Lesson Video: Arbitrary Roots of Complex Numbers | Nagwa Lesson Video: Arbitrary Roots of Complex Numbers | Nagwa

# Lesson Video: Arbitrary Roots of Complex Numbers Mathematics

In this video, we will learn how to use de Moivreβs theorem to find the πth roots of a complex number and explore their properties.

16:51

### Video Transcript

In this video, weβll learn how to use De Moivreβs theorem to find the πth root of a complex number and explore their properties. By this point, you should feel comfortable finding the πth roots of unity. And this lesson looks to extend these concepts into finding the πth root of any complex number.

Weβll also consider the relationship between πth roots of a complex number and the roots of unity, before looking at a geometrical interpretation and application for these roots. Letβs begin by recalling De Moivreβs 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 ππ over π plus π sin of π plus two ππ over π. And these are for integer values of π between zero and π minus one.

In this video, weβll use this theorem for both roots in polar and exponential form. So letβs 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 Moivreβs theorem for roots uses the polar and exponential form of a complex number instead of the algebraic form. So weβll need to begin by calculating the modulus and argument of a complex number thatβs 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. Itβs the square root of the sum of the squares of these two parts. Thatβs 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. Thatβs 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, weβll 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 Moivreβs 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 ππ 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 itβs 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 weβve expressed them in exponential form.

For part two, weβre 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 πth roots of a complex number, much like the πth 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. Letβs 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, weβre going to need to take the sixth root of both sides. And to do this, weβll need to apply De Moivreβs 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 ππ 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, weβll leave the Argand diagram in place. Itβs going to be useful to us in a moment. In a similar way, we could use De Moivreβs 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, letβs 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 itβs 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.

Letβs 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 πth 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. Letβs 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 weβre dealing with a pentagon, letβs 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 letβs 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 thatβs the equivalent to finding the fifth root of π§ one to the power of five.

Now we could use De Moivreβs 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 πth 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, weβll 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. Thatβs 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, itβs 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 weβre trying to find the coordinates, weβll need to represent this in algebraic form. And to convert from exponential to algebraic form, we first convert it to polar form. Thatβs 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 20ππ. 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 πth roots of complex numbers. Weβll 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?

Letβs begin with part one. To solve this equation, weβll subtract 16 from both sides to get π§ to the power of eight equals negative 16. And then weβll find the eighth roots of both sides. But to apply De Moivreβs theorem for roots, negative 16 is going to need to be expressed in exponential or polar form.

Letβs write it in exponential form. Its modulus is 16. And since itβs 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 ππ. Applying De Moivreβs theorem and we see that the roots are given by 16 to the power of one-eighth times π to the π plus two ππ 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. Itβs 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, itβs 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, weβve learnt that we can use De Moivreβs theorem to find arbitrary roots of complex numbers. Weβve seen that if π§ one is one of the πth 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 thatβs when π is the primitive root of unity. Weβve 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.

Attend sessions, chat with your teacher and class, and access class-specific questions. Download the Nagwa Classes app today!

Windows macOS Intel macOS Apple Silicon