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Current time:0:00Total duration:11:53

in this video we're going to hopefully understand why the complex form of a comp or why the exponential form of a complex number is actually useful so let's say we wanted to find so let's say we want to solve the equation X to the third power is equal to one so we want to find all of the real and/or complex roots of this equation right over here this is the same thing as X to the third minus one is equal to zero so we're looking for all of the real and complex roots of this and there are ways to do this without exponential form of a complex number but what the technique we're going to see in this video can be applied if this was X to the fifth the minus 1 or X to the 13th minus one and it's also going to show us kind of the the patterns that emerge when you start looking at things on an Argand diagram so to do this let's think about let's think about the exponential representation of one so let's just say Z is equal to 1 1 is a complex number it's a real number real not all real numbers are also complex numbers they're a subset they're part there in the complex plane they just don't have an imaginary they just don't have an imaginary part so let's draw this on an Argand diagram so that's my real axis this is my imaginary axis this is my imaginary axis and so this is the real this is the imaginary and if I wanted to represent a Z equals 1 it only has a real part so let me just draw one all around negative 1 negative 1 z Z looks like this Z would look like Z would look like that a position vector that just goes to 1 0 one way to view it this is the same thing is equal to 1 plus 1 plus 0i now let's put this in exponential form well it's magnitude is pretty straightforward the magnitude of Z is just the length of this vector or it's the absolute value of 1 well that's that's just going to be that's just going to be 1 now what's the argument what's the argument of Z what's the angle that this vector makes with the positive real axis what's on the positive real axis it's a real number so it has no angle so it's the argh of Z is zero so that might not be too interesting so far we just wrote we just figured out that 1 could be equal to 1 is equal to 1 times e to the 0 I 0 I and this is kind of obvious either the 0 this is going to be 0 0 times i 0 e to the 0 is going to be 1 times 1 is equal to 1 not a big deal there but what is neat is that this argument you could view it as 0 radians or you could go all the way around and add 2 pi to it and get to the same point go all the way around and add 2 pi and get to the same point so the argument of our complex number or of the number one really could also be an angle of 2 pi or an angle of 4 pi or an angle of 6 pi or an angle of 8 pi so we can write 1 we can write 1 is we can also write it as 1 times e I won't write the one anymore 1 times e to the 2 pi I or 1 times e to the 4 to the 4 PI I and the reason why this is interesting is in this equation this equation right here can be written in multiple ways it can be written as X to the third is equal to 1 X to the third is equal to 1 it could be written as X to the third is equal to e to the 2 pi e to the 2 pi I or it could be written as X to the third is equal to e to the 4 pi e to the 4 PI I and this is interesting and we're going to see this in a second let's take both sides of all these equations to the 1/3 power to solve for X so to the 1/3 we're going to take that to the 1/3 we can do that same thing over here we're just taking everything to the 1/3 power to solve for the X's in each of these equations to the 1/3 power so this first equation over here becomes X is equal to 1 to the 1/3 power which is just equal to 1 now what's the second equation become this second equation X is equal to e to the well it's just going to be the 2 PI over 3 I power e to the 2 PI over 3 I power and then this equation over here is going to be so x is going to be equal to obviously the three to the one-third that just becomes X to the 1 X is going let me do that same blue X over here is going to be equal to e to the 4 PI over 3 the 4 PI over 3 I so let's think about let's think about this for a little bit what is this so we could so immediately what's the argument here so let me let me write let me these are three different roots we call this X 1 X 2 and X 3 so these are three different numbers one of the roots is one that's pretty clear over here one is one of the cube roots of itself but these are other numbers and these are going to be complex numbers so let's let's visualize these numbers a little bit so what is the argument so for all of these so the magnitude of X 2 the magnitude of X 2 is still clearly 1 it's the coefficient out in front of the e it's clearly 1 the magnitude of X 3 let me do that same color the magnitude of X 3 is also clearly going to be 1 but what is the argument of X 2 what is Phi what is the argument well it's 2 PI over 3 it's 2 PI over 3 so how would we draw how would we draw X 2 so the angle is 2 PI over 3 I always it's easier for me to visualize in degrees so 2 pi is 360 degrees 360 degrees divided by 3 is 120 degrees so this is going to be 120 degrees is 60 short of so it's going to look like this it's going to look like this just like this this is so this angle right here it's argument is going to be 100 120 degrees which is the same thing as 2 pi as 2 PI over 3 and it's going to have the exact same length let me do the same color so this is x1 so that is this green color right over here x2 is this magenta one right over here and they all have the same magnitude so we really just rotated we rotated 120 degrees and what about x3 what's its argument what's X three's argument its argument is 4 pi 4 PI over 3 that's the same thing as 720 degrees over 3 if we were to put it into degrees and so 3 goes into 720 three goes into 720 three goes into 720 was it 240 720 240 right I should have known that so 240 degrees we're going to go 180 degrees and then go another 60 degrees so it's going to be right over here it's going to be right over here so let me draw it like this it's going to be right over here so this is going to be that angle right over there is 4 PI over 3 radians 4 PI over 3 radians which is equal to which is equal to 240 degrees and once again it has the same magnitude so we just saw is when I take the cube roots of this real number I'm essentially taking the entire I guess we could call it the entire circle or the entire at 360 degrees or the entire two pi radians and I'm dividing it into into angle into 3 essentially this is 1/3 then we have another 120 degrees and then we have another 120 degrees and so you kind of see the pattern of where all of the roots are and in case you're still not satisfied you're like well you know you said you would find complex roots yeah you know I I'm not used to this or this is actually being complex numbers I watched you I actually want it to be in the form a plus bi we can easily figure it out from this right over here so X 2 X 2 is going to be equal to it's going to be equal to the cosine the cosine of 2 pi over 3 plus I times the sine of 2 pi over 3 sine of 2 pi over three and if you look at this over here we can figure out what those things are going to be this is the angle right over here if this angle right here over here is sixty degrees which it is because this up here is 30 degrees and this the hypotenuse or the length is one then this over here square root of three over two and that this distance right over here is negative one half so x2 is going to be equal to is going to be equal to cosine of two pi over 3 is negative negative one half did I do that right yet negative one half plus I times sine of 2 pi over 3 that's this height over here which is square root of 3 over 2i so that's x2 and we can do the exact same thing with x3 x3 is going to be equal to its x value or I should say its real value is going to be the exact same thing it's going to be negative 1/2 and then it's why it's it's imaginary value so this angle right over here is also good this just from the negative real axis down to the vector is going to be negative 60 degrees so this height right over here is going to be negative square root of 3 over 2 so it's negative 1/2 minus the square root of 3 over 2i so using this technique we were able to find the three roots the three complex roots of one this is one of them this is another one and of course one is one of them as well where did we do that one is one of them as well you could use this exact same technique if we were finding the fourth roots we would take the three hundred the two pi radians or the 360 degrees and divide it into 4 and so it actually be this it would be I it would be negative one it would be negative I and we know if you take I to the fourth you get one if you take negative I to the fourth you get one and if you take negative one to the fourth you get one and if you take one to the fourth you get one and so you could do this you could find the eighth roots of one using this technique now the other question that might be popping in your brain is why did I stop at e to the 4 PI why didn't I go on why didn't I go on and say well this is equal to e to the 6 PI I and look another root and so if I did that if I did that if I said X to the third let's say I wanted to find a fourth root here maybe X to the third is equal to e to the 6 PI I and I take both sides of this equation to the 1/3 power so it get X is equal to taking this to the 1/3 a good e to the 2 pi I would get e to the 2 pi I well what's either the 2 pi e to the 2 pi I will just get us back to 1 so when I added 2 pi again it just gets us back to this root again and if I added if I took E to the 6 PI if I took either the 8 pi to get this root again so you're going to get only 3 roots if you're taking if you're taking if you're taking well if you're finding if you're finding the third roots of something anything beyond that it just becomes redundant