### Video Transcript

Find the sum of the sixth roots of
unity.

When we think about the sixth roots
of unity, theyβre the values of π§ such that π§ to the sixth power equals one. And we use a formula for finding
the πth roots of unity. The polar coordinates will be cos
of two ππ over π plus π sin of two ππ over π, where π is the integer values
between zero and π minus one.

As weβre dealing with the sixth
roots of unity, our π-value equals six. And our sixth roots will come from
π-values between zero and five. When π equals zero, we have cos of
zero plus π sin of zero. And we know that this equals
one. For π equals one, we have cos of
two π over six plus π sin of two π over six. We can simplify that argument to π
over three. And at this point, letβs just leave
this root in polar form and find the root for π equals two, which is cos of four π
over six plus π sin of four π over six. That argument simplifies to two π
over three. Again, weβll leave that there.

Then we have the cos of three π
over three plus π sin of three π over three. The cos of π plus the π sin of π
equals negative one. For our fifth root, we have the cos
of eight π over six plus the π sin of eight π over six. We know that this argument
simplifies to four π over three but that four π over three is outside the range of
the principal argument.

We can find an equivalent value
inside the range by subtracting two π. When we do that, we get negative
two-thirds π so that we have cos of negative two-thirds π plus π sin of negative
two-thirds π as our fifth root. And weβll move on to our final
sixth root, where π equals five, which is cos of 10π over six plus π sin of 10π
over six. We can simplify the argument to
five π over three. However, itβs outside the range of
the principal argument, which means again weβll subtract two π from five π over
three to get a new value of negative π over three, and our final root as cos of
negative π over three plus π sin of negative π over three.

Here we have the sixth roots of
unity listed in polar form. To sum these values, letβs rewrite
them in algebraic form. The one stays the same. The second root becomes one-half
plus the square root of three over two π. The third root is negative one-half
plus the square root of three over two π. The fourth root is the same,
negative one. The fifth root is negative one-half
minus the square root of three over two π, and the final root one-half minus the
square root of three over two π.

Our final step will be summing
these values. When we add all sixth of these
terms, we see that weβre adding one plus negative one, which equals zero. One-half plus negative one-half
equals zero. Again, negative one-half plus
one-half equals zero. From there, we have the square root
of three over two π minus the square root of three over two π. When we combine those values, we
get zero. And again, our final term is
positive square root of three over two π minus the square root of three over two
π, which equals zero. The sum of the sixth roots of unity
equals zero.

However, the process of calculating
the πth roots of unity and then summing them is not something that you would do
over and over again. And thatβs because we have a
principle that tells us the sum of the πth roots of unity for π is greater than
one is zero. By this principle, we can say that
the sum of the sixth roots of unity will be equal to zero. And weβve proven this is true by
our calculations.