# Question Video: Finding 𝑛th Roots of Unity Mathematics

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 𝜋𝑘 over 𝑛 plus 𝑖 sin of two 𝜋𝑘 over 𝑛, when 𝑘 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 𝑘 takes integer values from zero through to five. The first root is found when 𝑘 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 𝑘 be equal to one. This is cos of two 𝜋𝑘 over six plus 𝑖 sin of two 𝜋𝑘 over six. The argument here simplifies to 𝜋 by three. And we could also write this in exponential form as 𝑒 to the 𝜋 by three 𝑖. When 𝑘 is equal to two, our root is cos of four 𝜋 by six plus 𝑖 sin of four 𝜋 by six. And this argument simplifies to two 𝜋 by three.

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

And we have the sixth roots of unity. In exponential form, they are one, 𝑒 to the 𝜋 by three 𝑖, 𝑒 to the two 𝜋 by three 𝑖, negative one, 𝑒 to the negative two 𝜋 by three 𝑖, and 𝑒 to the negative 𝜋 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.