### Video Transcript

Find the ninth roots of unity.

To find the ninth roots of unity,
weβre trying to solve all the values of π§ such that π§ to the ninth power equals
one. And we know a formula for
calculating πth roots of unity will be cos of two ππ over π plus π times sin of
two ππ over π which is equal to π to the two ππ over π times π power, for
π-values beginning at zero and ending with π minus one. Since weβre dealing with the ninth
root, our π is equal to nine. And to find the ninth roots of
unity, weβll start with π equals zero. Our first term becomes cos of zero
plus π sin of zero, which would be π to the zero power, which is equal to one.

Next, we have π equals one. If we plug in one for π and nine
for π, we can simplify this argument to two-ninths π, which gives us π to the
two-ninths ππ power. Now for π equals two, we get
four-ninths π, which will give us π to the four-ninths ππ power. For π equals three, our argument
is six-ninths π, which we can reduce to two-thirds π, which will give us π to the
two-thirds ππ power. Next, we have π equals four. When we plug in four for π and
nine for π, we get eight-ninths π, which gives us π to the eight-ninths π times
π power.

At this point when we get to π
equals five, we get cos of ten-ninths π plus π sin of ten-ninths π. The problem here is that ten-ninths
π is outside the range for the principal argument. We want the principal argument to
be greater than negative π and less than or equal to π. And that means we need to take
ten-ninths π and subtract two π, which we write as eighteen-ninths π. When we do this subtraction, we get
negative eight-ninths π. We want to substitute that value in
in place of the ten-ninths π, which will give us π to the negative eight-ninths
ππ.

Notice the interesting pattern
here. The fourth root was π to the
eight-ninths ππ, while the fifth root was π to the negative eight-ninths
ππ. Continuing on, we have π to the
sixth. And our argument is twelve-ninths
π, which simplifies to four-thirds π. But again, this is outside the
range of the principal argument, which means we would need to subtract two π from
the four-thirds π. When we do that, we get negative
two-thirds π so that we have π to the negative two-thirds ππ.

And again we have this pattern; π
equals three gave us π to the positive two-thirds ππ, and π equals six gave us
π to the negative two-thirds ππ. Using this pattern for π equals
seven, we can expect that it would be equal to π to the negative four π over nine
π. If we plugged in π equals seven,
we get a 14 over nine π. And by subtracting two π, we get
negative four-ninths π. We can use this pattern to solve
for π equals eight. When π equals eight, we have a
root of π to the negative two-ninths ππ. From here we wanna list out these
roots in order beginning with π equals zero, which would give us the list that
looks like this.