Complex cube roots?

Question

How do I get the complex cube root of 8

Answer 1

Let's solve the equation x³ = 8.
We have
x³-8 = 0
(x-2)(x²+2x + 4) = 0
So the real root is x = 2
and the complex cube roots are
x = ½(-2 + i√12) = -1 + i√3
and
x = -1 -i√3.

Answer 2

Find the three cube roots of 8.

All the cube roots will lie in the complex plane on a circle with magnitude 8^(1/3) = 2. They will be equally spaced around the circle. So you have:

2[cos(0) + isin(0)] = 2 + 0 = 2
2[cos(2π/3) + isin(2π/3)] = -1 + i√3
2[cos(4π/3) + isin(4π/3)] = -1 - i√3

Answer 3

1)z^3 =8

z^3-8= 0

(z-2)(z^2+2z+4) = 0

z=2 and the other two you get with quadratic formula

2)Alternative method:

z^3 = 8 = 8( cos 0 + i sin 0)

so by Moivre method

z_i = cube root 8( cos ( 2k pi/3) + i sin (2k pi/3)),

where k varies from 0 to 2

that is
z_1 = 2
z_2 = 2( cos (pi/3) + i sin(pi/3))
z_3 = 2( cos ( 2 pi/3) + i sin(pi/3))

3) yet another proof
Let e be a cube root unit of 1,
that is e^3=1, e different than 1 which implies
(e-1)(e^2+e+1)=0

than z^3 = 8 implies

z = 2 e^k, where k varies from 0 to 2

but e is solution of e^2+e+1=0 that you solve with quadratic.

Answer 4

RE:
Complex cube roots?
How do I get the complex cube root of 8

Answer 5

Of course there are 3 of them...

The general formula for the nth roots of 1 are

w_k = e^(2k*pi*i/n), for k=0,1,2,...,n-1.

Thus, the 3 cube roots of 1 are
1, e^(2i*pi/3), and e^(4i*pi/3).

To get cube roots of 8, simply multiply cube roots of 1 by the real cube root of 8, namely 2:

2, 2*e^(2i*pi/3), and 2*e^(4i*pi/3).

Answer 6

cis(x) = cos(x)+i sin(x)

8^(1/3) = 2cis(0) = 2,
2cis(2pi/3) = 2(-1/2 + isqrt(3)/2) = -1+sqrt(3)i,
2cis(4pi/3) = -1-sqrt(3)i