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.
