Video: Finding 𝑛th Roots of Unity

Find the sixth roots of unity.

02:28

Video Transcript

Find the sixth roots of unity.

Finding the sixth roots of unity is the same as solving the equation 𝑧 to the power of six equals one. We’ll once again use the general formula for the 𝑛th roots of unity. They are cos of two πœ‹π‘˜ over 𝑛 plus 𝑖 sin of two πœ‹π‘˜ over 𝑛, when π‘˜ takes integer values from zero to 𝑛 minus one. In this example, we’re looking to find the sixth roots of unity. So 𝑛 is six. And π‘˜ takes integer values from zero through to five. The first root is found when π‘˜ is equal to zero. This is cos of zero plus 𝑖 sin of zero, which is one.

Now actually, we just saw that, on an Argand diagram, the points which represent the 𝑛th roots of unity form a regular 𝑛-gon with a vertex at the point one, zero. That’s this root. For the second root, we let π‘˜ be equal to one. This is cos of two πœ‹π‘˜ over six plus 𝑖 sin of two πœ‹π‘˜ over six. The argument here simplifies to πœ‹ by three. And we could also write this in exponential form as 𝑒 to the πœ‹ by three 𝑖. When π‘˜ is equal to two, our root is cos of four πœ‹ by six plus 𝑖 sin of four πœ‹ by six. And this argument simplifies to two πœ‹ by three.

When π‘˜ is equal to three, we have cos of six πœ‹ over six plus 𝑖 sin of six πœ‹ over six, which is negative one. And then, when π‘˜ is equal to four, we have cos of eight πœ‹ over six plus 𝑖 sin of eight 𝑖 over six. Now here, the argument simplifies to four πœ‹ by three. And this is outside of the range for the principal argument. We therefore subtract two πœ‹ from four πœ‹ by three to get negative two πœ‹ by three. And in exponential form, our fifth root is 𝑒 to the negative two πœ‹ by three 𝑖. Finally, when π‘˜ is equal to five, we get cos of 10πœ‹ over six plus 𝑖 sin of 10πœ‹ over six. This time, the argument simplifies to five πœ‹ by three, which is once again outside of the range for the principal argument. Five πœ‹ by three minus two πœ‹ is negative πœ‹ by three. And we therefore see that, in exponential form, our final root is 𝑒 to the negative πœ‹ by three 𝑖.

And we have the sixth roots of unity. In exponential form, they are one, 𝑒 to the πœ‹ by three 𝑖, 𝑒 to the two πœ‹ by three 𝑖, negative one, 𝑒 to the negative two πœ‹ by three 𝑖, and 𝑒 to the negative πœ‹ by three 𝑖. We could plot the sixth roots of unity on an Argand diagram. And we’d see that the points representing these roots form the vertices of a regular hexagon inscribed in a unit circle as shown.

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