Powers of Complex Numbers (DeMoivre's Theorem) - Online Tutor, Practice Problems & Exam Prep

To raise a complex number in polar form to a power, apply de Moivre's theorem: raise the modulus \( r \) to the power \( n \) and multiply the angle \( \theta \) by \( n \). For roots, use the formula \( r^{1/n} \) and adjust the angle by \( \frac{\theta + 360^\circ k}{n} \) for \( k = 0, 1, \ldots, n-1 \). This process reveals multiple solutions, essential for understanding complex number operations and their applications in various mathematical contexts.

Powers Of Complex Numbers In Polar Form (DeMoivre's Theorem)

So in recent videos, we've been talking about how you can take complex numbers in polar form and how you can do different operations on them. So let's say I had this complex number right here in polar form, and I wanted to multiply it by itself. Well, we learned how to multiply complex numbers by taking the product of the \( r \) values and adding the angles. But it turns out there's another way that we could go about doing this, which is a much simpler approach. What you could do instead is take this complex number and raise it to a power of 2. And raising complex numbers to powers is something we're going to be talking about in this video. This is a skill you're going to need to have, and there's a very intuitive shortcut to do this. So let's just jump right into an example.

So if you wanted to take a complex number and raise it to some kind of power, what you want to do is, whatever that power is, we'll call it \( n \), you're going to take the \( r \) value out in front and you're going to raise it to the \( n \)th power, and then you're going to multiply this \( n \) by the angle theta. So by doing this, you'll be able to complete this operation, which we call de Moivre's theorem. De Moivre's theorem is just the equation that we named for this situation. So let's go ahead and do this. First, I'm going to take this \( r \) and I'm going to raise to the \( n \). Now I can see that the \( r \) we have is 3, so \( 3^2 \) would be \( 3^2 \), and then this is all going to be multiplied by the cosine of \( n \) times our angle. So we're multiplying the angles here. So we're going to have \( n \) which is 2 multiplied by 15 degrees, And then what we're going to do is have \( i \) times the sine of this whole thing as well. So doing this we're going to have +isin(n × 15°). So this right here would be the operation and that's how we can apply De Moivre's theorem to this example. So let's go ahead and simplify things. Now, \( 3^2 \), that's the same thing as 9. And then what I can do is recall that we talked about in previous videos, we can take this whole cosine and sine thing and we can reduce it to just cis. This is an abbreviated version. So you can see this is the same thing as \( 9 \) cis, and then we have \( 2 \times 15 \), which is 30 degrees. So this right here would be the solution to our problem, and that's how you can raise complex numbers to a power. Now as you can see, it gives us the exact same result as if we were to just multiply this number by itself. So this is the shortcut to doing these types of problems.

Now this does beg actually an interesting question, which is why would we necessarily need this shortcut? I mean, yeah, it might be slightly faster, but wouldn't it be easier just to remember one equation like this? Well, there are actually a lot of situations where this shortcut is actually going to be needed to solve the problem or is going to be way simpler than trying to just multiply things out. And to understand when this happens, well, let's take a look at this example down here. So in this example, we're asked to evaluate the following expression. And here we have this complex number where we're given the angle in radians, and notice that we're now raised to the 3rd power. So if I were to try and take this complex number and multiply it by itself 3 times in a row, that would be really tedious. We'd have so much going on. But using this de Moivre's theorem shortcut, we can actually do this very quick. So let's try this. So what I'm first going to do is take 4, the \( r \) value, I'm going to raise it to a power of \( n \). So it's going to be \( 4^3 \), and then this cosine and sine will become cis, and then we're going to have \( n \times \) our angle. So that means that we're going to have this, this \( n \) value 3 multiplied by our angle \( \frac{\pi}{6} \). Now \( 4^3 \) that's the same thing as 64 and then that's going to be cis and \( 3 \times \frac{\pi}{6} \), well, this 3 is actually going to reduce with that 6 because 6 is the same thing as 2 times 3, so we can just cancel a 3. So that means we're going to have cis \( \frac{\pi}{2} \). And this right here is the solution to this example. So that is how you can solve these types of problems. As you can see, de Moivre's theorem allowed us to do this very quickly when we just applied this shortcut, and we're able to get our answers. So hope you found this video helpful. Thanks for watching, and let's get some more practice.

Given $z=3CiS\left(\frac{4\pi}{5}\right)$, calculate $z^5$.

• A. $z^5=243CiS\left(4\pi\right)$
• B. $z^5=243CiS\left(32\pi\right)$
• C. $z^5=243CiS\left(100\right)$
• D. $z^5=243CiS\left(4\right)$

Powers Of Complex Numbers In Polar Form (DeMoivre's Theorem) Example 1

This problem is really interesting because we are asked to perform multiple operations. We are given 2 complex numbers both in polar form, \( z_1 \) and \( z_2 \). We are asked to find \( z_1 \) divided by \( z_2 \), all raised to the 3rd power. So we're going to need to combine multiple skills that we've learned to solve this and to get our final solution. Let's just go ahead and go over this example to see what happens. Now the way that I'm going to do this is to break things up into 2 steps. My first step is going to be to deal with everything inside these parentheses, which is finding this quotient or basically dividing. Now my second step is going to be, once I've divided these 2, to take whatever result that I get and to raise it to the 3rd power. So I'll deal with the division first and the power second. So let's go ahead and start with step 1.

For step 1, I need to find what \( z_1 \) over \( z_2 \) is. Now to do this, we have discussed what these rules are for division. What you want to do is divide the \( r \) values and then subtract the angles. Now going up here, I can see that our \( r \) values are 4 and one half. So for \( z_1 \) divided by \( z_2 \), we're going to have 4 divided by 1 half, and then that's going to be cis and then the angles subtracted. Now, I can see that this angle here is 25 degrees. So gonna have 25 degrees minus the angle for \( z_2 \) which I can see is 10 degrees. So this right here is what we end up getting when we plug everything in. Now at this point it just turns into reducing everything and simplifying. Now, 4 divided by 1 half, we can flip this fraction and bring it to the top. That's the same thing as 4 times 2. And 4 times 2 is going to give us 8. So we're going to have 8 cis, and by the way this is also like dividing 4 by 0.5 that should give you 8 as well. So we're gonna have 8 cis, and then we have 25 minus 10, which is 15 degrees. So this is what happens when we divide \( z_1 \) and \( z_2 \), and that is what we end up getting. Now this is not the final solution because recall that this is just the first step. We've dealt with the first step here. Now what we need to do next is we need to raise this to the power of 3. So I'm gonna do that over here.

So we'll have \( z_1 \) divided by \( z_2 \), and then that's all gonna be raised to the 3rd power. And to do this, well, we figured out what \( z_1 \) divided by \( z_2 \) is. So what I can do is apply De Moivre's theorem to figure out what this is raised to the third power. So we're going to have 8, and recall we need to take the \( r \) value and raise it to the 3rd. So we'll have \( 8^3 \), and that's going to be cis. Then we need to take whatever value we have here and multiply it by the angle of 15 degrees that we calculated. So we're gonna have this number, 3 multiplied by 15 degrees. So this is what we have. Now from here, once again, our solution is going to be found by simplifying. Now first off, I need to deal with this \( 8^3 \). \( 8^3 \) is 8 times 8 times 8, which you can put on a calculator. This should come out to 512. So we're gonna have 512, and that's going to be cis 3 times 15. 3 times 15 is 45 degrees, and this is what we get. So this right here would be the final solution to the problem. That's going to be \( z_1 \) divided by \( z_2 \), all raised to the 3rd power. So, that is how you can solve these types of situations where you have multiple operations on a complex number in polar form. You first want to deal with whatever is inside the parentheses, and then deal with any kind of exponents or anything else outside of the parentheses. Hope you found this video helpful. Thanks for watching.

Complex Roots

There. So up to this point, we've talked a lot about operations you can do on complex numbers written in polar form. Now in this video, we're going to be taking a look at how you can find the roots of complex numbers, such as the square root or the cube root. Now, something I'm going to warn you about: the process for doing this is actually quite tedious. And before we get into things, I may recommend that you check with your professors, teachers, or instructors to see if this is something you're actually going to need to do. Because oftentimes in courses, this will actually be taught at the end, so it's not something you might necessarily need to know. But if this is something you are going to need to know for your course, don't sweat it because we're going to go over some examples and situations that will hopefully make this process seem super straightforward. Let's just go ahead and get right into things.

Now we've talked about how to deal with real numbers. Right? And if we have some number, let's say, 2, and we raise it to the 3rd power, we get 8. Now, it's actually possible to reverse this procedure by taking the cube root. So, if you cube 2, taking a cube root will get you back to 2. This is something we should know by now. Now if we're dealing with a complex number, we learned in the last video how you can take a complex number and raise it to a power using de Moivre's theorem. So in theory, by taking this complex number and raising it to the 13 power or cube rooting it, it should get us back to where we started. But let's actually see if this is true.

So, let's go over to this complex number here, and let's see if we can apply de Moivre's theorem to this exponent that we have. So I'm going to take this 8, this is the r value, and I'm going to raise it to the 13 power. So we're going to have 813, and then we'll have cis in front of this, and keep in mind cis is just a combination of the sines and cosines. It's cosine theta plus i sine theta. So we're going to have cis, and then it's going to be 13 of the angle that we have started with 45 degrees. Now, as you can see, 813 is going to give you 2. And 13 of 45 degrees is 15 degrees. So it looks like we actually did get back to where we started.

Well, it turns out this is actually not the only solution, and that's what makes dealing with complex numbers in polar form very tedious when you're trying to find these roots. Because when it comes to these numbers, you would actually have multiple roots that form when you do this. So what's actually going to happen is you're going to have cis 13 of 45 degrees, but then we need to take this whole thing, and we need to add 360 degrees times k. And this will actually give you the full result.

Now, of course, this begs an interesting question. What's with this 360 degrees, and what exactly is k? Well, it turns out k is actually a number that depends on whatever number you have here. So if we say that 3 here is n, k actually depends on this n value, which we're going to talk about more in a moment here. And 360 degrees, that just accounts for a full rotation. So it's possible that you'll see 360 degrees or you'll see 2π. In this case, since we were dealing with degrees, 360 degrees is what we dealt with. But if we had radians, we would say 2π. And you're gonna take that multiplied by k, and this will give you all possible solutions when dealing with roots of complex numbers.

So what the equation looks like is something like this. You'll have r to the one over n, which is what we have right there, but then it will be cis θk. Now finding these θk's is the tedious part, but if you recognize θk, you just gonna be 1 over n, which is what we have there, times θ plus 2πk or 360 degrees times k, then you can find these solutions whenever you have them. And the k values are going to go from 0 all the way up to n minus 1, and they will be the integer values that you have. So in this case, we can see that n is 3 and 3 minus 1 is 2, so that means our k values are going to be 0, 1, and 2.

So if you take this 0 and you replace the k with it and then solve that, then you_take 1 and replace it with k, then you take 2 and replace it with k, you'll get 3 different solutions and that will be the final solution to your problem. Now if I simplify this a bit further, I can write this 2 to the 13 as 2, so it's going to be 2, and then we'll have cis 13 of 45 degrees plus 360 degrees, and that 360 is gonna be multiplied by k. And this right here would be the solution to what we have.

Now to actually see what all of these solutions are going to be, to write them out to make sure we have this down, let's actually try an example where we're dealing with the same numbers we have here. So in this example, we're asked to find the cube root of 8cis 45 degrees. And in these types of problems, your first step should be to figure out what this r to the one over n is. Well, we actually already figured that out up here. We said that it's 2 because r to the one over n, in this case, is going to be 8, and since we're finding a cube root, it's 8 to the 1 third. 8 to the 1 third is 2. So that means that every single z value that we have, every solution that we're going to have is going to have 2 as the r value, and that's our first step.

Now our next step is going to be to set up all of the different z solutions, which remember it's r to the one over n cis each of these thetas. So in this case when we have that zk is r to the one over n cis each of these thetas. So the different θk's, well we said that k is going to be 0, 1, and 2. So down here if k is going to be 0, 1, and 2, then those are going to be all the z's and θk's that we have. So we'll have z₀, z₁, and z₂, and then for here we'll have θ₀, θ₁, and θ₂. Now, of course, our last step is going to be to figure out what all these θk's are, and this is really the tricky part when dealing with complex roots. But we're going to go ahead and do this by just going over each θ that we have and solving it.

Now this is the equation for θk, and what I see is that our first θk that we have is θ₀. Now one over n is the same thing as 1 third in this problem and we won't have 1 third of our angle θ. Our angle θ is 45 degrees then that's going to be plus 360 degrees times our first k value, which is 0. So I can see that we have 0 here, so we're just gonna multiply this 360 degrees by 0. Now anything times 0 is just going to be 0. So all we're going to end up having is that θ₀ is 1 third of 45 degrees, and 1 third of 45 is 15. So the first angle we have is 15 degrees, and that's the first solution right there. Now let's find our second solution. Our second solution is going to be 1 third of 45 degrees plus, and they're going to have 360 times the next k value. The next k value is going to be 1, so we're going to multiply that by 1. Now the way that I can do this is I could combine things here, or I could actually just distribute this 1 third to do this by hand. And that's what I'm gonna do, because I think that's gonna be a little easier. So we're going to have 1 third of 45 degrees, which is 15 degrees. 360 times 1 is just 360, and 1 third of 360 is a 120. Now 120 degrees plus 15 degrees is a 135 degrees, so that's going to be θ₁, and that's our second solution right there. Now our last solution is going to be θ₂, and for θ₂, we're going to have 1 third of 45 degrees plus 360 degrees times our last k value, which is 2. So we're gonna multiply that by 2. Now again, I'm going to take this 1 third here, I'm going to distribute it into the parentheses. So going to have a 1 third of 45 degrees, which is 15 degrees, that's going to be plus, and I'm going to distribute this 1 third to the 360 degrees. Because 1 third of 360 is a 120 degrees, and that's still going to be multiplied by this 2 though. Now from here, we're going to have 15 degrees plus a 120 times 2 which is 240 degrees. So 15 degrees plus 240 degrees, and at 15 plus 240 that's going to be 255 degrees. So our last solution has an angle of 255 degrees, and this right here is z₀, z₁, and z₂_, and those are all our solutions and our 3rd step. So that is how you can solve these types of problems. As you can see, you get multiple solutions and multiple roots. But as long as you go through this process of finding each of the θ values and then finding each of the corresponding z values, solving these problems becomes pretty straightforward. So that is how you can take roots of complex numbers. Let's go ahead and get some more practice with this.

Complex Roots Example 1

Let's give this problem a try. So in this problem, we are told if \( z \) is equal to \( 1024 \times \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) \), calculate the 5th root of \( z \). And recall that the 5th root of \( z \) is the same thing as \( z^{1/5} \). Now how we can solve this problem is by recognizing that we're dealing with a complex root. So for complex roots, what I'm going to do is write out everything that we know about these. Now I know that if I want to find \( Z^{1/n} \), that's going to equal the \( r \) value, the number out in front, to the \( 1/n \) power, and that's going to be times cis where cis is a combination of the cosines and sines it's gonna be cis \( \theta_k \). Now \( \theta_k \) is the tedious part of the problem where you have to calculate each of these angles, but I know that \( \theta_k \) is going to be \( 1/n \), and that's going to be that times our angle \( \theta \) plus \( 2\pi \times k \). Now since I can see that we have radians in this problem, I'm using radians, so \( 2\pi \). Now for the \( k \) values, the \( k \) values are going to start at 0 and it's gonna go 1, 2, all the way up to \( n-1 \). So those are gonna be the \( k \) values, this is \( \theta_k \), this is each of the \( z \) values, and that's the ultimate setup. Now what I can do is figure out in this problem what the \( r \) values are going to be and what the \( k \) values are going to be because that's going to help me a lot. So if I go over here, I can see that the \( r \) value is 1024. If I want to find \( r^{1/n} \), in this case, \( 1/n \) is \( 1/5 \), so you can see that this is gonna be equal to \( 1024^{1/5} \). Now this is something you can put into your calculator because this is gonna be pretty tedious to try and figure this out, but the 5th root of 1024 turns out to just be 4. So this is going to be the \( r \) value we use throughout this problem. Now next, what I'm going to do is figure out what our \( k \) values are because \( k \) values go from 0 all the way up to \( n-1 \). Well, I can see that \( n-1 \) is going to be 5 minus 1 this number of the denominator, and 5 minus 1 is 4. So that means our \( k \) values are going to be 0, 1, 2, 3, and 4, and these are the \( k \) values we're going to use in this problem. So as we have the \( r \) and we have the \( k \), so we can actually set up what this problem is going to look like. I can see that we have 1, 2, 3, 4, 5 \( k \) values, which means we're going to have 5 solutions in this problem. So we're going to have \( z_0 \), we're gonna have \( z_1 \), \( z_2 \). We're gonna have the \( z \) for each of these \( k \) values, so we're going to have go over here we'll have a \( z_3 \), and then we'll have \( z_4 \). And these are going to be all the solutions we have in this problem. Now for \( z_0 \), this is going to be this pattern right here. So we're going to have \( r^{1/n} \) which we said was 4. So we have \( 4 \) cis, and again, this is just the cosines and sines, and that's going to be cis some angle, and then for \( z_1 \) we're going to have \( 4 \) cis some angle, and this pattern just repeats. So here we're gonna have \( 4 \) cis some angle, and you can keep doing this for each of these solutions we get. Because the pattern stays the same, the \( r \) value stays the same, but it's just the angle inside that's going to change. Because what we're trying to do now from here is figure out what this \( \theta_k \) is, and that's really the tricky part of solving t