This lecture, however, opens the discussion to all the roots of complex numbers (remember real numbers are complex numbers, but not every complex number is real).

Nth Roots

In this lesson, we present two theorems (without proof) then use the theorems to find all the complex number roots (real or non-real) of a complex number. We start with a definition.

For example, 5 is a 4th root of 625 because 5 ⁴ = 625. Of course, –5 is a 4 ^th root of 625 because (–5) ⁴ = 625. Indeed, there are two more 4th roots of 625, namely, 5i and –5i. The fact that 625 has four 4th roots is not coincidence.

In algebra—where we were keeping it real—we concentrated on the real roots of real numbers.

This lecture, however, opens the discussion to all the roots of complex numbers (remember real numbers are complex numbers, but not every complex number is real).

In algebra, we let ⁴ 625 denote the positive real 4th root of 625, and we let  ⁴ 625 denote the positive and negative real 4th roots of 625. Moreover, we let fractional exponents represent roots, so that 625

 ⁴ 625 . Now that we expand the discussion to include all the roots, we let

4 625 denote the principle 4th root of 625.

To denote all the nth roots w of a complex number z   a bi , we write w ⁿ  z . Using this notation, the mathematical expression ⁴ 625 denotes the principle root, 5, while the

mathematical statement r ⁴  625 denotes all four fourth-roots: 5, 5i, –5, and –5i.

The complex number w is an nth root ( n  ) of the complex number z if w ⁿ  z . Primary Objective: Students divide circles into equal sectors by computing the roots of a complex number.

Rolle’s Theorem: The complex number z   a bi has exactly n nth roots.

Our principle nth root of complex number z   a bi is the root with the argument

whose radian measure is smallest in magnitude, selecting arguments in quadrant I over

arguments in quadrant IV if the magnitudes are equal.

Whew! Notation is tricky, but squaring away the notation prepares us for the theorem that computes all the nth roots of a number.

The careful reader will note that the arguments of the roots divide (or chop) the circle of radius

n r into n equal sectors; hence, the theorem’s name.

Example Exercise 1 Roots

We need the fifth roots of ¹ ₂  ₂ ³ i . This complex number sits on the unit circle. Write the complex number in trigonometric form.

   

1 3

2  2 i   1 cis ^ 3  cis ^ 3

Compute the moduli of the roots.

5 1 1  Divide the argument by 5.

3 1

3 5 3 5 15

5

        

The arguments of the roots divide a circle into 5 equal sectors, so each argument cuts away ² ₅ ^ radians or 72° of the circle. Create a sequence of 5 arguments all ² ₅ ^  ⁶ ₁₅ ^ radians apart beginning with ₁₅ ^ .

7 13 19 25 5

15 ^ , 15 ^ , 15 ^ , 15 ^ , 15 ^  3 ^

List the five 5th roots of ¹ ₂  ₂ ³ i .

  15 ,   ⁷ 15 ,   ¹³ 15 ,   ¹⁹ 15 ,   ⁵ 3

cis ^ cis ^ cis ^ cis ^ cis ^ Circle Chop Theorem:

Let z     a bi r cis  . Let n  . Let k  0, 1, 2, , n  1 . Then, complex number w is an nth root of z if

 ² 

n

n n

w  r cis  ^   ^ k

Compute w: w ⁵   ¹ _{2 2} ³ i

Example Exercise 2 Roots

We need the seventh roots of 64 2   64 2  i . Write in trigonometric form.

  ³ 4

64 2 64 2 i 128 cis ^

    

Compute the moduli of the roots.

7 64  2 Divide the argument by 7.

3

4 3 3 1 3

4 7 4 7 28

7

        

The arguments of the roots divide a circle into 7 equal sectors, so each argument cuts away ² ₇ ^ radians of the circle. Create a sequence of 7 arguments all ² ₇ ^  ⁸ ₂₈ ^ radians apart beginning with

3 28

 .

3 11 19 27 35 43 51

28 ^ , 28 ^ , 28 ^ , 28 ^ , 28 ^ , 28 ^ , 28 ^

List the seven 7th roots of  64 2  64 2  i .

  ³ 28   ¹¹ 28   ¹⁹ 28   ²⁷ 28   ³⁵ 28   ⁴³ 28   ⁵¹ 28

2 cis ^ , 2 cis ^ , 2 cis ^ , 2 cis ^ , 2 cis ^ , 2 cis ^ , 2 cis ^ Graph the roots.

Graph the solutions to x ⁷   64 2  64 2  i in the complex plane.

Example Exercise 3 Roots

Add 1 to both sides of the equation.

6 6

1 0 1 x

x

 

 Write 1 in trigonometric form.

 

1 1   cis 0 Compute the moduli of the roots.

6 1 1  Divide the argument by 6.

0 0 6 

The arguments of the roots divide a circle into 6 equal sectors, so each argument cuts away ^ ₃ radians or 60° of the circle. Create a sequence of 6 arguments all ^ ₃ radians apart. Begin with 0.

2 4 5

3 3 3 3

0, , ^{ } , ,  ^ , ^ List the six 6th roots of 1.

  0 ,   3 ,   ² 3 ,   ,   ⁴ 3 ,   ⁵ 3

cis cis ^ cis ^ cis  cis ^ cis ^

Convert to standard form.

3 3 3 3

1 1 1 1

2 2 2 2 2 2 2 2

1,  ,   , 1,    , 

Mathematicians refer to these numbers as the sixth roots of unity.

Solve: x ⁶   1 0

#1 trigonometric form: cis   4 ^π , cis   ⁵ 4 ^π ; standard form: ₂ ²  ₂ ² i ,   ₂ ² ₂ ² i

#3 standard form: 3    i , 1 3 , i  3  i , 1  3 i

#5 2 2

2



#7) Compute & graph the cubed-roots of any complex number with a modulus of 64. For example, ³ 64 cis   4 ^π :

#9)  ³ ₅

Application Exercise

Practice Problems

#1) Compute the square roots of i . #2) Compute the cube roots of 4 4 3i  .

#3) Identify all the solutions to x ⁴   8 8 i 3  0 . #4) Identify all the solutions to x ³   1 0 .

#5) Determine exact value of sin   ⁵ 8 ^π . #6) Graph ^{y  sin 3}   ^{x .}

#7) Chop a radius 4 circle into 3 equal sectors. #8) Compute the fourth roots of 16.

The terminal side of θ is in quadrant III, and sin θ   ⁴ ₅ .

#9) What is the value of cosθ ? #9) What is the value of tan θ ? Powers of i possess an interesting periodic property.

2

3 2

4 2 2

5

1 1

1 1 1

i i i

i i i i i

i i i

i i



 

      

      



Thus, every natural number power of i can be expressed as one of the numbers: i ,

 1 ,  i , or 1. Use this fact and the pattern above to simplify i ²²⁸ .