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, like 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 or 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're 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'll get 8. Now it's actually possible to reverse this procedure by taking the cube root. So if you cube it, 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 is 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 power, that's 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 is 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 going to 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 r1n, which is what we have right there, but then it will be cis theta k. Now finding these theta k's is the tedious part, but if you recognize theta k you just are going to be 1n, which is what we have there times theta plus 2πk or 360 degrees 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 farther I can write this 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 going to 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 8cis45° degrees. And in these types of problems, your first step should be to figure out what this r1n is. Well, we actually already figured that out up here. We said that it's 2 because r13, 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 r1ncis theta k. So in this case when we have that zk is r1ncis each of these thetas. So the different theta k'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 theta k's that we have. So we'll have z0, z1, and z2, and then for here we'll have theta0, theta1, theta2. Now of course our last step is going to be to figure out what all these theta k'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 theta that we have and solving it. Now this is the equation for theta k, and what I see is that our first theta k that we have is theta 0. Now one over n is the same thing as 13 in this problem and we won't have 13 of our angle theta. Our angle theta 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 going to 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 theta 0 is 13 of 45 degrees, and 13 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 13 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 13 to do this by hand. And that's what I'm going to do, because I think that's going to be a little easier. So we're going to have 13 of 45 degrees, which is 15 degrees. 360 times 1 is just 360, and 13 of 360 is a 120. Now 120 degrees plus 15 degrees is a 135 degrees, so that's going to be theta 1, and that's our second solution right there. Now our last solution is going to be theta 2, and for theta 2, we're going to have 13 of 45 degrees plus 360 degrees times our last k value, which is 2. So we're going to multiply that by 2. Now again, I'm going to take this 13 here, I'm going to distribute it into the parentheses. So going to have 13 of 45 degrees, which is 15 degrees, that's going to be plus, and I'm going to distribute this 13 to the 360 degrees. Because 13 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. 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.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
11. Graphing Complex Numbers
Powers of Complex Numbers (DeMoivre's Theorem)
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