### Video Transcript

In this lesson, weβll learn how to use De Moivreβs theorem to find the πth root of a
complex number and explore their properties. By this point, you should feel comfortable finding the πth roots of unity. And this lesson looks to extend these concepts into finding the πth root of any
complex number.

Weβll also consider the relationship between πth roots of a complex number and the
roots of unity, before looking at a geometrical interpretation and application for
these roots. Letβs begin by recalling De Moivreβs theorem for roots. It says that, for a complex number of the form π cos π plus π sin π, the 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 these are for integer values of π between zero and π minus one.

In this video, weβll use this theorem for both roots in polar and exponential
form. So letβs look at an example of how to use the formula to solve an equation involving
finding the roots of a complex number.

1) Solve π§ to the power of five equals 16 root two plus 16 π root two. 2) By plotting the solutions on an Argand diagram, or otherwise, describe the
geometric properties of the solutions.

For part one, we need to solve an equation that involves finding the roots for a
complex number written in algebraic form. Remember though, De Moivreβs theorem for roots uses the polar and exponential form of
a complex number instead of the algebraic form. So weβll need to begin by calculating the modulus and argument of a complex number
thatβs denoted π§ to the power of five.

The real part of this complex number is 16 root two. And similarly, its imaginary part is also 16 root two. So the modulus is fairly straightforward to calculate. Itβs the square root of the sum of the squares of these two parts. Thatβs the square root of 16 root two squared plus 16 root two squared, which is
simply 32. So the modulus of π§ to the power of five is 32.

In exponential form, this is the value of π. Its argument is also fairly straightforward. The complex number has both positive real and imaginary parts. So it must lie in the first quadrant on the Argand diagram. This means we can use the formula arctan of π divided by π, where π is the
imaginary part and π is the real part, to find the argument of π§ to the power of
five. Thatβs arctan of 16 root two over 16 root two.

Well, in fact, 16 root two divided by 16 root two is one. So we need to find the arctan of one. And this is a value we should know by heart. We know that tan of π by four is one. So the arctan of one must be π by four. And the argument of π§ to the power of five is π by four. And we can say, in exponential form, we can write this equation as π§ to the power of
five equals 32π to the π by four π.

To solve this equation, weβll need to find the fifth root of both sides. Now the fifth root of π§ to the power of five is simply π§. And we can say that the fifth root of 32π to the π by four π is 32π to the π by
four π to the power of one over five. Comparing this to the formula for De Moivreβs theorem, we see that π, the modulus,
is 32. π, the argument, is π by four. And π must be equal to five, which means π is going to take the values of zero,
one, two, three, and four.

Applying this theorem with π equals five, we get π§ equals 32 to the power of
one-fifth times π to the π by four plus two ππ over five π. 32 to the power of one-fifth is two. And substituting π is equal to zero into the equation, we see that the first
solution must be two π to the π by 20π.

Our second solution is two π to the nine π by 20π. When π is equal to two, we get two π to the 17π by 20π. Substituting π is equal to three into our equation and then subtracting two π from
the argument so that itβs within the range for the principal argument, we see that
the fourth solution is two π to the negative three π by four π. And similarly, the final solution is two π to the negative seven π by 20π.

And there we have the five solutions to the equation π§ to the power of five equals
16 root two plus 16π root two. And weβve expressed them in exponential form.

For part two, weβre going to need to plot these solutions on an Argand diagram. Now one way to do this would be to convert these numbers back into their algebraic
form. Once we know their real and imaginary parts, we can plot them fairly easily on the
Argand diagram. Alternatively, we could spot that their modulus is two and then use their arguments
to plot the roots. Either way, we see that they form the vertices of a regular pentagon, inscribed in a
circle of radius two, whose centre is the origin.

Geometrically, we can say that the πth roots of a complex number, much like the πth
root of unity, form the vertices of a regular polygon with π sides, a regular
π-gon. Now there are further relationships between these arbitrary roots and the roots of
unity. Letβs have a look in more detail at these properties.

1) Find the solutions to the equation π§ to the power of six equals 125π to the two
π by three π. What are their geometrical properties? 2) State the sixth roots of unity. And 3) What is the relationship between the sixth roots of unity and the solutions to
the equation π§ to the power of six equals 125π to the two π by three π?

Here we have an equation involving finding the roots of a complex number. To solve this equation, weβre going to need to take the sixth root of both sides. And to do this, weβll need to apply De Moivreβs theorem for roots. That tells us that the solutions to this equation are given by 125 to the power of
one-sixth times π to the power of two π over three plus two ππ over six π,
where π takes values from zero through to five.

Substituting these values of π into our formula and then subtracting multiples of
two π where necessary from the argument to express the argument within the range
for the principal argument. And we see that our solutions to the equation are root five π to the π by nine π,
root five π to the four π by nine π, root five π to the seven π by nine π,
root five π to the negative eight π by nine π, root five π to the negative five
π by nine π, and root five π to the negative two π by nine π.

As expected, when we plot these on an Argand diagram, we see that they form the
vertices of a regular hexagon. And this hexagon is inscribed within a circle whose centre is the origin and whose
radius is root five.

Now as we go ahead and answer part two and three of this question, weβll leave the
Argand diagram in place. Itβs going to be useful to us in a moment. In a similar way, we could use De Moivreβs theorem to find the sixth roots of
unity. Or we can simply recall that 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.

So to find the relationship between the sixth roots of unity and the solutions to our
equation, letβs recall the geometric interpretation of the sixth roots of unity. The sixth roots of unity are represented geometrically by the vertices of a regular
hexagon. This time, that hexagon is inscribed within a unit circle. And once again, its centre is the origin. And we can see that we can transform the sixth roots of unity to the roots of our
equation by a dilation scale factor root five and a counterclockwise rotation by π
by nine radians.

Another way of thinking about this is itβs the same as multiplying them by a complex
number whose modulus is the square root of five and whose argument is π by nine, In
other words, root five π to the π by nine π. This means if we call the sixth roots of unity one, π, π squared all the way
through to π to the power of five, then the roots of our equation can be expressed
as root five π to the π by nine π, π times root five π to the π by nine π,
all the way through to π to the power of five times root five π to the π by nine
π. And these results would have stood had we used any of the other roots of π§ to the
power of six equals 125π to the two π by three π.

Letβs look to generalise this. If π§ one is a root of the equation π§ to the power of π minus π€ equals zero and
one, π, π squared, all the way through to π to the power of π minus one are the
πth roots of unity, the roots of π§ to the power of π minus π€ equals zero are π§
one, π§ one multiplied by π, π§ one multiplied by π squared, all the way through
to π§ one multiplied by π to the power of π minus one.

We can think about this geometrically. We know that multiplication by a complex number whose modulus is one represents a
counterclockwise rotation by the argument of that complex number. So if we start at a root of π§ to the power of π minus π€ equals zero, each rotation
by an angle of two π by π will map the vertex at this root onto the other vertices
representing the other roots. Letβs have a look at an example of this geometric interpretation.

Find the coordinates of the vertices of a regular pentagon centred at the origin with
one vertex at three, three.

Since weβre dealing with a pentagon, letβs see how we can link this with the fifth
root of a complex number. We know that, on an Argand diagram, the fifth roots of unity form a regular
pentagon. That pentagon is inscribed within a unit circle whose centre is the origin. And one of the vertices lies at the point whose cartesian coordinates are one,
zero. So letβs consider the cartesian plane to be an Argand diagram containing our regular
pentagon.

We can transform this pentagon to a regular pentagon centred at the origin with a
vertex at π§ one by multiplying each of the fifth roots of unity by π§ one. And thatβs the equivalent to finding the fifth root of π§ one to the power of
five.

Now we could use De Moivreβs theorem or simply recall that the fifth roots of unity
are one, π, π squared, π cubed, and π to the power of four, where π is π to
the two π by five π. Since one of the vertices of our pentagon lies at the point three, three, which
represents the complex number three plus three π, we can say that π§ one is equal
to three plus three π.

Now we know that if π§ one is a root to the equation π§ π minus π€ equals zero and
one π all the way through to π to the power of π minus one are the πth roots of
unity, then the roots of π§ to the power of π minus π€ equals zero are π§ one, π§
one times π, π§ one times π squared, all the way through to π§ one times π to the
power of π minus one. So we can find the coordinates of all the vertices of our regular pentagon by
multiplying our π§ one by the fifth roots of unity.

Before we do that though, weβll need to write it in exponential form. The modulus of π§ one is the square root of the sum of the squares of the real and
imaginary parts. Thatβs the square root of three squared plus three squared, which is three root
two. And since both its real and imaginary parts are positive, we know it lies in the
first quadrant. So its argument is the arctan of three divided by three, which is π by four. And we can say that π§ one is equal to three root two π to the π by four π.

The rest of the roots and therefore the other vertices of our pentagon will be given
by π§ one times π, π§ one times π squared, and all the way through to π§ one times
π to the power of four. To find π§ one times π, itβs three root two π to the π by four π times π to the
two π by five π. And remember, to multiply complex numbers in exponential form, we multiply their
moduli and then we add their arguments. This means our second root is three root two π to the 13π by 20π.

Now since weβre trying to find the coordinates, weβll need to represent this in
algebraic form. And to convert from exponential to algebraic form, we first convert it to polar
form. Thatβs three root two times cos 13π by 20 plus π sin of 13π by 20. Distributing the parentheses, we see this is the same as three root two cos of 13π
by 20 plus three root two π sin of 13π by 20. And we can see that the second vertex of our pentagon will lie at the point with
cartesian coordinates three root two cos of 13π by 20, three root two sin of 13π
by 20.

We repeat this process with the third vertex. And we subtract two π from the argument so that we can express the argument within
the range for the principal argument. And we see that the third solution is three root two π to the negative 19 over
20ππ. Once again, representing this in polar form and distributing the parentheses, we find
the coordinates here are three root two cos of negative 19π by 20, three root two
sin of negative 19π by 20.

We can repeat this process for π§ one times π cubed and π§ one times π to the power
of four. And we find the vertices of the pentagon to lie at the point whose cartesian
coordinates are three, three; three root two cos 13π by 20, three root two sin 13
π by 20, three root two cos of negative 19 π by 20, three root two sin of negative
19π by 20. We have three root two cos of negative 11π by 20, three root two sin of negative
11π by 20, and three root two cos of negative three π by 20, three root two sin of
negative three π by 20.

And there are some further interesting geometric properties of the πth roots of
complex numbers. Weβll consider one more example.

1) Find the roots of π§ to the power of eight plus 16 equals zero. 2) The complex numbers representing the roots of π§ to the power of eight plus 16
equals zero are each squared to form the vertices of a new shape. What is the area of the shape?

Letβs begin with part one. To solve this equation, weβll subtract 16 from both sides to get π§ to the power of
eight equals negative 16. And then weβll find the eighth roots of both sides. But to apply De Moivreβs theorem for roots, negative 16 is going to need to be
expressed in exponential or polar form.

Letβs write it in exponential form. Its modulus is 16. And since itβs a purely real number, which would lie on the negative real axis of an
Argand diagram, its argument is π radians. And therefore, the solutions to our equation are the eighth roots of 16π to the
ππ. Applying De Moivreβs theorem and we see that the roots are given by 16 to the power
of one-eighth times π to the π plus two ππ over eight π, where π takes values
from zero through to seven.

So our roots whose arguments are expressed in the range for the principal argument
are root two π to the π by eight π, root two π to the three π by eight π, all
the way through to root two π to the negative π by eight π. And since these are the eighth roots of a complex number, it follows that they will
form the vertices of a regular octagon inscribed in a circle whose radius is root
two and centre is the origin.

And what about part two? Well, to square a complex number in exponential form, we square its modulus and we
double its argument. Notice how our eight roots have diminished to just four. These roots represent the vertices of a square this time, inscribed in a circle
radius two units. Its area can be found by using the Pythagorean theorem. And that will generate a side length of two root two units. So its area is two root two squared. Itβs eight square units.

Now did you predict what might happen when we squared the roots? It actually makes a lot of sense that the number of roots would halve. Essentially, itβs a little like simply finding the fourth roots of our original
equation, which we know would form the vertices of a square.

In this video, weβve learnt that we can use De Moivreβs theorem to find arbitrary
roots of complex numbers. Weβve seen that if π§ one is one of the πth roots of a complex number, its other
routes are given by π§ one π, π§ one π squared, all the way through to π§ one π
to the power of π minus one. And thatβs when π is the primitive root of unity. Weβve also seen that we can use the geometric properties of the roots of complex
numbers to help us find the coordinates of regular polygons plotted in the
plane.