### Video Transcript

In this video, weβll learn how to
find the cubic roots of unity and explore their properties. Weβll begin by learning what we
mean by the cubic roots of unity and how to calculate them. Weβll learn how to calculate
products and negative powers of these roots of unity and their sum and
difference. Finally, weβll learn how to
simplify expressions using the properties of these roots. Consider the equation π§ cubed
equals one.

Find all the values of π§ for
which π§ cubed equals one.

Here we have an equation π§
cubed equals one, where π§ is a complex number. There are a number of ways we
can solve this equation. One is to recall De Moivreβs
theorem for roots. The other is to rearrange this
equation and solve π§ cubed minus one equals zero. To do this, we would need to
spot one root and use the factor theorem, followed by either polynomial long
division or equating coefficients to find the other roots. Letβs look at how we might
solve this using De Moivreβs theorem.

Weβll use De Moivreβs theorem
for a complex number written in polar form. Thatβs π cos π plus π sin
π, where π is the modulus and π is the argument of the complex number in
radians. This says we can calculate π§
to the power of one over π by finding π to the power of one over π multiplied
by cos of π plus two ππ over π plus π sin π plus two ππ over π when π
is equal to zero all the way through to π minus one. Weβll begin then by expressing
the number one in polar form. Its real part is one. And its imaginary part is
zero. So itβs a fairly easy number to
represent in polar form.

If we represent one on an
Argand diagram, we see it can be represented by the point whose Cartesian
coordinates are one, zero. The modulus of this number is
the length of the line segment that joins this point to the origin. Thatβs clearly one. The argument is the measure of
the angle that this line segment makes with the positive real axis. And thatβs measured in a
counterclockwise direction. So we can see that itβs
zero. In polar form then, one is the
same as one multiplied by cos of zero plus π sin of zero. And if we substitute this back
into our equation, we see that π§ cubed is therefore equal to one times cos of
zero plus π sin of zero.

Since weβre going to be solving
this equation, we need to calculate the value of π§. And to do this, we find the
cube root of each side of the equation. Now if we compare this to De
Moivreβs theorem, we can see that π, in our case, is three. So π§ must be equal to one to
the power of one-third multiplied by cos of zero plus two ππ over three plus
π sin zero plus two ππ over three. And of course, we can simplify
this somewhat. We know that one to the power
of a third is simply one. And our argument zero plus two
ππ over three can be simply written as two ππ over three.

Weβre now going to apply the
final part of De Moivreβs theorem. Since our value of π is three,
weβre going to substitute π is equal to zero, one, and two into this
equation. If we substitute π is equal to
zero in, we get π§ is equal to cos of zero plus π sin of zero. Now, cos of zero is one. And π sin of zero is zero. So the first solution to our
equation is π§ is equal to one. We then substitute π is equal
to one. And the argument becomes two π
multiplied by one over three, which is simply two π by three. And our second solution is cos
of two π by three plus π sin of two π by three.

Our final solution is found
when π is equal to two. We get π§ is equal to cos of
four π by three plus π sin of four π by three. Now the argument of this
solution is outside of the range for the principal argument. And this is π is greater than
negative π and less than or equal to π. We can add and subtract
multiples of two π to four π by three so we can express this solution with its
principal argument.

Four π by three minus two π
is negative two π by three. And we can see that the cubic
roots of one are π§ is equal to one. π§ is equal to cos of two π by
three plus π sin of two π by three. And π§ is equal to cos of
negative two π by three plus π sin of negative two π by three. These are the cubic roots of
unity, so-called because theyβre all possible values for the cube root of
one.

We can actually express these
in algebraic form as well. The first solution is still
simply one. The second solution is negative
a half plus root three over two π. And the third solution is
negative a half minus root three over two π.

Weβre now going to look at an
example which will demonstrate the properties of products of the cubic roots of
unity.

Let π§ one equal π to the π
two π by three and π§ two equal π to the negative π two π by three be the
complex cubic roots of unity. 1) Evaluate π§ one squared. How does this compare with π§
two? 2) Evaluate π§ two squared. How does this compare with π§
one?

Notice how π§ one and π§ two
are the complex solutions to π§ cubed equals one, written in exponential
form. This means we can use De
Moivreβs theorem to evaluate π§ one squared. This says that, for a complex
number of the form π§ is equal to ππ to the ππ, π§ to the power of π is
equal to π to the power of π times π to the πππ. Remember, π is the modulus and
π is the argument. We can see that the modulus of
π§ one is simply one. And the argument of π§ one is
two π by three. So π§ one squared is one
squared times π to the π two π by three times two.

One squared is one. And two π by three multiplied
by two is four π by three. So we can see that π§ one
squared is equal to π to the π four π by three. The argument of π§ one squared
is outside of the range for the principal argument. So weβll subtract two π. And we find that the principal
argument of π§ one squared is negative two π by three. And we can now see that π§ one
squared is equal to π§ two.

Letβs repeat this process for
question two. Letβs begin by making a
prediction. We saw that π§ one squared is
equal to π§ two. So it might seem to follow that
π§ two squared will be equal to π§ one. But letβs check. Once again, the modulus of π§
two is one. But this time, its argument is
negative two π by three. Negative two π by three
multiplied by two is negative four π by three. Once again, the argument of π§
two squared is outside of the range for the principal argument.

This time, weβll add two
π. Remember, weβre allowed to add
or subtract any multiple of two π to achieve an argument thatβs within the
range for the principal argument. This time, the argument of π§
two squared is two π by three. And we now see that π§ two
squared is equal to π§ one as we predicted. So π§ one squared is equal to
π§ two. And π§ two squared is equal to
π§ one, where π§ one and π§ two are the complex cubic roots of unity.

And in fact, we can extend this
idea for higher powers of the complex cube roots of unity.

We can say that, for a positive
integer π, π§ one to the power of π is equal to one when any zero moduli
three. In other words, thereβs a remainder
of zero when π is divided by three. Itβs equal to π§ one when π is one
moduli three. Thereβs a remainder of one when π
is divided by three. And itβs equal to π§ two when π is
two moduli three. And this provides us with a useful
way of forming a definition of the cubic roots of unity.

So our definition, there are three
cubic roots of unity. Lowercase π is used to represent
the primitive root. Thatβs the root with the smallest
strictly positive argument, an argument of two π by three. π is therefore π to the π two π
by three in exponential form. In polar form, itβs cos two π by
three plus π sin two π by three. And in algebraic form, itβs
negative a half plus root three over two π. And all three roots can be defined
as one, π, and π squared. We can represent the cyclic
property with regards to multiplication of the cubic roots of unity as shown. Letβs now consider the properties
of the cubic roots of unity when raised to a negative exponent.

Let π be the primitive cubic
roots of unity. 1) Find π to the power of
negative one. How is this related to the
other cubic roots of unity? 2) Find π to the power of
negative two. How is this related to the
other cubic roots of unity?

Letβs begin by writing π in
its exponential form. Itβs π to the π two π by
three. This means that π to the power
of negative one is π to the π two π by three to the power of negative
one. And if we apply the laws of
exponents, we see that π to the power of negative one is equal to π to the
negative π two π by three. And thatβs of course the same
as π squared.

Letβs repeat this process for
part two. This time, π to the power of
negative two is π to the π two π by three, all to the power of negative
two. And again, applying the laws of
exponents, we see that π to the power of negative two is equal to π to the
negative π four π by three. The argument for this complex
number is outside of the range for the principal argument. So we add two π. And we see that π to the power
of negative two is equal to π to the π two π by three, which is equal to
π. And since π to the power of
negative one is the reciprocal of π, we can see that the cubic roots of unity
also form a cycle under division.

We can even extend the visual
representation of the cubic roots of unity by representing them on an Argand
diagram. We can see that theyβre evenly
spaced about the origin. In fact, they form the vertices
of an equilateral triangle inscribed within a unit circle. But letβs have a look at this
carefully.

The visual representation of π
and π squared on our Argand diagram is as a reflection in the real axis or the
horizontal axis. And if we recall, we know that
the complex conjugate of a number is represented by a reflection in the
horizontal axis. So this means that π squared
must be equal to the conjugate of π.

So weβve seen the properties of the
cubic roots of unity under multiplication and division. But what about addition and
subtraction?

Let π be the primitive cubic
root of unity. 1) Find π plus π squared. 2) Find π minus π
squared. 3) What is π plus one and how
is it related to the other roots of unity? 4) What is π squared plus one
and how is it related to the other roots of unity?

To answer part one, we could
try writing π and π squared in algebraic form and finding their sum that
way. Alternatively, we recall that
π squared is the same as the conjugate of π. And this means that π plus π
squared is equal to π plus the conjugate of π. This has its own property. We know that the sum of a
complex number and its conjugate is two times the real part of that complex
number. So π plus the conjugate of π
is two times the real part of π.

Well, the real part of the
primitive cubic root of unity is negative one-half. And two times negative a half
is negative one. So we can see that π plus π
squared is equal to negative one. This also means that π squared
plus π plus one is equal to zero. Notice that these are the three
cubic roots of unity. And weβve shown that their sum
is equal to zero.

Letβs repeat this process for
part two. Once again, we express π
squared as the conjugate of π. But this time, the difference
between a complex number and its conjugate is two π times the imaginary part of
that complex number. The imaginary part of π is
root three over two. So the difference between π
and π squared is π root three or root three π. And we can use what weβve
calculated here to work out π plus one for part three and π squared plus one
for part four. We saw that π squared plus π
plus one is equal to zero. So letβs subtract π squared
from both sides of this equation. When we do, we see that π plus
one is equal to negative π squared. Similarly, we can also deduce
that π squared plus one is equal to negative π.

So the cubic roots of unity
have three really important properties. We know that π squared is
equal to the conjugate of π. We know that the sum of the
three cubic roots is equal to zero. And we know that π minus π
squared is equal to π root three.

Letβs have a look at a question
which will demonstrate how to apply these properties.

Evaluate nine minus π squared
plus nine π to the power of four all squared plus six plus six π squared plus
six π to the power of four all squared.

Letβs begin by using the
multiplication cycle for the cubic roots of unity to replace π to the power of
four. π to the power of four is the
same as π squared multiplied by π and then multiplied by π again. So we can say that π to the
power of four must be the same as π. And we can rewrite our
expression as shown. Weβre next going to factorise
each expression. In the first parentheses, weβre
going to look to take out a factor of nine and in the second parentheses, a
factor of six. And why do we want to do
this? Well, we know that the sum of
π squared and π and one is zero. And we can rearrange this to
show that one plus π is equal to negative π squared.

So our expression can be
simplified to nine multiplied by negative π squared minus π squared all
squared plus six times zero all squared. And of course, six times zero
all squared is simply zero. So this simplifies further to
negative 10π squared squared. Negative 10 squared is 100. And π squared squared is π to
the power of four. But we already saw that π to
the power of four is simply π. So our expression is simply
100π.

In this video, weβve learned that
there are three cubic roots of unity, denoted one, π, and π squared. And we call π the primitive cubic
root of unity. Itβs π to the π two π by three
or, in polar form, cos two π by three plus π sin of two π by three or, in
algebraic form, negative one-half plus root three over two π. Both positive and negative powers
of π form a closed cycle. For any integer value of π, π to
the power of three π is one. π to the power of three π plus
one is π. And π to the power of three π
plus two is π squared.

Weβve also seen that they have
three important properties. And that is the sum of the three
cubic roots of unity is zero. π squared is equal to the
conjugate of π. And π minus π squared is equal to
π root three. And we can use these properties to
simplify more complicated looking expressions.