Explainer: Arbitrary Roots of Complex Numbers

In this explainer, we will learn how to use de Moivreโ€™s theorem to find the ๐‘›th roots of a complex number and explore their properties.

We will be interested in the solutions to equations of the form ๐‘ง=๐‘ค๏Š. Finding solutions to equations in this form is equivalent to finding the ๐‘›th roots of ๐‘ค. Recall, when taking the square root of a positive real number ๐‘Ž, that we have two possible solutions, โˆš๐‘Ž and โˆ’โˆš๐‘Ž. Likewise, when taking the ๐‘› root of a complex number, we have ๐‘› distinct roots. Using de Moivreโ€™s theorem for roots, we can determine all possible distinct solutions.

De Moivreโ€™s Theorem for Roots

For a complex number ๐‘ง=๐‘Ÿ(๐œƒ+๐‘–๐œƒ)cossin, the ๐‘›th roots are given by ๐‘Ÿ๏€ฝ๏€ฝ๐œƒ+2๐œ‹๐‘˜๐‘›๏‰+๐‘–๏€ฝ๐œƒ+2๐œ‹๐‘˜๐‘›๏‰๏‰๏Ž ๏‘ƒcossin for ๐‘˜=0,1,โ€ฆ๐‘›โˆ’1. Equivalently, we can express this in exponential form as ๐‘Ÿ๐‘’๏Ž ๏‘ƒ๏‘ต๏Žฉ๏Žก๏‘ฝ๏‘€๏‘ƒ๏ƒ for ๐‘˜=0,1,โ€ฆ๐‘›โˆ’1.

Example 1: Roots of a Complex Number

  1. Solve ๐‘ง=16โˆš2+16๐‘–โˆš2๏Šซ.
  2. By plotting these solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions.

Answer

Part 1

We begin by finding the modulus and argument of 16โˆš2+16๐‘–โˆš2 so we can express it in polar form. The modulus is given by ๏„ž๏€ป16โˆš2๏‡+๏€ป16โˆš2๏‡=โˆš1,024=32๏Šจ๏Šจ. Since the complex number lies in the first quadrant, we can find the argument by evaluating the inverse tangent of the imaginary part of ๐‘ง divided by the real part: arctanarctan๏€ฟ16โˆš216โˆš2๏‹=(1)=๐œ‹4. Hence, we can rewrite the equation as ๐‘ง=32๐‘’.๏Šซ๏ƒ๏‘ฝ๏Žฃ Applying de Moivreโ€™s theorem, the roots are given by 32๐‘’๏Ž ๏Žค๏‘ฝ๏Žฃ๏Žฉ๏Žก๏‘ฝ๏‘€๏Žค๏ƒ for ๐‘˜=0,1,โ€ฆ4. Hence, considering each value of ๐‘˜, the solutions are 2๐‘’,2๐‘’,2๐‘’,2๐‘’,2๐‘’.๏‘ฝ๏Žก๏ŽŸ๏Žจ๏‘ฝ๏Žก๏ŽŸ๏Ž ๏Žฆ๏‘ฝ๏Žก๏ŽŸ๏Žฆ๏‘ฝ๏Žก๏ŽŸ๏Žข๏‘ฝ๏Žฃ๏ƒ๏ƒ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ

Part 2

Below is a plot of these complex numbers on an Argand diagram.

Clearly, we can see that the roots lie at the vertices of a regular pentagon inscribed in a circle of radius 2 at the origin.

Example 2: Expressions Involving ๐‘›th Roots

Find the possible values of 1โˆš3๏€ฝ(๐‘–)+(๐‘–)๏‰.๏Ž ๏Žข๏Ž ๏Žข๏Šฑ

Answer

Letting ๐‘ค=๐‘–๏Ž ๏Žข, we can rewrite this expression as 1โˆš3๏€น๐‘ค+๐‘ค๏….๏Šฑ๏Šง

From the definition of ๐‘ค, we can see that |๐‘ค|=|||๐‘–|||=|๐‘–|=1๏Ž ๏Žข๏Ž ๏Žข, from which we can conclude that ๐‘ค=๐‘ค๏Šฑ๏Šงโˆ—. Consequently, we can rewrite the expression as 1โˆš3(๐‘ค+๐‘ค).โˆ—

Using the properties of conjugates, we have 1โˆš3(2(๐‘ค))=2โˆš3(๐‘ค).ReRe

We now calculate the ๐‘ค. Expressing ๐‘– in exponential form, we have ๐‘–=๐‘’๏‘ฝ๏Žก๏ƒ. Using de Moivreโ€™s theorem, its cube roots are given by ๐‘’๏‘ฝ๏Žก๏Žฉ๏Žก๏‘ฝ๏‘€๏Žข๏ƒ for ๐‘˜=0,1,2and. Substituting in the values of ๐‘˜ and expressing each argument in the principal range, the three possible cube roots of ๐‘– are given by ๐‘ค=๐‘’,๐‘ค=๐‘’,๐‘ค=๐‘’.๏Šง๏ƒ๏Šจ๏ƒ๏Šฉ๏Šฑ๏ƒ๏‘ฝ๏Žฅ๏Žค๏‘ฝ๏Žฅ๏‘ฝ๏Žก

We can now substitute each value of ๐‘ค in the simplified expression: 2โˆš3(๐‘ค)Re. For ๐‘ค๏Šง, we have 1โˆš3๏€น๐‘ค+๐‘ค๏…=2โˆš3๏€ผ๐‘’๏ˆ=2โˆš3๏€ป๐œ‹6๏‡=2โˆš3๏€ฟโˆš32๏‹=1.๏Šฑ๏Šง๏ƒRecos๏‘ฝ๏Žฅ

Similarly, for ๐‘ค๏Šจ, we can write 1โˆš3๏€น๐‘ค+๐‘ค๏…=2โˆš3๏€ฝ๐‘’๏‰=2โˆš3๏€ผ5๐œ‹6๏ˆ=2โˆš3๏€ฟโˆ’โˆš32๏‹=โˆ’1.๏Šฑ๏Šง๏ƒRecos๏Žค๏‘ฝ๏Žฅ

Finally, for ๐‘ค๏Šฉ, we have 1โˆš3๏€น๐‘ค+๐‘ค๏…=2โˆš3๏€ผ๐‘’๏ˆ=2โˆš3๏€ปโˆ’๐œ‹2๏‡=0.๏Šฑ๏Šง๏Šฑ๏ƒRecos๏‘ฝ๏Žก

Therefore, the three possible values of the expression are โˆ’1, 0, and 1.

The expression in the previous example is the one that you get when applying Tartagliaโ€™s method for solving cubics to the equation ๐‘ฅโˆ’๐‘ฅ=0๏Šฉ. Trying to understand expressions like this led the mathematician Rafael Bombelli to formalize the rules of arithmetic of complex numbers. In fact, later mathematicians showed that when using general methods to solve cubic equations with real roots, the use of complex numbers is unavoidable.

Example 3: Relation between Arbitrary Roots and Roots of Unity

  1. Find the solutions to the equation ๐‘ง=125๐‘’๏Šฌ๏ƒ๏Žก๏‘ฝ๏Žข. What are their geometrical properties?
  2. State the 6th roots of unity.
  3. What is the relationship between the 6th roots of unity and the solutions to the equation ๐‘ง=125๐‘’๏Šฌ๏ƒ๏Žก๏‘ฝ๏Žข?

Answer

Part 1

Applying de Moivreโ€™s theorem, the roots of the equation are given by 125๐‘’๏Ž ๏Žฅ๏Žก๏‘ฝ๏Žข๏Žฉ๏Žก๏‘ฝ๏‘€๏Žฅ๏ƒ for ๐‘˜=0,1,โ€ฆ,5. Substituting in each value of ๐‘˜ and simplifying, we have โˆš5๐‘’,โˆš5๐‘’,โˆš5๐‘’,โˆš5๐‘’,โˆš5๐‘’,โˆš5๐‘’.๏‘ฝ๏Žจ๏Žฃ๏‘ฝ๏Žจ๏Žฆ๏‘ฝ๏Žจ๏Žก๏‘ฝ๏Žจ๏Žค๏‘ฝ๏Žจ๏Žง๏‘ฝ๏Žจ๏ƒ๏ƒ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ

Plotting these roots on an Argand diagram, we see that they lie at the vertices of a regular hexagon centered at the origin, inscribed in a circle of radius โˆš5.

Part 2

In a similar way, we can find the 6th roots of unity by substituting ๐‘˜=0,1,โ€ฆ,5 into ๐‘’๏Žก๏‘ฝ๏‘€๏Žฅ๏ƒ. Hence, the 6th roots of unity are given by 1,๐‘’,๐‘’,โˆ’1,๐‘’,๐‘’.๏‘ฝ๏Žข๏Žก๏‘ฝ๏Žข๏‘ฝ๏Žข๏Žก๏‘ฝ๏Žข๏ƒ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ

Part 3

Geometrically, we can transform the 6th roots of unity to the roots of ๐‘ง=125๐‘’๏Šฌ๏ƒ๏Žก๏‘ฝ๏Žข through a dilation of scale factor โˆš5 and a counterclockwise rotation by ๐œ‹9. This can be represented by multiplying by a complex number with modulus โˆš5 and argument ๐œ‹9, that is, by โˆš5๐‘’๏‘ฝ๏Žจ๏ƒ. Hence, if we represent the 6th roots of unity by 1, ๐œ”, ๐œ”๏Šจ, ๐œ”๏Šฉ, ๐œ”๏Šช, and ๐œ”๏Šซ, the roots of ๐‘ง=125๐‘’๏Šฌ๏ƒ๏Žก๏‘ฝ๏Žข can be expressed as โˆš5๐‘’,๐œ”โˆš5๐‘’,๐œ”โˆš5๐‘’,๐œ”โˆš5๐‘’,๐œ”โˆš5๐‘’,๐œ”โˆš5๐‘’.๏‘ฝ๏Žจ๏‘ฝ๏Žจ๏‘ฝ๏Žจ๏‘ฝ๏Žจ๏‘ฝ๏Žจ๏‘ฝ๏Žจ๏ƒ๏ƒ๏Šจ๏ƒ๏Šฉ๏ƒ๏Šช๏ƒ๏Šซ๏ƒ

Notice that there is nothing special about โˆš5๐‘’๏‘ฝ๏Žจ๏ƒ; we could have equally used any other of the roots of ๐‘ง=125๐‘’๏Šฌ๏ƒ๏Žก๏‘ฝ๏Žข and got the same result.

The result of the previous question generalizes to the ๐‘›th root of any complex number as stated in the box below.

Relationship between the ๐‘›th Root of an Arbitrary Complex Number and the ๐‘›th Roots of Unity

If ๐‘ง๏Šง is one root of the equation ๐‘งโˆ’๐‘ค=0๏Š, and 1,๐œ”,๐œ”,โ€ฆ,๐œ”๏Šจ๏Š๏Šฑ๏Šง are the ๐‘›th roots of unity, the roots of ๐‘งโˆ’๐‘ค=0๏Š are given by ๐‘ง,๐‘ง๐œ”,๐‘ง๐œ”,โ€ฆ,๐‘ง๐œ”.๏Šง๏Šง๏Šง๏Šจ๏Šง๏Š๏Šฑ๏Šง

This result can be thought about in multiple ways. The first is the one presented in the previous example, in which we consider the transformation that maps the ๐‘›th roots of unity to the ๐‘› roots of ๐‘ง=๐‘ค๏Š. Alternatively, we can consider it in a purely algebraic way: given an equation ๐‘ง=๐‘ค๏Š, we can think of its roots as being ๐‘ง=โˆš๐‘คโˆš1๏‘ƒ๏‘ƒ, in which case we need only to find one of the ๐‘›th roots of ๐‘ค, and then we can use our knowledge of the ๐‘›th roots of unity to find the others. Another way to think about this result is the fact that multiplication by any complex number with modulus one represents a counterclockwise rotation by the argument of the complex number. Hence, starting with one of the roots of ๐‘ง=๐‘ค๏Š, we successively rotate it by an angle of 2๐œ‹๐‘› and hence sweep out a regular ๐‘›-gon where each vertex represents an additional root of ๐‘ง=๐‘ค๏Š.

Example 4: Coordinates of Regular Polygons at the Origin

Find the coordinates of the vertices of a regular pentagon centered at the origin with one vertex at (3,3). Give your answer as exact Cartesian coordinates.

Answer

If we consider the Cartesian plane to be an Argand plane, then we know that the fifth roots of unity form a regular pentagon centered at the origin with one vertex at ๐‘ง=1. We can transform this pentagon to a regular pentagon centered at the origin with one vertex at ๐‘ง๏Šง by multiplying each of the fifth roots of unity by ๐‘ง๏Šง. This is equivalent to finding the fifth roots of ๐‘ง๏Šซ๏Šง. Using de Moivreโ€™s theorem, we know that the fifth roots of unity are 1, ๐œ”, ๐œ”๏Šจ, ๐œ”๏Šฉ, and ๐œ”๏Šช, where ๐œ”=๐‘’.๏Žก๏‘ฝ๏Žค๏ƒ

The point (3,3) can be represented by the complex number ๐‘ง=3+3๐‘–๏Šง. We can express this in exponential form as ๐‘ง=3โˆš2๐‘’๏Šง๏ƒ๏‘ฝ๏Žฃ. Therefore, the vertices of a pentagon centered at the origin with one vertex at ๐‘ง๏Šง will be ๐‘ง๏Šง, ๐‘ง๐œ”๏Šง, ๐‘ง๐œ”๏Šง๏Šจ, ๐‘ง๐œ”๏Šง๏Šฉ, and ๐‘ง๐œ”๏Šง๏Šช. We now calculate each of these in turn and express it in polar form to then covert it to coordinates in the Cartesian plane. Starting with ๐‘ง๐œ”๏Šง, we have ๐‘ง๐œ”=3โˆš2๐‘’๐‘’=3โˆš2๐‘’.๏Šง๏ƒ๏ƒ๏ƒ๏‘ฝ๏Žฃ๏Žก๏‘ฝ๏Žค๏Ž ๏Žข๏‘ฝ๏Žก๏ŽŸ

Expressing this in polar form, we have ๐‘ง๐œ”=3โˆš2๏€ผ๏€ผ13๐œ‹20๏ˆ+๐‘–๏€ผ13๐œ‹20๏ˆ๏ˆ.๏Šงcossin

Therefore, the coordinates of the point are ๏€ผ3โˆš2๏€ผ13๐œ‹20๏ˆ,3โˆš2๏€ผ13๐œ‹20๏ˆ๏ˆcossin.

Considering ๐‘ง๐œ”๏Šง๏Šจ, we have ๐‘ง๐œ”=3โˆš2๐‘’๐‘’=3โˆš2๐‘’=3โˆš2๐‘’.๏Šง๏Šจ๏ƒ๏ƒ๏ƒ๏Šฑ๏ƒ๏‘ฝ๏Žฃ๏Žฃ๏‘ฝ๏Žค๏Žก๏Ž ๏‘ฝ๏Žก๏ŽŸ๏Ž ๏Žจ๏‘ฝ๏Žก๏ŽŸ

Expressing this in polar form, we have ๐‘ง๐œ”=3โˆš2๏€ผ๏€ผโˆ’19๐œ‹20๏ˆ+๐‘–๏€ผโˆ’19๐œ‹20๏ˆ๏ˆ๏Šง๏Šจcossin which has coordinates ๏€ผ3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ๏ˆcossin.

Similarly, ๐‘ง๐œ”=3โˆš2๐‘’๐‘’=3โˆš2๐‘’.๏Šง๏Šฉ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ๏‘ฝ๏Žฃ๏Žฃ๏‘ฝ๏Žค๏Ž ๏Ž ๏‘ฝ๏Žก๏ŽŸ Expressing this in polar form, we have ๐‘ง๐œ”=3โˆš2๏€ผ๏€ผโˆ’11๐œ‹20๏ˆ+๐‘–๏€ผโˆ’11๐œ‹20๏ˆ๏ˆ๏Šง๏Šฉcossin which has coordinates ๏€ผ3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ๏ˆcossin.

Finally, ๐‘ง๐œ”=3โˆš2๐‘’๐‘’=3โˆš2๐‘’.๏Šง๏Šช๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ๏‘ฝ๏Žฃ๏Žก๏‘ฝ๏Žค๏Žข๏‘ฝ๏Žก๏ŽŸ Expressing this in polar form, we have ๐‘ง๐œ”=3โˆš2๏€ผ๏€ผโˆ’3๐œ‹20๏ˆ+๐‘–๏€ผโˆ’3๐œ‹20๏ˆ๏ˆ๏Šง๏Šชcossin which has coordinates ๏€ผ3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ๏ˆcossin.

Therefore, the coordinates of the regular pentagon are (3,3),๏€ผ3โˆš2๏€ผ13๐œ‹20๏ˆ,3โˆš2๏€ผ13๐œ‹20๏ˆ๏ˆ,๏€ผ3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’19๐œ‹20๏ˆ๏ˆ,๏€ผ3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’11๐œ‹20๏ˆ๏ˆ,๏€ผ3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ,3โˆš2๏€ผโˆ’3๐œ‹20๏ˆ๏ˆ.cossincossincossincossin

Considering the transformations that map the ๐‘›th roots of unity to another regular polygon, we use the same technique to find the coordinates of any regular polygon in the plane.

Example 5: Coordinates of Regular Polygons

Find the coordinates of the vertices of a regular hexagon centered at (โˆ’1,2) with one vertex at the origin. Give your answer as exact Cartesian coordinates.

Answer

Considering the Cartesian plane to be an Argand plane, we consider the transformations that map the sixth roots of unity onto the given hexagon. Since the roots of unity are centered at the origin, we first consider the translation which takes the given hexagon to one centered at the origin. Since its center is at (โˆ’1,2), the mapping that takes it to a polygon centered at the origin is represented by subtracting the complex number ๐‘ค=โˆ’1+2๐‘–. This mapping takes the vertex that was at the origin to ๐‘ง=1โˆ’2๐‘–๏Šง. We can therefore map the sixth roots of unity to the vertices of this polygon by multiplying each one by ๐‘ง๏Šง. Combining these two mappings, we can map the sixth roots of unity to the given hexagon by applying the map ๐œ”โ†ฆ๐‘ง๐œ”+๐‘ค๏‡๏Šง๏‡. Notice we are adding ๐‘ค now since we are mapping a polygon centered at the origin to one centered at ๐‘ค.

The sixth roots of unity are given by 1, ๐œ”, ๐œ”๏Šจ, ๐œ”๏Šฉ, ๐œ”๏Šช, and ๐œ”๏Šซ, where ๐œ”=๐‘’.๏‘ฝ๏Žข๏ƒ

The table represents the calculations of applying this map to each of the sixth roots of unity.

๐‘˜๐œ”๏‡๐‘ง๐œ”๏Šง๏‡๐‘ง๐œ”+๐‘ค๏Šง๏‡
111โˆ’2๐‘–0
212+โˆš32๐‘–1+2โˆš32+โˆš3โˆ’22๐‘–โˆ’1+2โˆš32+โˆš3+22๐‘–
3โˆ’12+โˆš32๐‘–โˆ’1+2โˆš32+โˆš3+22๐‘–โˆ’3+2โˆš32+โˆš3+62๐‘–
4โˆ’1โˆ’1+2๐‘–โˆ’2+4๐‘–
5โˆ’12โˆ’โˆš32๐‘–โˆ’1โˆ’2โˆš32+โˆ’โˆš3+22๐‘–โˆ’3โˆ’2โˆš32+โˆ’โˆš3+62๐‘–
612โˆ’โˆš32๐‘–1โˆ’2โˆš32+โˆ’โˆš3โˆ’22๐‘–โˆ’1โˆ’2โˆš32+โˆ’โˆš3+22๐‘–

Therefore, the coordinates of the vertices are (0,0),๏€ฟโˆ’1+2โˆš32,โˆš3+22๏‹,๏€ฟโˆ’3+2โˆš32,โˆš3+62๏‹,(โˆ’2,4),๏€ฟโˆ’3โˆ’2โˆš32,โˆ’โˆš3+62๏‹,๏€ฟโˆ’1โˆ’2โˆš32,โˆ’โˆš3+22๏‹.

Example 6: Geometric Applications of Complex Numbers

A small robot is programmed to travel ๐‘Ž units forward and then turn to the left by an angle of 2๐œ‘. If it does this ๐‘› times, how far will it be from the starting point? Give you answer in exact form.

Answer

Let the position of the robot be given by ๐‘ง=๐‘Ž+๐‘๐‘–. Using this convention, moving forward by ๐‘Ž units at an angle of ๐œƒ to the left is modeled by adding ๐‘Ž๐‘’๏ƒ๏ผ. Hence, starting from the origin, after ๐‘› iterations, the robotโ€™s position will be given by ๐‘ง=๐‘Ž+๐‘Ž๐‘’+๐‘Ž๐‘’+โ‹ฏ+๐‘Ž๐‘’.๏Šจ๏ŽŠ๏ƒ๏Šช๏ŽŠ๏ƒ๏Šจ(๏Š๏Šฑ๏Šง)๏ŽŠ๏ƒ

Since for any complex number ๐‘ค, ๐‘คโˆ’1=(๐‘คโˆ’1)๏€น๐‘ค+๐‘ค+โ€ฆ+๐‘ค+1๏…๏Š๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šจ, we can rewrite ๐‘ง as ๐‘ง=๐‘Ž๐‘’โˆ’1๐‘’โˆ’1.๏Šจ๏Š๏ŽŠ๏ƒ๏Šจ๏ŽŠ๏ƒ

Factoring out ๐‘’๏Š๏ŽŠ๏ƒ from the numerator and ๐‘’๏ŽŠ๏ƒ from the denominator, we can express ๐‘ง as ๐‘ง=๐‘Ž๐‘’๏€น๐‘’โˆ’๐‘’๏…๐‘’๏€น๐‘’โˆ’๐‘’๏….๏Š๏ŽŠ๏ƒ๏Š๏ŽŠ๏ƒ๏Šฑ๏Š๏ŽŠ๏ƒ๏ŽŠ๏ƒ๏ŽŠ๏ƒ๏Šฑ๏ŽŠ๏ƒ

Since sin๐œƒ=12๐‘–๏€น๐‘’โˆ’๐‘’๏…๏ƒ๏ผ๏Šฑ๏ƒ๏ผ, we can express the numerator and denominator in terms of sine as follows: ๐‘ง=๐‘Ž2๐‘–(๐‘›๐œ‘)2๐‘–๐œ‘๐‘’.sinsin(๏Š๏Šฑ๏Šง)๏ŽŠ๏ƒ

Finally, simplifying gives ๐‘ง=๐‘Ž(๐‘›๐œ‘)๐œ‘๐‘’.sinsin(๏Š๏Šฑ๏Šง)๏ŽŠ๏ƒ

This is the position of the robot after ๐‘› iterations. Given that it started at the origin, the distance from the starting point is just the modulus ๐‘ง. Therefore, taking the modulus, we have |๐‘ง|=|||๐‘Ž(๐‘›๐œ‘)๐œ‘๐‘’|||=|๐‘Ž||||(๐‘›๐œ‘)๐œ‘|||||๐‘’||.sinsinsinsin(๏Š๏Šฑ๏Šง)๏ŽŠ๏ƒ(๏Š๏Šฑ๏Šง)๏ŽŠ๏ƒ

Since ||๐‘’||=1(๏Š๏Šฑ๏Šง)๏ŽŠ๏ƒ, the distance of the robot from its starting point after ๐‘› iterations is given by |๐‘Ž||||(๐‘›๐œ‘)๐œ‘|||sinsin.

Example 7: Geometric Applications of the ๐‘›th Roots of a Complex Number

  1. Find the roots of ๐‘ง+16=0๏Šฎ.
  2. The complex numbers representing the roots of ๐‘ง+16=0๏Šฎ are each squared to form the vertices of a new shape. What is the area of this shape?

Answer

Part 1

We begin by expressing โˆ’16 in exponential form. Clearly, its modulus is 16 and since it is a negative real number, its argument is ๐œ‹. We can express it in polar form as 16๐‘’๏Ž„๏ƒ. Hence, we can rewrite the equation as ๐‘ง=16๐‘’.๏Šฎ๏Ž„๏ƒ

Applying de Moivreโ€™s theorem, the roots are given by 16๐‘’๏Ž ๏Žง๏‘ฝ๏Žฉ๏Žก๏‘ฝ๏‘€๏Žง๏ƒ for ๐‘˜=0,1,โ€ฆ,8. Hence, considering each value of ๐‘˜, the solutions are โˆš2๐‘’,โˆš2๐‘’,โˆš2๐‘’,โˆš2๐‘’,โˆš2๐‘’,โˆš2๐‘’,โˆš2๐‘’,โˆš2๐‘’.๏‘ฝ๏Žง๏Žข๏‘ฝ๏Žง๏Žค๏‘ฝ๏Žง๏Žฆ๏‘ฝ๏Žง๏Žฆ๏‘ฝ๏Žง๏Žค๏‘ฝ๏Žง๏Žข๏‘ฝ๏Žง๏‘ฝ๏Žง๏ƒ๏ƒ๏ƒ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ

Plotting these points on an Argand diagram, we have the vertices of a regular octagon.

Part 2

If we square each of the complex numbers representing the vertices of the hexagon, we get the following set of four complex numbers: 2๐‘’,2๐‘’,2๐‘’,2๐‘’.๏‘ฝ๏Žฃ๏Žข๏‘ฝ๏Žฃ๏Žข๏‘ฝ๏Žฃ๏‘ฝ๏Žฃ๏ƒ๏ƒ๏Šฑ๏ƒ๏Šฑ๏ƒ

Plotting these on an Argand diagram, we find they represent the vertices of a square inscribed in a circle of radius 2.

Using the Pythagorean theorem, we find that the side length of the square is 2โˆš2. Hence, its area is 8.

Key Points

  1. Using de Moivreโ€™s theorem, we can find arbitrary roots of complex numbers.
  2. If ๐‘ง๏Šง is one of the ๐‘›th roots of a complex number, its other roots are given by ๐‘ง,๐‘ง๐œ”,๐‘ง๐œ”,โ€ฆ,๐‘ง๐œ”,๏Šง๏Šง๏Šง๏Šจ๏Šง๏Š๏Šฑ๏Šง where ๐œ” is a primitive ๐‘›th root of unity.
  3. Using the geometric properties of the roots of complex numbers, we can find the coordinates of regular polygons plotted in the plane.

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