Intro to complex analysis
Exponential form to find complex roots Using exponential form to find complex roots
⇐ Use this menu to view and help create subtitles for this video in many different languages.
You'll probably want to hide YouTube's captions if using these subtitles.
- In this video we are hopefully going to understand why the
- complex form of the - why the *exponential* 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.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
|
Have something that's not a question about this content? |
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
abuse
- disrespectful or offensive
- an advertisement
not helpful
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
wrong category
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site
Share a tip
Suggest a fix
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.