### Video Transcript

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.