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where we laughed less where if we left off in the last video we had somewhat simplified our polynomial P of X from the way that it was originally described in the problem but it still doesn't seem like we're any closer to finding the roots at least we were able to reduce the degree from a 47th degree polynomial to the product of a 1st degree in a twenty third degree polynomial but it still doesn't seem like we are any closer to finding the roots especially the roots that matter and here we're gonna have to there's another little bit of an aha moment here that this part this part in blue which is really just the sum of all of the degrees of X from X to the zero all the way to X 223 this actually looks a lot it is actually if you view the common ratio as X as a finite geometric series and I've done it in other videos there is a way to find the sum of a finite geometric series and we'll use a similar technique right here to simplify this a little bit to help us find the roots so let's just call this whole thing s so let's just call this s is equal to X to the 23 plus X to the 22nd plus all the way down to X plus 1 so that's this thing over here now let's multiply s times X so x times s is going to be equal to and actually let me do it like this because this will help us a little bit I'm gonna write it up here and it's gonna look a little bit many messy x times s is going to be so actually let me I'll just write it over here x times s is equal to let me save some space actually x times s is equal to x times X - 23 is X to the 24th power x times X to the 22nd it's going to be X to the 20 X to the 23rd power and then X there's going to be an X to the 21st when you multiply it by X you're gonna get X to the 22nd all the way down to when you multiply x times X you're going to get plus x squared and then when you put multiply the one times actually going to get X now what I want to do let's subtract this from that let's subtract it let's subtract it I won't put the sign out there so on the left-hand side of the equation you have s minus XS so s minus XS is equal to what happens on the right-hand side of the equation well you're going to have this guy - this guy these are going to cancel out these are going to cancel out everything is going to cancel out except you're going to have a 1 minus X to the 24th so you're gonna have 1 minus X to the 24th now let's just multiply both sides by negative 1 because this will just give it a slightly easier form for us so this is the same thing if we multiply both sides times negative 1 is X s minus s is equal to X to the 24th minus 1 we can factor out an S here I'll do it over here so if we factor out an X from the left-hand side we get s times X minus 1 is equal to X to the 24th minus 1 or we could say that the sum the sum is equal to X to the 24th minus 1 over X minus 1 so this whole thing in blue over here can be written can be rewritten as this right over here is X to the 24th minus 1 over X to the minus X over X minus 1 now let's be clear here what we really care about for this problem is to figure out the roots that have the roots whose squares have complex parts we already we already know that one of the roots is 0 because if you set this equal to 0 X could be equal to 0 0 squared is still 0 does not have a complex part so it really doesn't matter so for our problem and the other thing that we we want to keep in mind is that we don't want to double count roots let's reread the problem a little bit because what we're gonna have is a bunch of I guess you could view them as double roots you're going to have you're only going to have 23 dis drinked distinct roots from this part of the polynomial and they say let Z 1 Z 2 and Z R be the distinct zeros the distinct zeros of P sub X so what we only care about right now are these 23 zeros we don't have to think about those 23 zeros twice we're not going to add their imaginary parts twice we can kind of substitute P of X with we really just care I'll just come up let's call it Q of X let's call Q of X we really just care about the zeros of Q of X which is just X to the 23 plus X to the 22nd plus all the way down to x squared plus X plus 1 that's this inside part the fact that it squared that just doubles the roots but we just care about the distinct roots so we really just care about the sum of the imaginary part of the absolute value of the imaginary parts of the squares of the roots of this guy right over here and we just figured out that this guy can be rewritten as this over here or or even better let's define let's define instead of defining Q of X like this let's define Q of X as this times as this times X as this times X let me let me write it a little bit as this times X minus 1 let's just define Q of X here so this Q of X is going to have all of the same distinct roots that matter as this P of X but it's also going to have the root X is equal to 1 which is clearly not a root here if you put x equals 1 here you're just kind of you're actually going to get 24 so it's not going to be equal to 0 but what makes this better easier to find the roots of is that this thing right over here is equivalent to its equivalent to this part in yellow is X to the 24th minus 1 over X minus 1 times X minus 1 times X minus 1 which is equal to X to the 24th minus 1 so we've simplified the problem a actually draw a huge amount what we've said is the roots that we care about are actually the roots of this polynomial right here except for X is equal to 1 we just kind of added that there to simplify it which is equal to the roots of this polynomial here except for 1 because we added 1 as a root so we just care about the roots of this polynomial except for 1 and the roots of this polynomial or well I won't say easy but it's a lot more straight it's a it's much more straightforward to calculate if we set this equal to zero if we set it equal to zero to figure out the roots we just get X to the 24th add one to both sides is equal to one so we're going to have remember this is going to have twenty-four roots one of them is going to be one we what we want to ignore the one so we care about the 23 non one roots and we care about squaring them and then taking the absolute value of their imaginary parts so we essentially need to think about all of the 24th roots of one and then ignore one so let's think about how to do that a little bit and this is going to delve a little bit into complex into complex into the complex plane so just as a refresher how do we find all of the all of the roots of one the 24th roots of one and just remember we can rewrite one so let's draw a little complex plane over here so let's draw a complex plane so this is this is the real axis and this is the imaginary axis one is sitting right over here it only has a real part it is one but we know that we can rewrite this we can rewrite one we can rewrite one as as let me write it this way it could be e to the I times zero actually getting it so this is e to the zero obviously equal one or we can add 2 pi to it if we're dealing with radians so one is equal to e to the I 0 which is also equal to e to the 2 pi I which but you can keep adding because if you add 2 pi it goes around the complex plane once if you add another 2 pi it goes around again so it's also plus it also equals it also equals e to the 4 pi it also equals e to the 6 pi e to the 6 pi I it also equals e to the 8 pi I think you get the general idea so if you want to find the 24th roots you want to just raise all of these things to the 20 to the 1 over 24 1 over 24 so essentially you're going to take this divided by 24 divided by 24 divided by 24 divided by 24 divided by 24 and then keep going you can actually go all the way well I'll I'll depict it over here so what is this the first root is e to the 0 over 24 that's just e to the 0 that's just 1 that's the root that we can ignore right because we added that root to Q of X to simplify a little bit now what's this root over here what's in this root over here this is e to the PI over 12 I eat ooh the PI over 12 I which remember is the same thing as cosine of PI over 12 that's the real part plus I sine of PI over 12 I sine of PI over 12 and if we were to draw that we have an angle of PI over 12 and if you wanted to think in degrees that would be 15 degrees so it would look something like this it would look something like this right over here so we do care about that root right over here now what would this root be well this root is going to be PI over 6 it's going to have twice the angle so it's going to look like this so this root right here is going to look like this this root over here and notice there PI over 12 apart this root over here this is PI over so 6 this is PI over 4 so it's going to look like this PI over 4 that's 45 degrees so it's going to look like this so notice each of the roots we're just adding we are just multiples of this 4 I shouldn't say multiples are powers of this first root over here we're just increasing the degree by PI over 12 and remember we just care about the imaginary parts of their squares so let's just first of all just think about the roots that matter so one way to think about it is we could literally just draw a unit circle here and divide it into twenty-four sections so let me draw that let me draw it a little bit let me redraw a little bit so I don't overload this graph over here actually I'm running out of time in this video I might continue it in the next so we could draw a unit circle here they're all going to have magnitude one so we're gonna have we could draw a unit circle over here one of the roots is one but we're gonna have twenty-four roots so we can draw it we can divide it into twenty-four sections so we have four sections right here each of them we're gonna have sick roots in each section so we're gonna have one PI over 12 2 PI over 12 PI over 4 and then you have the one symmetric to that then you have the one symmetric to that let me make sure I'm doing this right you're gonna have the debate and then you're going to have and then you're going to have PI over 2 so these are the roots that are actually let me just continue this in the next video I don't want to when I start rushing this is kind of a new concept in tree I'll continue this in the next video