Lesson: \( n \)-th Roots

Question 2

1 point


De Moivre’s Theorem can just as easily be used to find the roots of a number instead of the powers of a number, since it makes it easy to see that ((r-to-the-(1/n))(e-to-the-(i(theta)/n)))-all-to-the-nth-power would equal ((r-to-the-(1/n))-to-the-n)(e-to-the-(n(i(theta)/n))), which equals r(e-to-the-(i(theta))). So, this means that (r-to-the-(1/n))(e-to-the-(i(theta)/n)) is an nth root of r(ei(theta)).

Now, I say an nth root because there could be more than one. At first glance, it may not look like the formula for the nth root could yield more than one result, but the secret is in the fact that theta is not unique. We can replace theta with (theta + 2(pi)k) for any integer k, and we would still get that r(e-to-the-(i(theta))) = r(e-to-the-(i(theta + 2(pi)k))). So this means that (r-to-the-(1/n))(e-to-the-(i(theta + 2(pi)k)/n)) is also an nth root of r(e-to-the-i(theta)).

Let’s look at this process in an example. Let’s find the fourth roots of 8 – (8(root 3))i. To solve this, we first need to find a polar form for 8 – (8(root 3))i. We start by calculating the modulus equals the square root of (8-squared + (8(root 3))-squared), which equals the square root of 256, which is 16. Next, we need to find a theta such that cosine(theta) = 8/16, which is 1/2, and sine(theta) = -8(root 3)/16, which is –(root 3)/2. Now since the cosine is positive and the sine is negative, we want theta to be in the fourth quadrant, so we can use theta = 5(pi)/3. In fact, we can use theta = 5(pi)/3 + 2(pi)k for any integer k.

Let’s look at the roots we get using the various values for theta. Well, if we set theta = 5(pi)/3, then our (r-to-the-(1/4))(e-to-the-(i(theta)/4)) becomes 2(e-to-the-(i(5(pi)/12))). Now if we add 2(pi) to 5(pi)/3, we get the new theta value of 11(pi)/3, and this gives us a new value for our fourth root: 2(e-to-the-(i(11(pi)/12))). If we add yet another multiple of 2(pi), so now we’re looking at 5(pi)/3 + 4(pi), which will equal 17(pi)/3, then our fourth root value will become 2(e-to-the-(i(17(pi)/12))). If we keep going, next we’ll look at 5(pi)/3 + 6(pi), which is 23(pi)/3. This produces a fourth root value of 2(e-to-the-(i(23(pi)/12))). Adding the next multiple of 2(pi) gives us 5(pi)/3 + 8(pi), which is 29(pi)/3. This gives us a fourth root value of 2(e-to-the-(i(29(pi)/12))).

Okay, now let’s take a closer look at the last value we calculated. Since 29(pi)/12 is more than 2(pi), we can subtract 2(pi) from it, getting 5(pi)/12. Well, this is the same argument found in the first root we calculated. And from this point on, if we were to continue adding multiples of 2(pi) to theta, we would end up repeating the roots, in order, that we have already found. I’ll just show you quickly that if we were to add 10(pi), getting us 35(pi)/3, then this would create 2(e-to-the-(i(35(pi)/12))), which is the same as 2(e-to-the-(i(11(pi)/12))). Adding 12(pi) to get us 41(pi)/3 would simply give us the same root as when we used 17(pi)/3, and so on, eventually even getting to the fact where we’ve repeated back to our 29(pi)/12, which is, again, our 5(pi)/12.

So at this point, we see that the roots start cycling through yet again. Let’s take a closer look at why this is happening. So we’re starting with theta = 5(pi)/3, and then we are adding 2(pi)k, which we could think of as 6(pi)k/3, to get a value of theta_k = (5 + 6k)(pi) all divided by 3. Then when we find the fourth root, we divide theta_k by 4, giving us (5 + 6k)(pi)/12. Well, we could rewrite this as 5(pi)/12 + k(pi)/2.

From this description, it is easy to see that when k is a multiple of 4, say k = 4d, then the argument for our root is 5(pi)/12 + 4d(pi)/2, which equals 5(pi)/12 + (2(pi))d, which will always be equivalent to 5(pi)/12. If k is 1 more than a multiple of 4, we’ll say that k = 4d + 1, then the argument for our root is 5(pi)/12 + (4d + 1)(pi)/2, which equals 5(pi)/12 + (2(pi))d + pi/2, which will always be equivalent to 11(pi)/12. We can get rid of the (2(pi))d. Continuing this process, we’ll note that if k is 2 more than a multiple of 4, so k is 4d + 2, then the argument for our root will be 5(pi)/12 + (4d + 2)(pi)/2, which we can write as 5(pi)/12 + (2(pi))d + pi, and getting rid of our (2(pi))d, we see that this is always going to be equivalent to 17(pi)/12. And lastly, if k is 3 more than a multiple of 4, say k = 4d + 3, then the argument for our root is 5(pi)/12 + (4d + 3)(pi)/2, which we can write as 5(pi)/12 + (2(pi))d + 3(pi)/2, and all these will always be equivalent to 23(pi)/12.

We have now looked at all possible integer values of k: k must either be a multiple of 4, 1 more than a multiple of 4, 2 more than a multiple of 4, or 3 more than a multiple of 4. So we’ve seen that the fourth roots of 2(e-to-the-(i(5(pi)/3))) are listed in the following list. As all these are, in fact, different, we see that there are 4 different fourth roots of 8 – (8(root 3))i.

Learning from this example, we’ll find that any non-zero complex number will have n distinct nth roots. We’ll see this by going through a similar argument to the one pursued in this example. For if m is an integer, we can write m as k more than a multiple of n, say m = dn + k. And then when we add 2(pi)m to our argument theta, we will get theta + 2(pi)m will equal to theta + 2(dn + k)(pi). Then when we divide theta by n to find the argument for the nth root, we’ll get a theta/n + (2d(pi) + 2k(pi))/n, but since the 2d(pi) is a multiple of 2(pi), we can remove it from the argument without changing the complex number, getting the argument of theta/n + 2k(pi)/n.

As such, it does not matter what m is, per se—it simply matters how much m differs from a multiple of n. That k part’s going to be important, not the d, since we need only consider k values of 0 to (n – 1) to cycle through all possible integers m, since m either will be a multiple of n, or it could be 1 more than a multiple of n, or 2 more than a multiple of n, all the way up to (n – 1) more than a multiple of n. We find that no matter what multiple of 2(pi) we add to theta, our list of arguments for the nth root of r(e-to-the-(i(theta))) will be (theta/n + 2k(pi)/n) for k = 0 to (n – 1), and this means we’ve proven the following theorem.

Let z be a non-zero complex number. Then the n distinct nth roots of z = r(ei(theta)) are w-sub-k = (r-to-the-(1/n))(e-to-the-(i(theta + 2(pi)k)/n)) for k = 0 through (n – 1).

Now with the theorem in hand, let’s look at an example of finding the roots. For example, we can find all the cube roots of 27. Now, in the real numbers, 27 has only one cube root: 3. But as a complex number, 27 must have three distinct roots, one of which is still 3. So let’s put the number 27 into polar form, and use our theorem to find all the cube roots. Since 27 is a positive real number, we know that its modulus is 27 and its argument is 0. So this means that the cube roots of 27 are, the first one will be (27-to-the-(1/3))(e-to-the-(i(0)/3)). This equals 3(e-to-the-(i(0))). The second will be (27-to-the-(1/3))(e-to-the-(i(0 + 2(pi)(1))/3)), so this equals 3(e-to-the-(i(2(pi)/3))). And our third root will be (27-to-the-(1/3))(e-to-the-(i(0 + 2(pi)(2))/3)), which equals 3(e-to-the-(i(4(pi)/3))). And we can, of course, note that our very first root 3(e-to-the-(i(0))) = 3. But we also look at the process, that we added a 0 multiple of 2(pi), we added 1 multiple of 2(pi), and we added 2 multiple of 2(pi), and we stopped at 2 because it equals 3 – 1.

So let’s look at another example. Let’s find all the fifth roots of 1 + i. Now, another way you might see this question phrased is simply, find (1 + i)-to-the-(1/5), but that will still mean to find all of the roots. So we already know that (root 2)(e-to-the-(i((pi)/4))) is a polar form for 1 + i. So we can simply plug into our formula to find the fifth roots. First, by setting k = 0, we’ll get the root ((root 2)-to-the-(1/5))(e-to-the-(i((pi)/4)/5)), which equals (2-to-the-(1/10))(e-to-the-(i(pi)/20)). Next, we’ll set k = 1. This gives us ((root 2)-to-the-(1/5))(e-to-the-(i((pi)/4 + 2(pi)(1))/5)), which equals (2-to-the-(1/10))(e-to-the-(i(9(pi)/20))). Setting k = 2, we’ll get ((root 2)-to-the-(1/5))(e-to-the-(i((pi)/4 + 2(pi)(2))/5)), which equals (2-to-the-(1/10))(e-to-the-(i(17(pi)/20))). Setting k = 3, we’ll get ((root 2)-to-the-(1/5))(e-to-the-(i((pi)/4 + 2(pi)(3))/5)), which equals (2-to-the-(1/10))(e-to-the-(i(25(pi)/20))). And lastly, we set k = 4 to get ((root 2)-to-the-(1/5))(e-to-the-(i((pi)/4 + 2(pi)(4))/5)), which equals (2-to-the-(1/10))(e-to-the-(i(33(pi)/20))).

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