The portal has been deactivated. Please contact your portal admin.

Question Video: Finding the Cubic Roots of Unity Mathematics

Find all the values of 𝑧 for which 𝑧³ = 1.

04:05

Video Transcript

Find all the values of 𝑧 for which 𝑧 cubed equals one.

Here we have an equation 𝑧 cubed equals one, where 𝑧 is a complex number. There are a number of ways we can solve this equation. One is to recall De Moivre’s theorem for roots. The other is to rearrange this equation and solve 𝑧 cubed minus one equals zero. To do this, we would need to spot one root and use the factor theorem, followed by either polynomial long division or equating coefficients to find the other roots. Let’s look at how we might solve this using De Moivre’s theorem.

We’ll use De Moivre’s theorem for a complex number written in polar form. That’s π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ, where π‘Ÿ is the modulus and πœƒ is the argument of the complex number in radians. This says we can calculate 𝑧 to the power of one over 𝑛 by finding π‘Ÿ to the power of one over 𝑛 multiplied by cos of πœƒ plus two πœ‹π‘˜ over 𝑛 plus 𝑖 sin πœƒ plus two πœ‹π‘˜ over 𝑛 when π‘˜ is equal to zero all the way through to 𝑛 minus one. We’ll begin then by expressing the number one in polar form. Its real part is one. And its imaginary part is zero. So it’s a fairly easy number to represent in polar form.

If we represent one on an Argand diagram, we see it can be represented by the point whose Cartesian coordinates are one, zero. The modulus of this number is the length of the line segment that joins this point to the origin. That’s clearly one. The argument is the measure of the angle that this line segment makes with the positive real axis. And that’s measured in a counterclockwise direction. So we can see that it’s zero. In polar form then, one is the same as one multiplied by cos of zero plus 𝑖 sin of zero. And if we substitute this back into our equation, we see that 𝑧 cubed is therefore equal to one times cos of zero plus 𝑖 sin of zero.

Since we’re going to be solving this equation, we need to calculate the value of 𝑧. And to do this, we find the cube root of each side of the equation. Now if we compare this to De Moivre’s theorem, we can see that 𝑛, in our case, is three. So 𝑧 must be equal to one to the power of one-third multiplied by cos of zero plus two πœ‹π‘˜ over three plus 𝑖 sin zero plus two πœ‹π‘˜ over three. And of course, we can simplify this somewhat. We know that one to the power of a third is simply one. And our argument zero plus two πœ‹π‘˜ over three can be simply written as two πœ‹π‘˜ over three.

We’re now going to apply the final part of De Moivre’s theorem. Since our value of 𝑛 is three, we’re going to substitute π‘˜ is equal to zero, one, and two into this equation. If we substitute π‘˜ is equal to zero in, we get 𝑧 is equal to cos of zero plus 𝑖 sin of zero. Now, cos of zero is one. And 𝑖 sin of zero is zero. So the first solution to our equation is 𝑧 is equal to one. We then substitute π‘˜ is equal to one. And the argument becomes two πœ‹ multiplied by one over three, which is simply two πœ‹ by three. And our second solution is cos of two πœ‹ by three plus 𝑖 sin of two πœ‹ by three.

Our final solution is found when π‘˜ is equal to two. We get 𝑧 is equal to cos of four πœ‹ by three plus 𝑖 sin of four πœ‹ by three. Now the argument of this solution is outside of the range for the principal argument. And this is πœƒ is greater than negative πœ‹ and less than or equal to πœ‹. We can add and subtract multiples of two πœ‹ to four πœ‹ by three so we can express this solution with its principal argument.

Four πœ‹ by three minus two πœ‹ is negative two πœ‹ by three. And we can see that the cubic roots of one are 𝑧 is equal to one. 𝑧 is equal to cos of two πœ‹ by three plus 𝑖 sin of two πœ‹ by three. And 𝑧 is equal to cos of negative two πœ‹ by three plus 𝑖 sin of negative two πœ‹ by three. These are the cubic roots of unity, so-called because they’re all possible values for the cube root of one.

We can actually express these in algebraic form as well. The first solution is still simply one. The second solution is negative a half plus root three over two 𝑖. And the third solution is negative a half minus root three over two 𝑖.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.