### Video Transcript

If π is a primitive sixth root of
unity, which of the following expressions is equivalent to π plus π squared plus
π cubed? Option (A) π to the fourth power
plus π to the fifth power plus π to the sixth power. Option (B) one minus π to the
fourth power minus π to the fifth power. Option (C) one. Option (D) negative one times one
plus π to the fourth power plus π to the fifth power. Or option (E) one-half times π
squared plus π to the fourth power plus π to the sixth power.

In this question, weβre told that
π is a primitive sixth root of unity. And we need to use this information
to determine an equivalent expression to π plus π squared plus π cubed. Since π is a primitive sixth root
of unity and the expression weβre given involves summing powers of π, we can start
by recalling a very useful result about the sums of powers of roots of unity. We recall for any integer π
greater than one and if π§ is a primitive πth root of unity, then one plus π§ plus
π§ squared β and we sum all the way up to π§ to the power of π minus one β will be
equal to zero. Weβre going to want to use this
fact to find something equivalent to the expression given to us in the question.

First, weβre told that π is a
primitive sixth root of unity, so weβll set our value of π equal to six. And then using π as our primitive
root of unity, we get one plus π plus π squared plus π cubed plus π to the
fourth power plus π to the power of five is equal to zero. And of course, we want to use this
to find an expression for π plus π squared plus π cubed. And we can see that this already
appears inside of our expression. So weβll rearrange our equation to
make this the subject. Weβll do this by subtracting one
from both sides of our equation. Weβll subtract π to the fourth
power from both sides of our equation. And weβll subtract π to the fifth
power from both sides of our equation.

Doing this, we get that π plus π
squared plus π cubed is equal to negative one minus π to the fourth power minus π
to the fifth power. And now we can simplify this even
further. We can notice all three terms on
the right-hand side of our equation share a factor of negative one. So weβll take out the shared factor
of negative one. This gives us negative one times
one plus π to the fourth power plus π to the fifth power. And then we can see this is exactly
the same as the answer given in option (D). So we could just stop here. However, there is one thing worth
pointing out.

We might be wondering why this
identity is true in the first place. And one way to see that this is
true is to look at this geometrically. Weβre going to plot this onto an
Argand diagram. First, weβre going to want to use
another fact about roots of unity. If π§ is an πth root of unity,
then the modulus of π§ is going to be equal to one. And in fact, we can directly show
this is true. First, if π§ is an πth root of
unity, then π§ to the πth power will be equal to one. Weβre then going to take the
modulus of both sides of this equation. This gives us the modulus of π§ to
the πth power will be equal to the modulus of one. And of course, the modulus of one
is just equal to one. And the modulus of π§ to the πth
power is equal to the modulus of π§ all raised to the πth power.

Now we need to use the fact that
the modulus of any number is always nonnegative. This is because the modulus
represents the distance of our point from the origin on our Argand diagram, and
distances can never be negative. So we have a nonnegative number
raised to the πth power is equal to one. This means that the modulus of π§
must be equal to one. So letβs think about what weβve
shown. Weβve shown for any πth root of
unity, the modulus of this number has to be equal to one. And if their modules is equal to
one, their distance from the origin on our Argand diagram is equal to one. In other words, all πth roots of
unity lie on the unit circle centered at the origin on our Argand diagram.

And weβre going to need to use one
more piece of information. We know how to generate all of the
πth roots of unity. In particular, we can write them in
polar form. Weβll write them in the form the
cos of two ππ over π plus π times the sin of two ππ over π, where π is an
integer between zero and π minus one. So we know all of our roots of
unity lie on the unit circle. And we know one more thing; we know
all of their arguments. Their arguments are all in the form
two ππ over π, where π is an integer between zero and π minus one. We can then plot these onto our
diagram. Weβll use the value of π equal to
six. When π is equal to six, our
arguments are going to be multiples of two π divided by six, which we can simplify
to give us π by three.

So in this case, weβre going to get
six points on our unit circle spaced evenly because their arguments all differ by π
by three. Weβll get a sketch which looks
something like this. At this point, thereβs several
different choices for our primitive roots of unity. Weβll choose the primitive roots of
unity to be when π is equal to one. However, it wouldnβt actually
matter which of our primitive roots of unity we chose, we would get the same
answer. The only fact weβre going to use is
that the primitive roots of unity can be used to generate all of the other roots of
unity.

Now we want to add together all of
these roots of unity on our Argand diagram. Now, we could do this directly from
our formula for our roots of unity, and this would work. However, we can also do this
geometrically. We recall we can add together
points on our Argand diagram by instead considering vectors on our Argand
diagram. For example, by adding together the
two vectors shown on our Argand diagram, we would find the value of one plus π. However, thereβs a trick we can use
to make these calculations easier. Weβll start by moving our vector
from the origin to π to instead point from π to the fifth power to one.

Next, weβll do something very
similar to our vector pointing from the origin to one. Weβre going to move it so it points
from π to the fourth power to π to the fifth power. This gives us a different way of
looking at one plus π. Itβs exactly the same as the vector
which goes from π to the fourth power to one. But remember, we want to find
something equivalent to summing all of these values together. So weβre going to need to add on
more vectors to our diagram. Letβs start by adding the vector
from the origin to π squared. Once again, we can simplify this by
moving our vector. Weβll move this to point from one
to π. Next, we want to include π cubed
in our sum, so we need to add the following vector. And one way of doing this is to
move our vector to point from π to π squared.

And now weβre starting to see a
pattern. Doing exactly the same thing we did
before, we can add the final two roots of unity into our diagram. And then after moving our vectors,
we would end up with a diagram which looks like the following. The vectors end up forming a
perfect hexagon inside of our unit circle. And in fact, we can directly show
this is true just from the fact that their arguments differ by π by three, and they
lie on the unit circle. Now we can ask the question, what
happens when we add all of these roots of unity together? We know we can do this by just
adding our vectors together. Well, if we start at one and then
we add all of our vectors together, we see we just end up at one. And if we start and end at the same
point, then the horizontal and vertical components of our vector must be equal to
zero.

In other words, because this vector
is equal to zero, the sum of all of these roots of unity must also be equal to
zero. This gives us a geometric reason
why the sum of all of these roots of unity is equal to zero. And we were able to use this result
to show if π is a primitive sixth root of unity, then π plus π squared plus π
cubed will be equal to negative one times one plus π to the fourth power plus π to
the fifth power, which was given as option (D).