Lesson Video: Cube Roots of Unity | Nagwa Lesson Video: Cube Roots of Unity | Nagwa

# Lesson Video: Cube Roots of Unity Mathematics

In this video, we will learn how to identify the cubic roots of unity using de Moivreβs theorem.

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### Video Transcript

In this video, weβll learn how to find the cubic roots of unity and explore their properties. Weβll begin by learning what we mean by the cubic roots of unity and how to calculate them. Weβll learn how to calculate products and negative powers of these roots of unity and their sum and difference. Finally, weβll 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 Moivreβs 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. Letβs look at how we might solve this using De Moivreβs theorem.

Weβll use De Moivreβs theorem for a complex number written in polar form. Thatβs 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 ππ over π plus π sin π plus two ππ over π when π is equal to zero all the way through to π minus one. Weβll begin then by expressing the number one in polar form. Its real part is one. And its imaginary part is zero. So itβs 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. Thatβs clearly one. The argument is the measure of the angle that this line segment makes with the positive real axis. And thatβs measured in a counterclockwise direction. So we can see that itβs 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 weβre 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 Moivreβs 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 ππ over three plus π sin zero plus two ππ 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 ππ over three can be simply written as two ππ over three.

Weβre now going to apply the final part of De Moivreβs theorem. Since our value of π is three, weβre 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 theyβre 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.

Weβre 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 Moivreβs theorem to evaluate π§ one squared. This says that, for a complex number of the form π§ is equal to ππ to the ππ, π§ to the power of π is equal to π to the power of π times π to the πππ. 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 weβll 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.

Letβs repeat this process for question two. Letβs 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 letβs 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, weβll add two π. Remember, weβre allowed to add or subtract any multiple of two π to achieve an argument thatβs 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, thereβs a remainder of zero when π is divided by three. Itβs equal to π§ one when π is one moduli three. Thereβs a remainder of one when π is divided by three. And itβs 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. Thatβs 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, itβs cos two π by three plus π sin two π by three. And in algebraic form, itβs 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. Letβs 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?

Letβs begin by writing π in its exponential form. Itβs 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 thatβs of course the same as π squared.

Letβs 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 theyβre evenly spaced about the origin. In fact, they form the vertices of an equilateral triangle inscribed within a unit circle. But letβs 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 weβve 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 weβve shown that their sum is equal to zero.

Letβs 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 weβve 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 letβs 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.

Letβs 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.

Letβs 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. Weβre next going to factorise each expression. In the first parentheses, weβre 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, weβve 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. Itβs 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.

Weβve 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.