably get one, because this has been a common common factor here. 111 So the values we would get would be one Well, it's like this is this is one. We would get negative one. We would get negative I and positive I we would get negative I and positive I and so So let's look at this pattern here. Well, if we add, we add if we add the sun's here, this is equal to one and so negative one plus one is equal to zero. This is also equal to zero, and this is also equal to zero. So any value for end is not equal to one. So if let's say if and does not equal to one, then these some the some of and fruits will be zero will be zero. So this is this is our solution.

So this is asking us about the 10th roots of one. So first asked us about the cube root of one and asked us to find the exact values. Now we've done that in another problem, so you can find that one too. See how you do it. But here are the exact values of the cube roots of warning. Right. All three of those numbers, if you cube them the equal. Exactly one. And the nice thing about it is they have this beautiful symmetry as to where they are on the complex plane course. Here's one. And here's, uh, minus 1/2 plus route through over to I. And here's minus 1/2 minus +33 over to I. So they come in this lovely, threefold symmetric pattern there. Now what they're asking about is what happens when you add them. If you take all three of those numbers and you add them together, what do you get? Well, you might notice that the plus retriever to I and the minus retriever too why are gonna cancel each other out, and then we have two copies of negative 1/2 which together make negative one, and that exactly cancels with the positive one. So those all add together 20 All right. Next we have the fourth roots of one. So this time there's four of them, and again they come this lovely wheel of, ah, symmetry. Over here there's four of them equally spaced across the unit circle. And they're simply one I negative one and negative I all of those race to the fourth power is one. And indeed, it's true that if you add all of these together, this canceled with this and this cancer with this. So once again you get zero. Let's check with sixth roots of one. So again, in another section, you confined exactly how to find the exact values of all of these. But I've just listed them out here to begin with beforehand. So again they come in a beautiful symmetric pattern on the on the complex plane. And if we add them all together Oh, my gosh, I always do that. Here we go. If we had them all together, we get well, let's see this 1/2 and this negative 1/2 can cancel each other this negative 1/2 and this 1/2 can cancel each other. These two can cancel these two, and then this one can cancel this negative one. So altogether we get zero. Once again, it adds up to zero. And finally, let's check the eighth roots of one. There's eight of them, all equally spaced around the unit circle like that. All of these numbers race of the eighth power is one. If we add them together, we can see that this one cancels with the negative one. The I cancels with the negative. I, these two positive guys cancel with these two negative guys and this one where it's one negative, one positive cancers with one negative, the other positive. That was supposed to say that. Sorry. Ah, for no other way around Supposed to say that. There you go. Those canceled The point is they all under each other? They all exactly can't see each other every time. So no matter what you're looking at, the end fruits of one will always add up to zero. Let's Ah, let's do one more to see why it's always like that. Well, you can see that if we're talking about, let's say Arnold, uh, 12th Roots of one. Well, he's gonna be one here is gonna be one on the other side. And then there's gonna be one. Here is gonna be one on the other side. One here, one here, one here, one here, one here, one here, one here, one here. And these air literally negatives of each other. Right? This will cancel with this. This will cancel with this. This will cancel with this. This will cancel. With this, they're all gonna cancel each other out so that together they all add up to be zero, and that will happen every time.

Here in this problem we have to choose the correct option and we are given if one, Coma Alpha one comma alpha two up to Alpha and -1 other, any troops of unity and if omega is a non real fifth root of unity then Omega -91 into Omega -92 up to omega minus alpha and minus one. So he'll one comma alpha one comma effort to up to alpha and minus one. The any troops of unity. This satisfies that Extra the power and -1 is equal to zero. Now we have x to the power and minus one is equals two, X minus one and do x minus alpha one into x minus alpha two up to x minus alpha and minus one Further. We have extra the power and -1 upon X -1 is equal to x minus alpha one into x minus alpha two have to x minus alpha and minus one. Therefore, As is equals two, Omega -91 Into Omega -92. Up to omega minus Alpha and -1 Which is equals two. Omega to the power and -1 upon omega minus one. Now if And as equals to 5K, then Omega to the power five K Is equal to one. It gives us S is equal to zero. If and is equal to five K plus one, then omega to the power five k plus one is equal to omega, It gives us S is equal to one. Now if N is equal to five key plus two Oh make a to the power five caters to is equal to omega square. We to give us S is equal to Omega Bless one. If n is equal to five P plus three, then omega to the pavel five k plus three as equals to omega cube, which gives us S is equal to one plus omega plus omega square, and if N is equal to five K plus fool, it means omega to the bubble. Five K plus four is equal to omega to the power fall, which gives us S. is equal to -0mega to the power for hence option A and D is the correct answer.


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