Exponential form to find complex roots Using exponential form to find complex roots
Exponential form to find complex roots
- In this video we are hopefully going to understand why the
- complex form of the - why the <i>exponential</i> form of a complex number is actually useful.
- So lets say we wanted to find- so lets say we want to solve the
- equation x to the third power is equal to 1.
- 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 are 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- the technique we are gonna see in this video can be applied if
- this is x to the fifth minus one or x to the thirteenth minus one,
- and its also gonna show us kind of the- the patterns that emerge
- when you start looking at things on an argand diagram.
- So to do this, lets think about, lets think about the exponential representation of 1.
- So lets just say z is equal to 1. 1 is a complex number it is a real number,
- a real numb-, all real numbers are also complex numbers, they are a subset.
- They're, they are in the complex plane, they just dont have an imaginary,
- They just don't have an imaginary part.
- So lets 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 the imaginary, and if I wanted to represent z equals 1,
- it only has a real part. So let me just draw 1 all around:
- -1, -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 as equal to 1 plus, 1 plus zero i.
- Now, lets put this in exponential form. Well its magnitude is
- pretty straightforward, the magnitude of z is just the length of this vector,
- or its the absolute value of 1, and well, that's- that's just going to be,
- that's just going to be 1. Now whats the argument? Whats the argument of z?
- Whats the angle that this vector makes,with the positive real axis?
- Whats on the positive real axis? Its a real number.
- So it has no angle. So its the arg of z is 0.
- So that might not be to 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 0i.
- 0i. And this is kind of obvious. e to the 0, this is just going
- to be 0, 0 times i is 0, e to the 0 is just 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 2pi to it and get to the same point.
- You can go all the way around, and add 2pi and get to the same point.
- So the argument of our complex number, or the number 1, really,
- could also be an angle of 2pi, or an angle of 4pi, or an angle of 6pi,
- or angle of 8pi, so we can write 1, we can write 1 as-
- we could also write it as 1 times e, I won't write the 1 anymore,
- 1 times e, to the 2pi i, or 1 times e to the 4- to the 4pi 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 2pi i, e to the 2pi i.
- Or, it could be written as x to the third is equal to e to the 4pi i.
- e to the 4pi i. And this is interesting, we are gonna see this in a second,
- lets take both sides of all these equations to the 1/3 power to solve for x.
- So to the 1/3 , we are gonna take that to the 1/3, we are
- gonna do that same thing over here, we are 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 whats this second equation become? This second
- equation, x is equal to e to the- well this is just going to be
- the 2pi over 3i power. e to the 2pi over 3i power.
- And then this equation over here is going to be, so x is going to be equal to,
- obviously the 3 to the 1/3, that just becomes x to the 1,
- x is going to be- let me do that same blue- x over here
- is going to be equal to e to the 4 pi over 3. The 4pi over 3i.
- So lets think about- lets think about this for a little bit.
- What is this- so we could- so immediately, whats the argument here?
- So let me- let me write- let me- these are 3 different roots
- let me call them x sub 1, x sub 2, and x sub 3.
- So these are three different numbers. One of the roots is 1,
- that's pretty clear over here. 1 is one of the cubed roots of
- itself, but these are other numbers, and these are going
- to be complex numbers. So lets- lets visualize these numbers a little bit.
- So what is the argument? So for all of these- so the magnitude of
- x sub 2- the magnitude of x sub 2 is still clearly 1.
- Its the coefficient out in front of the e, its clearly 1.
- The magnitude of x sub 3 - we're gonna do that same color-
- the magnitude of x sub 3 is also clearly going to be 1.
- But what is the argument of x sub 2? What is φ?
- What is the argument? Well its 2pi over 3.
- Its 2pi over 3. So how we draw, how would we draw x sub 2?
- So the angle is 2pi over 3- I always, its easier for me to
- visualize in degrees- so 2pi is 360 degrees, 360 degrees
- divided by 3 is 120 degrees. So this is going to be 120 degrees
- is sixty short of- so its going to look like this.
- It is going to look like this.
- Just like this. This is- so this angle right here, its argument is going to be
- 100, 120 degrees which is the same thing as 2pi,
- as 2pi over 3, and its going to have the exact same length.
- So let me do this in the same color.
- So this is x sub 1, so that is this green color right over here.
- x sub 2 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 x sub 3?
- Whats its argument?
- Whats x sub 3's argument?
- Its argument is 4pi, 4pi 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, 3 goes into 720,
- 3 goes into 720- what is it?
- 240? 720...is uh 240. Right, I should have known that.
- So 240 degrees, we are gonna go 180 degrees, and then
- go another 60 degrees. So its going to be right over here.
- Its going to be right over here.
- So let me draw it like this:
- its going to right, over here.
- So this is going to be, that angle right over there is
- 4pi over 3 radians. 4pi over 3 radians, which is equal to-
- which is equal to 240 degrees. And once again, it has the
- same magnitude.
- So what we just saw is, that when I take the cubed roots
- of this real number, I'm essentially taking the entire
- - I guess we could call it the entire circle, or the entire 360 degrees,
- or the entire 2pi 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 are still not satisfied-
- you're just like, "well, you know that you said you would find complex roots or
- Yeah, you know, I- I'm not used to this, or this is actually
- being complex numbers. I want- I actually want it to be in the form a+bi."
- We can easilly figure this out from this over here.
- So x sub 2, x sub 2 is going to be equal to- its going to be equal to
- the cosine, the cos(2pi/3) + i*sin(2pi/3).
- Sine of 2pi over 3. 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 over here is 60 degrees, which it is
- because this up here is 30 degrees, and this- the hypotenuse
- or the length is 1, then this over here is the
- square root of 3/2 and then this distance right over here
- is -1/2. So x sub 2 is going to be equal to,
- is going to be equal to,
- cos(2pi/3) is -1/2. Did I do that right?
- Yep, -1/2 + i*sin(2pi/3), thats this height over here
- which is square root of 3/2i.
- So thats x sub 2, and we can do the exact same thing
- with x sub 3. x sub 3 is going to be equal to
- its x value, or I should say its real value is going to be
- the exact same thing, its going to be -1/2.
- And then it's y, its, um, its imaginary value,
- so this angle right over here, is also- its just from the
- negative real axis down to the real vector is going to be
- negative 60 degrees, so this height right over here is going to be
- negative square root 3/2. So its -1/2 minus the square root of 3/2i.
- So using this technique, we were able to find the three roots,
- the three complex roots of 1.
- This is one of them.
- This is another one.
- and of course 1 is one of them as well.
- Where did we do that?
- 1 is one of them as well.
- And you can use this exact same technique if you were finding the
- 4th root, we would take the 300- or 2pi radians, 360 degrees,
- and divide it into 4, and so it would actually be this:
- it would be i, it would be -1, and it would be -i.
- And we know, if you take i to the 4th, you get 1,
- if you take -i to the 4th, you get 1,
- and if you take -1 to the 4th you get 1,
- And if you take 1 to the 4th you get 1.
- So you could do this- you could find
- the 8th roots of 1 using this technique.
- Now the other question that might be popping into your mi-
- brain is why did I stop at e to the 4pi i?
- Why didn't I go on, why didn't I go on and say, well
- is equal to e to the 6pi i, and look for another root?
- And so if I did that, if I did that, if I said x to the 3rd, lets say
- I wanted to find a 4th root here maybe, x to the 3rd is
- equal to e to the 6pi i.
- And I take both sides of this equation to the 1/3 power.
- So I would get x is equal to-
- taking this to the 1/3, I would get e to the 2pi,
- I would get e to the 2pi i.
- Well whats e to the 2pi i?
- e to the 2pi i would just get us back to 1.
- So when I added 2pi again it just gets us
- back to this root again.
- And if I- if I took e to the 6pi- if I took e to the 8pi
- I would get this root again. So you're gonna get
- only three roots, if you're- if you're taking- if you're taking-
- well if- if you're- if you're finding the third root of something.
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