8 CHAPTER 1. SOME PREREQUISITE TOPICS

This requires rk = |z| and so r = |z|1/k and also both cos(kα) = cos t and sin(kα) = sin t.This can only happen if

kα = t +2lπ

for l an integer. Thus

α =t +2lπ

k, l ∈ Z

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

8 CHAPTER 1. SOME PREREQUISITE TOPICSThis requires r* = |z| and so r = |z|!/* and also both cos (ka) = cost and sin (kat) = sint.This can only happen ifka=t+210for / an integer. Thust+ 21a= LEZband so the k’" roots of z are of the formt+2l t+2lIz|1/* (cos (2) isin ( = *)).tezSince 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 (Cera) isin (GO *))where / = 0,1,2. Therefore, the roots are1 2 (a 5 .. (5 3 .. [3cos (=) +1sin (Z) ,COS (2) +1sin (Zn) ,COS (5") +1sin (5") .1\ - 1Thus 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 + SS) , and 3 (5 — SS) . Therefore, x? — 27 =(x — 3) [3 (SS) (3 (3-8) .Note also (x-3 (+ +i%3)) (x-3 (s-8)) =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 VW .-1 V3complex zeros, > +i > and > —i 3 These zeros are complex conjugates of each