# Question Video: Finding πth 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 π§ to the power of six equals one. Weβll once again use the general formula for the πth roots of unity. They are cos of two ππ over π plus π sin of two ππ over π, when π takes integer values from zero to π minus one. In this example, weβre looking to find the sixth roots of unity. So π 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 π sin of zero, which is one.

Now actually, we just saw that, on an Argand diagram, the points which represent the πth roots of unity form a regular π-gon with a vertex at the point one, zero. Thatβs this root. For the second root, we let π be equal to one. This is cos of two ππ over six plus π sin of two ππ over six. The argument here simplifies to π by three. And we could also write this in exponential form as π to the π by three π. When π is equal to two, our root is cos of four π by six plus π sin of four π by six. And this argument simplifies to two π by three.

When π is equal to three, we have cos of six π over six plus π sin of six π over six, which is negative one. And then, when π is equal to four, we have cos of eight π over six plus π sin of eight π 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 π to the negative two π by three π. Finally, when π is equal to five, we get cos of 10π over six plus π 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 π to the negative π by three π.

And we have the sixth roots of unity. In exponential form, they are one, π to the π by three π, π to the two π by three π, negative one, π to the negative two π by three π, and π to the negative π by three π. We could plot the sixth roots of unity on an Argand diagram. And weβd see that the points representing these roots form the vertices of a regular hexagon inscribed in a unit circle as shown.