### 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. Weβll once again use the general
formula for the πth roots of unity. They are cos of two ππ over π
plus π sin of two ππ over π, when π takes integer values from zero to π minus
one. In this example, weβre 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 πth roots of unity form a regular
π-gon with a vertex at the point one, zero. Thatβs this root. For the second root, we let π be
equal to one. This is cos of two ππ over six
plus π sin of two ππ 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 weβd see that the points
representing these roots form the vertices of a regular hexagon inscribed in a unit
circle as shown.