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### Course: Precalculus (Eureka Math/EngageNY)>Unit 1

Lesson 6: Topic C: Lessons 18-19: Exploiting the connection to trigonometry

# 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?
(57 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 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?
(21 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.
(17 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.
(27 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 ""?
(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.
(12 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?
(10 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.
(14 votes)
• At , 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)
• Where did he get e from? Is that just a random number he is using or is there a reason?
(6 votes)
• By definition e equals the value of (1+1/n)^n as n approaches infinity. If you want could plug in a large value of n such as 10,000. Doing this will give you approximate value of e (2.718...). e like pi is transcendental in other words it cannot be expressed as a fraction or using surds.

As it just so magically turns out e^(i*theta) = cis(theta)

Remember cis(theta) = cos(theta) + isin(theta)

This theorem can be proved using Taylor's Series. However this will involve studying calculus. The proof is given in AP Calculus BC in the last couple of lectures.

As a side note, you should be able complete all the exercises for this section despite not understanding the proof. So don't let that discourage you.
(3 votes)
• sal is speaking witchcraft man
(7 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)

## 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.