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## Precalculus

### Unit 3: Lesson 8

Multiplying and dividing complex numbers in polar form- Multiplying complex numbers in polar form
- Dividing complex numbers in polar form
- Multiply & divide complex numbers in polar form
- Taking and visualizing powers of a complex number
- Complex number equations: x³=1
- Visualizing complex number powers
- Powers of complex numbers
- Complex number polar form review

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# Complex number equations: x³=1

Sal finds all complex solutions to the equation x^3=1. Created by Sal Khan.

## Want to join the conversation?

- How does this work for z raised to a non-integer exponent?

For example: z^2.5 = 1

What could z be?(56 votes)- In general, for a^b = c, the complex and real solutions are exp() of any multiple of (2pi * 1/b)i. In this case, the answers are e^(2pi/2.5)i = e^(4pi/5)i, e^(8pi/5)i, e^(12pi/5), e^(16pi/5)i. The next one just simplifies into the first, so we are done. If you don't believe me, you can try making the 2.5th power of each of these and verify by yourself: They reduce to 1.(10 votes)

- So if the equation would be x^6=1, would there be six roots?(19 votes)
- Yes, there will be 6 roots, 2 reals and 4 imaginary. 1, (1/2) + (sqrt(3)/2)i, (-1/2) + (sqrt(3)/2)i, -1, (-1/2) - (sqrt(3)/2)i, (1/2) - (sqrt(3)/2)i. Just divide the degrees of a circle, 2pi, into 6 angles and plug into the form x = r (cos (theta) + i sin (theta)). (theta) = each angle you found.

Hope this helps!(33 votes)

- So I get the concept, and understand the math in the video. But what does it mean? What does taking the roots and drawing it on the argand diagram really prove?(20 votes)
- It shows the periodicity of the function. You can go 'round and 'round the circle and always end up on the same n points of the unit circle.(16 votes)

- If 0 = 2pi = 4pi (all in radians), then e^0i = e^2pi i = e^4pi i, then when you take the cube root of both sides, shouldn't they all equal each other? However, in the video, the cube roots are clearly different. Why is that so?(13 votes)
- As you point out, if you put e^(2*pi*i) in an advanced calculator it is = to 1, just like any e^((2n)*pi*i) where n is an integer. However, 0 != 2pi != 4pi. They are NOT equal. The reference angle measure at 0, 2pi, and 4pi is equal, not the measures themselves. So the difference is between the returned value of a reference angle and the angle itself.

Another way to think about it is this: Start facing North. Spin once, you are facing North. Spin a second time, you are facing North. All three times the result is the same, you are facing North. However, the number of times you have spun are 0, 1, and 2 and NOT equal. This allows a calculation that involves both the facing and the spin to use elements that are both equal and not.(25 votes)

- I know this is a really stupid question and against the point of this video, but why couldn't you just take the cube root of both sides at "0:18"?(9 votes)
- Well, think about this: x^2=1

If you take the square root of both sides, you get x=1. But x=-1 is also valid. Because you're taking the principal square root to get x=1. Same in this case, you would be taking the principal cube root if you would be x=1. but if you think about the non-principal cube roots, either you use the method of this video or you use factorisation.(11 votes)

- Is deMoivre's Theorem needed to find roots if we use the method describe in the video to find roots?(5 votes)
- z^5 =-243i

Solving for theta

1) 5⋅theta=270+k⋅360

theta=54+k⋅72

Remember that theta is strictly between 180-270

Therefore, we need to find the multiple of 90

90, degree that is strictly within the range of 180-54=126

and 270-54=216

multiple is simply 144

theta = 198

My questions are as follows:

1) How do you know 5⋅theta=270+k⋅360?

Where does the 270 come from I wanted to say either 180 or 90.

2) How do you get 144 from 126, 216 and even after theta=198

is 144 found from 360-216. and you pick 216 as the right value because it alone is between 180 and 270?(9 votes)

- Why is the magnitude of z notated as |z|? Isn't that absolute value notation?(3 votes)
- Because the absolute value is just a distance, and the magnitude is the length, or distance, of the complex number, the absolute value and the magnitude is the same thing.(12 votes)

- At9:10, How can Khan say its √3/2 ? Plz HELP ME!(7 votes)
- If you factor x^3-1 you get x-1 and x^2+x+1 which solves to x=1 and x=+/-(sqrt-3)/2 using the formula (-b+/-sqrt(b^2-4ac))/2a. The imaginary numbers are the same as Sal found.(2 votes)

- do you HAVE to divide the graph into 3 angles since the original problem was x^3=1?(3 votes)
- Yes, the power on x assures there will be maximum 3 roots, so 3 points/roots on the unit circle.(6 votes)

- How do we find out which number to add to (k*360)? sometimes we add 180 and sometimes 90..(5 votes)
- Not sure if this will still help or not since it's five months past, but the way I see it is on the number plane. Where the x is the real number and y is the imaginary (i). If the equation gives you something like x^3 = i. It means that the point on the plane is at (0,1). Thus giving you the 90 + k(360) [90 because of how it's a right triangle]. While if the equation is something like x^5 = -93, it gives you the point of (0,-93) which starting from the point (1,0) would give 270 degrees.(2 votes)

## Video transcript

In this video, we're going
to hopefully understand why the exponential
form of a complex number is actually useful. So let's 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 1 is equal to 0. So we're looking for all the
real and complex roots of this. And there are ways to do
this without exponential form of a complex number. But the technique we're
going to see in this video could be applied if this
was x to the fifth minus 1, or x to the 13th minus 1. And it's also going to
show us the patterns that emerge when you start looking
at things on an Argand diagram. So to do this, let's think about
the exponential representation of 1. So let's just say
z is equal to 1. 1 is a complex number. It's a real number. All real numbers are
also complex numbers. They're a subset. They're in the complex plane. 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. And so this is the real. This is 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. Negative 1. Negative 1. z looks like this. 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 0i. Now, let's put this
in exponential form. Well, its magnitude is
pretty straightforward. The magnitude of z is just
the length of this vector, or it's the absolute value of 1. That's just going to be 1. Now, what's the argument of z? What's the angle
that this vector makes with the
positive real axis? Well, it's on the
positive real axis. It's a real number. So it has no angle. So the arg of z is 0. So that might not be
too interesting so far. We just figured out that 1 is
equal to 1 times e to the 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 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. So the argument of our complex
number-- or of the number 1, 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
as 1 times e-- I won't write the 1
anymore-- 1 times e to the 2 pi i, or 1
times e to the 4 pi i. And the reason why
this is interesting is then this equation
right here can be written in multiple ways. It can be written as 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 i. Or it could be written
as x to the third is equal to 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 one-third
power to solve for x. So to the one-third. We're going to take
that to the one-third. We're going to do that
same thing over here. We're just taking everything
to the one-third power to solve for the x's in
each of these equations. To the one-third power. So this first equation over
here becomes x is equal to 1 to the one-third 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, this is going to be 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 3 to the one-third, that
just becomes x to the 1. Let me do that same blue. x over here is going to be equal
to e to the 4 pi over 3, i. So let's think about
this for a little bit. What is this? So immediately, what's
the argument here? So these are three
different roots. Let me call this x1, x2, and x3. So these are three
different numbers. One of the roots is 1. That's pretty clear over here. 1 is one of the cube
roots of itself. But these are other numbers. And these are going
to be complex numbers. So let's visualize these
numbers a little bit. So what is the argument? The magnitude of x2
is still clearly 1. It's the coefficient
out in front of the e. It's clearly 1. Let me do that same color. The magnitude of x3 is
also clearly going to be 1. But what is the argument of x2? What is phi? What is the argument? Well, it's 2 pi over 3. So how would we draw x2? So the angle is 2 pi over 3. 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. So this angle right
here, its argument is going to be
120 degrees, which is the same thing
as 2 pi over 3. And it's going to have
the exact same length. So let me do it
in 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 rotate it. We rotate it 120 degrees. And what about x3? What's its argument? What's x3's argument? Its argument is 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-- what is 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. So let me draw it like this. It's going to be
right over here. That angle right
over there is 4 pi over 3 radians, which
is equal to 240 degrees. And once again, it has
the same magnitude. So what 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 360 degrees or the
entire 2 pi radians-- and I'm dividing it
into three, essentially. This is one third. Then we have
another 120 degrees. And then we have
another 120 degrees. And so you see the pattern of
where all of the roots are. And in case you're
still not satisfied, you're just like, well, you said
you would find complex roots. Yeah, I'm not used
to this or this as actually being
complex numbers. I actually want it to be in the
form a plus bi-- we can easily figure it out from
this right over here. So x2-- it's going to be equal
to the cosine of 2 pi over 3 plus i times the
sine of 2 pi 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-- the hypotenuse,
or the length, is 1, then this over here is
square root of 3 over 2. And then this distance right
over here is negative 1/2. So x2 is going to be equal
to-- cosine of 2 pi over 3 is-- negative 1/2. Did I do that right? Yep, negative 1/2, plus i
times sine of 2 pi over 3. That's this height
over here, which is square root of 3 over 2, i. So that's x2. And we could 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, its imaginary
value, so this angle right over here-- 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 2, i. So using this technique,
we were able to find 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 could use this
exact same technique if we were finding
the fourth roots. We would take the 2 pi
radians, or the 360 degrees, and divide it into 4. And so it would
actually be this. It would be i. It would be negative 1. And it would be negative i. And we know if you take i
to the fourth, you get 1. If you take negative i
to the fourth, you get 1. And if you take negative 1
to the fourth, you get 1. And if you take 1 to
the fourth, you get 1. And so you can find
the eighth roots of 1 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 i? Why didn't I go
on and say, well, this is equal to e to the 6 pi
i and look for another root? And so 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 one-third power. Taking this to the one third,
I would get e to the 2 pi i. Well, what's e to the
2 pi i? e to the 2 pi i would just get us back to 1. So when I added 2 pi again, it
just gets us back to this root again. If I took e to the 6 pi,
if I took e to the 8 pi, I would get this root again. So you're going to get
only three roots if you're finding the third
roots of something. Anything beyond that, it
just becomes redundant.