14 CHAPTER 1. BASIC NOTIONS

and so the kth roots of z are of the form

|z|1/k(

cos(

t +2lπk

)+ isin

(t +2lπ

k

)), l ∈ Z.

Since the cosine and sine are periodic of period 2π, there are exactly k distinct numberswhich result from this formula. ■

Example 1.5.3 Find the three cube roots of i.

First note that i = 1(cos(

π

2

)+ isin

2

)). Using the formula in the proof of the above

corollary, the cube roots of i are

1(

cos((π/2)+2lπ

3

)+ isin

((π/2)+2lπ

3

))where l = 0,1,2. Therefore, the roots are

cos(

π

6

)+ isin

6

),cos

(56

π

)+ isin

(56

π

),cos

(32

π

)+ isin

(32

π

).

Thus the cube roots of i are

√3

2+ i(

12

),−√

32

+ i(

12

), and −i.

The ability to find kth roots can also be used to factor some polynomials.

Example 1.5.4 Factor the polynomial x3−27.

First find the cube roots of 27. By the above procedure using De Moivre’s theorem,

these cube roots are 3,3

(−12

+ i

√3

2

), and 3

(−12− i

√3

2

). Therefore, x3−27 =

(x−3)

(x−3

(−12

+ i

√3

2

))(x−3

(−12− i

√3

2

)).

Note also(

x−3(−12 + i

√3

2

))(x−3

(−12 − i

√3

2

))= x2 +3x+9 and so

x3−27 = (x−3)(x2 +3x+9

)where the quadratic polynomial x2+3x+9 cannot be factored without using complex num-bers.

Note that even though the polynomial x3 − 27 has all real coefficients, it has some

complex zeros,−12

+ i

√3

2and−12− i

√3

2. These zeros are complex conjugates of each

other. It is always this way. You should show this is the case. To see how to do this, seeProblems 17 and 18 below.

Another fact for your information is the fundamental theorem of algebra. This theoremsays that any polynomial of degree at least 1 having any complex coefficients always hasa root in C. This is sometimes referred to by saying C is algebraically complete. Gauss isusually credited with giving a proof of this theorem in 1797 but many others worked on it

14 CHAPTER 1. BASIC NOTIONSand so the k’" roots of z are of the formIz|1/* (cos (“2*) +isin (“2*)) ,lEZ.Since the cosine and sine are periodic of period 27, there are exactly k distinct numberswhich result from this formula.Example 1.5.3 Find the three cube roots of i.First note that i = 1 (cos ($) +isin ($)). Using the formula in the proof of the abovecorollary, the cube roots of i are1 (cos (Cena) ‘isin (GO *))where / = 0, 1,2. Therefore, the roots areT 2 (8 5 .. [5 3 .. [3cos (Z) +isin (Z) , COS (Zz) +isin (2x) , COS (57) +isin (5) .Thus the cube roots of i are 3 +i (5) 8 +i (5) , and —i.The ability to find k’” roots can also be used to factor some polynomials.Example 1.5.4 Factor the polynomial x° — 27.First find the cube roots of 27. By the above procedure using De Moivre’s theorem,—1 3 —1 3these cube roots are 3,3 (5 + ~) , and 3 (5 — SS) . Therefore, x? — 27 =Note also (x-3 (+ +i%3)) (x-3 (s-9)) =x? +3x+9 and sox —27 = (x—3) (x? +3x+9)where the quadratic polynomial x? + 3x-+9 cannot be factored without using complex num-bers.Note that even though the polynomial x°* — 27 has all real coefficients, it has some-1 V3 .-1 v3complex zeros, > +i— and > —i 3 These zeros are complex conjugates of eachother. It is always this way. You should show this is the case. To see how to do this, seeProblems 17 and 18 below.Another fact for your information is the fundamental theorem of algebra. This theoremsays that any polynomial of degree at least 1 having any complex coefficients always hasa root in C. This is sometimes referred to by saying C is algebraically complete. Gauss isusually credited with giving a proof of this theorem in 1797 but many others worked on it