### Video Transcript

In this video, weβll learn how to
find the πth roots of unity and explore their properties. Weβll begin by learning what we
mean by the πth roots of unity and find their general form. Weβll then calculate their sum and
find the properties of their reciprocal before discovering the application of the
πth roots of unity and their geometric properties.

If π§ is an πth root of unity,
then it satisfies the relation π§ to the power of π equals one.

We can use de Moivreβs theorem to
help us solve this equation and thus find the general form for the πth roots of
unity. de Moivreβs theorem for roots states that, for a complex number written in
polar form, π cos π plus π sin π, its πth roots are given by π to the power of
one over π times cos of π plus two ππ over π plus π sin of π plus two ππ
over π. And π takes integer values from
zero through to π minus one.

To solve the equation π§ to the
power of π equals one, weβll begin by writing the number one in polar form. One whose real part is one and
whose imaginary part is zero is a fairly easy number to write in polar form. If we represent one on an Argand
diagram, we see it can be represented by the points whose Cartesian coordinates are
one, zero. The modulus of this number, thatβs
π
in the general form for our complex number, is the length of the line segment
that joins this point to the origin. So its modulus must be one
unit. 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. We can see that its argument must
be zero.

And we can therefore say that one
is equal to one times cos of zero plus π sin of zero. The πth roots of unity, in other
words, the πth roots of one, are therefore given by one times cos of zero plus two
ππ over π plus π sin of zero plus two ππ over π for values of π between
zero, two, π minus one.

We can simplify this expression
somewhat. And we see that the πth roots of
unity are given by cos of two ππ over π plus π sin of two ππ over π, which in
exponential form is π to the two ππ over π π. And thatβs for values of π between
zero and π minus one. Itβs important to realise that this
is the definition of the πth roots of unity. It should be learned and recalled
when finding the πth roots of unity. Now that weβve derived this is
absolutely not necessary to use de Moivreβs theorem every time. In our next two examples, weβre
going to look at how to apply this formula to find the πth root of unity.

Find the cubic roots of unity and
plot them on an Argand diagram.

Finding the cubic roots of unity is
like saying what are the solutions to the equation π§ cubed equals one. To find them, we can use the
general formula for the πth roots of unity. Thatβs cos of two ππ over π plus
π sin of two ππ over π for integer values of π between zero and π minus
one. Since weβre finding the cubic roots
of unity, in this example, our value of π is three, which means π will take the
values zero, one, and two.

Letβs begin with the case when π
is equal to zero. This root is cos of zero plus π
sin of zero. Well, cos of zero is one. And sin of zero is zero. So the first root is one. And it makes complete sense if we
think about it that a solution to the equation π§ cubed equals one would be one.

Next, we let π be equal to
one. This root is cos of two π by three
plus π sin of two π by three. And in exponential form, thatβs π
to the two π by three π. Finally, we let π be equal to
two. The root here is cos four π by
three plus π sin of four π by three. Notice though that the argument for
this root 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. So our third and final root is cos
of negative two π by three plus π sin of negative two π by three. Or in exponential form, thatβs π
to the negative two π by three π.

Now that we have the cubic roots of
unity, we need to plot them on an Argand diagram. There are two ways we could
approach this problem. We could convert each number to
algebraic form. Thatβs one, negative a half plus
root three over two π, and negative a half minus root three over two π. These are plotted on the Argand
diagram as shown. Alternatively, we couldβve used the
modulus and argument of each root.

Either way, letβs notice that the
points that represent these complex numbers are the vertices of an equilateral
triangle. This triangle is inscribed in a
unit circle whose centre is the origin. Actually, an interesting
geometrical property of the πth roots of unity is that, on an Argand diagram, they
are all evenly spaced about the unit circle whose centre is the origin. They form a regular π-gon. It has a vertex at the point whose
Cartesian coordinates are one, zero.

Weβll investigate this property a
little bit further along in the video.

Find the sixth roots of unity.

Finding the sixth roots of unity is
the same as solving the equation π§ to the power of six equals one. Weβll once again use the general
formula for the πth roots of unity. They are cos of two ππ over π
plus π sin of two ππ over π, when π takes integer values from zero to π minus
one. In this example, weβre looking to
find the sixth roots of unity. So π is six. And π takes integer values from
zero through to five. The first root is found when π is
equal to zero. This is cos of zero plus π sin of
zero, which is one.

Now actually, we just saw that, on
an Argand diagram, the points which represent the πth roots of unity form a regular
π-gon with a vertex at the point one, zero. Thatβs this root. For the second root, we let π be
equal to one. This is cos of two ππ over six
plus π sin of two ππ over six. The argument here simplifies to π
by three. And we could also write this in
exponential form as π to the π by three π. When π is equal to two, our root
is cos of four π by six plus π sin of four π by six. And this argument simplifies to two
π by three.

When π is equal to three, we have
cos of six π over six plus π sin of six π over six, which is negative one. And then, when π is equal to four,
we have cos of eight π over six plus π sin of eight π 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 π to the negative two π by three π. Finally, when π is equal to five,
we get cos of 10π over six plus π 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 π to the negative π by three π.

And we have the sixth roots of
unity. In exponential form, they are one,
π to the π by three π, π to the two π by three π, negative one, π to the
negative two π by three π, and π to the negative π by three π. We could plot the sixth roots of
unity on an Argand diagram. And weβd see that the points
representing these roots form the vertices of a regular hexagon inscribed in a unit
circle as shown.

At this stage, itβs also worth
noting an extra definition. We say that a primitive root of
unity is a root which is not also a πth root of unity, where π is less than
π. So in this example, the primitive
roots are when π is equal to one and π is equal to five since the roots when π is
equal to two and π is equal to four are also cubic roots of unity. And when π is equal to zero and π
is equal to three, these are the square roots of unity. In other words, theyβre the square
root of one.

We will look in more detail at the
relationship between different roots of unity later in this video. Itβs useful to know that we often
use the symbol π to denote the primitive root of unity with the smallest strictly
positive argument. In the case of the sixth roots of
unity, this would be π to the π by three π. Interestingly, its other roots are
all powers of π. Now, itβs outside of the scope of
this video to explore this property more. But it is an interesting one that
you might wish to investigate. Now we have a couple of definitions
and the process that we need to take to find the πth roots of unity, letβs look at
the properties of the sum of the πth roots of unity.

Find the sum of the sixth roots of
unity.

Now, we already calculated the
sixth roots of unity. In polar form, they are as
shown. Weβre going to change these to
algebraic form. The second root is a half plus root
three over two π. The third root is negative a half
plus root three over two π. The fifth root is negative a half
minus root three over two π. And the final root is a half minus
root three over two π. And so their sum is as shown.

And we can find the sum of complex
numbers written in algebraic form by adding together their real parts and separately
adding their imaginary parts. Weβll begin with the real
parts. One plus negative one is zero. A half plus negative half is
zero. And negative a half plus a half is
also zero. And what about the imaginary
parts? Well, root three over two minus
root three over two is zero. And again, root three over two
minus root three over two is zero. And so the sum of the sixth roots
of unity is zero.

Now, this isnβt actually a process
that you would need to take each time. Itβs very much a means to an end
since we can actually generalize this. In general, the sum of the πth
roots of unity, when π is greater than one, is always zero. And this is another result that
needs to be learned and applied where necessary.

For our next example, weβre going
to look at how to find the reciprocal for the πth roots of unity.

Let π§ be an πth root of unity and
π be a positive integer. Which of the following is the
correct relationship between the reciprocal of π§ and π§? Is it a) the reciprocal of π§ is
equal to π§. Is it b) the reciprocal of π§ is
equal to negative π§. Is it c) the reciprocal of π§ is
equal to the negative conjugate of π§. Or is it d) the reciprocal of π§ is
equal to just the conjugate of π§.

To work out which of these is the
correct relationship, weβre first going to evaluate π§ to the power of negative one
or the reciprocal of π§. Since π§ is an πth root of unity,
we can say that π§ can be written as π to the ππ, where π is two ππ over π
and π takes integer values from zero to π minus one. This means π§ to the power of
negative one is the same as π to the ππ to the power of negative one, which is
the same as π to the negative ππ.

Next, we recall the property of the
conjugate of a complex number written in an exponential form. We know that the conjugate of ππ
to the ππ is ππ to the negative ππ. And this means that π§ to the power
of negative one is equal to the conjugate of π§ since we defined π§ to be π to the
ππ. And we can therefore see that the
reciprocal of π§ or π§ to the power of negative one is equal to the conjugate of
π§. The correct answer is d).

And this definition can be somewhat
extended a little bit. We can say that the reciprocal of
the πth root of unity is its complex conjugate. But that is also an πth root of
unity.

Weβre going to look at one more
definition. And then, we have an example of
that definition to the geometric properties of the πth roots of unity. Weβre going to begin by considering
how the πth roots of unity are related for different values of π.

What is the relationship of the
cubic roots of unity to the sixth roots of unity?

We have briefly considered this
idea. Weβve already seen that the cubic
roots are one, π to the two π by three π, and π to the negative two π by three
π. And we also saw that the sixth
roots of unity are one, π to the π by three π, π to the two π by three π,
negative one, π to the negative two π by three π, and π to the negative π by
three π. We can see that all the cubic roots
of unity are also sixth roots of unity. And we looked at this one when we
were discussing the concept of the primitive roots of unity.

Letβs extend this idea to a
definition. We can say that if π is equal to
the product of some other π and π, then the πth roots of unity are also the πth
roots of unity. And similarly, the πth roots of
unity must also be πth roots of unity. We can say that the common roots of
π§ to the power of π minus one equals zero and π§ to the power of π minus one
equals zero must be the roots of π§ to the power of π minus one equals zero, where
π is the greatest common divisor of π and π.

Letβs consider an example using the
applications of the properties of the πth roots of unity.

Two regular polygons are inscribed
in the same circle. The first has 1731 sides. And the second has 4039. If the two polygons have at least
one vertex in common, how many vertices in total will coincide?

Remember, the geometrical
interpretation of the πth roots of unity on an Argand diagram is as the vertices of
a regular π-gon inscribed within a unit circle whose centre is the origin. This means then that we can say
that, to solve this problem, we need to find the number of common roots of π§ to the
power of 1731 minus one equals zero and π§ to the power of 4039 minus one equals
zero. Remember, the common roots of π§ to
the power of π minus one equals zero and π§ to the power of π minus one equals
zero are the roots of π§ to the power of π minus one equals zero, where π is the
greatest common divisor of π and π.

So we know that the common roots of
our two equations are the roots of π§ to the power of π minus one equals zero,
where π is the greatest common divisor of 1731 and 4039. And this means if we can find the
value of π, the greatest common divisor of 1731 and 4039, that will tell us how
many common roots there actually are. As a product of their prime
factors, they can be written as three times 577 and seven times 577,
respectively. So their greatest common divisor
and the value of π is 577. And this means that as long as the
polygons have one vertex in common, they will actually have a total of 577 vertices
that coincide.

In this video, weβve seen that we
can find the πth roots of unity and express these in either polar or exponential
form. In exponential form, they are π to
the two ππ over π π, where π takes integer values from zero through to π minus
one. We saw that if we represent these
roots on an Argand diagram, the points that represent them from the vertices of a
regular π-gon inscribed within a unit circle whose centre lies at the origin.

We also saw that the sum of the
πth roots of unity is zero for values of π greater than one. And we saw that the reciprocal of
an πth root of unity is equal to the complex conjugate of that root. And that is also in itself an πth
root of unity. Finally, we saw that we can find
the common roots of π§ to the power of π minus one equals zero and π§ to the power
of π minus one equals zero by finding the roots of π§ to the power of π minus one
equals zero, where π is the greatest common divisor of π and π. And we considered briefly the
geometrical application of this fact.