How do you find all 6th roots of 64+0i (in both trig and standard form)?

Answer 1

Use de Moivre's Theorem

z = 64 + 0i

θ = 0

2(cis 0 / 6)

2(cis 360 / 6)

2(cis 720 / 6)

2(cis 1080 / 6)

2(cis 1440 / 6)

2(cis 1800 / 6)

Answer 2

2 + 0i = 2

2^6 = 64

2 is a 6th root of 64=64+0i

but there are 6 roots for a 6th degree equation

x^6 = 64

x^6-64 = 0

(x^3-8)(x^3+8) = 0

set each factor = 0 and solve for 6 solutions

3 for each factor

x^3-8 = (x-2)(x^2+2x+4) = 0

x-2 = 0, x= 2

x^2 +2x+4 = 0

x^2+2x +1 = 1-4 = -3

(x+1)^2 = -3

x+1 = + or - sqr(-3)

x =-1 +isqr3 or -1-isqr3

That's one real solution and two imaginary solutions, 6th roots of 64

x^3+8 = 0 gives another 3 solutions

x^3 = -8

x=-2 is a solution, (-2)^3 = -8

divide x^3+8 by x+2 to get another factor: x^2-2x+4

complete the square or use the quadratic formula to find 2 imaginary roots

x = 1+ or - isqr3

the 6 roots of 64 are

2,-2, 1+isqr3, 1-isqr3, -1+isqr3, 1-isqr3

2 real roots, 4 imaginary

2=2cis0

-2 ==2(cospi+isinpi) = 2cispi

1+isqr3 = 2(cospi/3 + isinpi/3) = 2cis(pi/3)

1-isqr3 = 2(cos(5pi/3) + isin(5pi/3)) = 2cis(5pi/3)

-1+isqr3 =2(cos(2pi/3) + isin(2pi/3)) = 2cis(2pi/3)

-1-isqr3 =2(cos(4pi/3) + isin(4pi/3)) 2cis(4pi/3)

6 roots are

2cis0, 2cispi, 2cis(pi/3), 2cis(5pi/3), 2cis(2pi/3), 2cis(4pi/3)